Gaussian Elimination - IIT Bombay

Gaussian Elimination

Carl Friedrich Gauss (1777-1855)

German mathematician and scientist,contributed to number theory, statistics, algebra, analysis,

differential geometry, geophysics, electrostatics,astronomy, optics

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Gaussian Elimination Method: This is a GEM of amethod to solve a system of linear equations.

Recall that a system of m linear equations in nunknowns x1, . . . , xn is of the form

a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2

. . . . . .

am1x1 + am2x2 + . . . + amnxn = bm

where aij (1 ≤ i ≤ m, 1 ≤ j ≤ n) and bi (1 ≤ i ≤ m)are known scalars.

Basic observation: Operations of three types onthese equations do not alter the solutions:

1. Interchanging two equations.2. Multiplying all the terms of an equation by a nonzero

scalar.3. Adding to one equation a multiple of another

equation.

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Gaussian Elimination Method: This is a GEM of amethod to solve a system of linear equations.Recall that a system of m linear equations in nunknowns x1, . . . , xn is of the form

a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2

. . . . . .

am1x1 + am2x2 + . . . + amnxn = bm

where aij (1 ≤ i ≤ m, 1 ≤ j ≤ n) and bi (1 ≤ i ≤ m)are known scalars.

Basic observation: Operations of three types onthese equations do not alter the solutions:



equation.

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a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2

. . . . . .

am1x1 + am2x2 + . . . + amnxn = bm

where aij (1 ≤ i ≤ m, 1 ≤ j ≤ n) and bi (1 ≤ i ≤ m)are known scalars.Basic observation: Operations of three types onthese equations do not alter the solutions:



equation.

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a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2

. . . . . .

am1x1 + am2x2 + . . . + amnxn = bm


1. Interchanging two equations.

2. Multiplying all the terms of an equation by a nonzeroscalar.

3. Adding to one equation a multiple of anotherequation.

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a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2

. . . . . .

am1x1 + am2x2 + . . . + amnxn = bm



scalar.

3. Adding to one equation a multiple of anotherequation.

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a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2

. . . . . .

am1x1 + am2x2 + . . . + amnxn = bm




equation.25/45

The above system of linear equations can be writtenin matrix form as follows. a11 . . . a1n

......

am1 . . . amn

x1

...xn

=

b1...

bm

(∗)

or in short as Ax = b, where

A =

a11 . . . a1n...

...am1 . . . amn

, x =

x1...

xn

, b =

b1...

bm

.

The m × n matrix A = (aij) is called the coefficientmatrix of the system. By a solution of (∗) we meanany choice of x1, x2, . . . , xn which satisfies all theequations in the system.If each bi = 0, then the system is said to behomogeneous. Otherwise it is called aninhomogeneous system.

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......

am1 . . . amn

x1

...xn

=

b1...

bm

(∗)


A =

a11 . . . a1n...

...am1 . . . amn

, x =

x1...

xn

, b =

b1...

bm

.

The m × n matrix A = (aij) is called the coefficientmatrix of the system. By a solution of (∗) we meanany choice of x1, x2, . . . , xn which satisfies all theequations in the system.

If each bi = 0, then the system is said to behomogeneous. Otherwise it is called aninhomogeneous system.

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......

am1 . . . amn

x1

...xn

=

b1...

bm

(∗)


A =

a11 . . . a1n...

...am1 . . . amn

, x =

x1...

xn

, b =

b1...

bm

.

The m × n matrix A = (aij) is called the coefficientmatrix of the system. By a solution of (∗) we meanany choice of x1, x2, . . . , xn which satisfies all theequations in the system.If each bi = 0, then the system is said to behomogeneous. Otherwise it is called aninhomogeneous system.

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All the known data in the system (∗) is captured in them × (n + 1) matrix

(A|b) :=

a11 . . . a1n...

...am1 . . . amn

∣∣∣∣∣∣∣b1...

bm

.

This is called the augmented matrix for the system.

Now the above three operations on the equations inthe linear system correspond to the followingoperations on the rows of the augmented matrix:

(i) interchanging two rows,(ii) multiply a row by a nonzero scalar,(iii) adding a multiple of one row to another.These are called elementary row operations.Gaussian Elimination Method consists of reducingthe augmented matrix to a simpler matrix from whichsolutions can be easily found. This reduction is bymeans of elementary row operations.

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(A|b) :=

a11 . . . a1n...

...am1 . . . amn

∣∣∣∣∣∣∣b1...

bm

.

This is called the augmented matrix for the system.Now the above three operations on the equations inthe linear system correspond to the followingoperations on the rows of the augmented matrix:

(i) interchanging two rows,(ii) multiply a row by a nonzero scalar,(iii) adding a multiple of one row to another.These are called elementary row operations.

Gaussian Elimination Method consists of reducingthe augmented matrix to a simpler matrix from whichsolutions can be easily found. This reduction is bymeans of elementary row operations.

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(A|b) :=

a11 . . . a1n...

...am1 . . . amn

∣∣∣∣∣∣∣b1...

bm

.

This is called the augmented matrix for the system.Now the above three operations on the equations inthe linear system correspond to the followingoperations on the rows of the augmented matrix:

(i) interchanging two rows,(ii) multiply a row by a nonzero scalar,(iii) adding a multiple of one row to another.These are called elementary row operations.Gaussian Elimination Method consists of reducingthe augmented matrix to a simpler matrix from whichsolutions can be easily found. This reduction is bymeans of elementary row operations.

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Example 1 (A system with a unique solution):

x − 2y + z = 52x − 5y + 4z = −3

x − 4y + 6z = 10.

The augmented matrix for this system is the 3× 4matrix 1 −2 1

2 −5 41 −4 6

∣∣∣∣∣∣5−310

The elementary row operations mentioned above willbe performed on the rows of this augmented matrix,

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Example 1 (A system with a unique solution):

x − 2y + z = 52x − 5y + 4z = −3

x − 4y + 6z = 10.

The augmented matrix for this system is the 3× 4matrix 1 −2 1

2 −5 41 −4 6

∣∣∣∣∣∣5−310

The elementary row operations mentioned above willbe performed on the rows of this augmented matrix,

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First we add -2 times the first row to the second row. Thenwe subtract the first row from the third row.

i.e.,

1 −2 10 −1 20 −2 5

∣∣∣∣∣∣5−13

5

The circled entry is the first nonzero entry in the firstrow and all the entries below this are 0. Such acircled entry is called a pivot. This next step is called‘sweeping’ a column. Here we repeat the process forthe smaller matrix:

viz.(−1 2−2 5

∣∣∣∣ −135

)⇒

(-1 20 1

∣∣∣∣∣ −1331

)Put back the rows and columns that has been cut outearlier:

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i.e.,

1 −2 10 −1 20 −2 5

∣∣∣∣∣∣5−13

5


viz.(−1 2−2 5

∣∣∣∣ −135

)⇒

(-1 20 1

∣∣∣∣∣ −1331


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i.e.,

1 −2 10 −1 20 −2 5

∣∣∣∣∣∣5−13

5


viz.(−1 2−2 5

∣∣∣∣ −135

)⇒

(-1 20 1

∣∣∣∣∣ −1331


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i.e.,

1 −2 10 −1 20 −2 5

∣∣∣∣∣∣5−13

5


viz.(−1 2−2 5

∣∣∣∣ −135

)

⇒

(-1 20 1

∣∣∣∣∣ −1331


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i.e.,

1 −2 10 −1 20 −2 5

∣∣∣∣∣∣5−13

5


viz.(−1 2−2 5

∣∣∣∣ −135

)⇒

(-1 20 1

∣∣∣∣∣ −1331

)

Put back the rows and columns that has been cut outearlier:

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i.e.,

1 −2 10 −1 20 −2 5

∣∣∣∣∣∣5−13

5


viz.(−1 2−2 5

∣∣∣∣ −135

)⇒

(-1 20 1

∣∣∣∣∣ −1331


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1 −2 10 -1 20 0 1

∣∣∣∣∣∣∣5−1331

(∗)

The matrix represents the linear system:

x − 2y + z = 5−y + 2z = −13

z = 31

These can be solved successively by backwardsubstitution

z = 31;y = 13 + 2z = 75;x = 5 + 2y − z = 124.

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1 −2 10 -1 20 0 1

∣∣∣∣∣∣∣5−1331

(∗)


x − 2y + z = 5−y + 2z = −13

z = 31


z = 31;y = 13 + 2z = 75;x = 5 + 2y − z = 124.

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1 −2 10 -1 20 0 1

∣∣∣∣∣∣∣5−1331

(∗)


x − 2y + z = 5−y + 2z = −13

z = 31


z = 31;

y = 13 + 2z = 75;x = 5 + 2y − z = 124.

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1 −2 10 -1 20 0 1

∣∣∣∣∣∣∣5−1331

(∗)


x − 2y + z = 5−y + 2z = −13

z = 31


z = 31;y = 13 + 2z = 75;

x = 5 + 2y − z = 124.

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1 −2 10 -1 20 0 1

∣∣∣∣∣∣∣5−1331

(∗)


x − 2y + z = 5−y + 2z = −13

z = 31


z = 31;y = 13 + 2z = 75;x = 5 + 2y − z = 124.

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We can continue Gaussian elimination to simplify theaugmented matrix further. This is called theGauss-Jordan Process.

Here, we ensure that all thepivots are equal to 1 and moreover all the otherentries in the column containing the pivot are 0. Inother words, we have 0’s not only below but alsoabove the pivot. numbers.

Recall

1 −2 10 −1 20 0 1

∣∣∣∣∣∣5−1331

(∗)

(1) multiply the second row throughout by −1,(2) add twice third row to the second(3) then subtract the third row from the first(3) add twice the second row to the first

This gives

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We can continue Gaussian elimination to simplify theaugmented matrix further. This is called theGauss-Jordan Process. Here, we ensure that all thepivots are equal to 1 and moreover all the otherentries in the column containing the pivot are 0. Inother words, we have 0’s not only below but alsoabove the pivot. numbers.

Recall

1 −2 10 −1 20 0 1

∣∣∣∣∣∣5−1331

(∗)


This gives

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Recall

1 −2 10 −1 20 0 1

∣∣∣∣∣∣5−1331

(∗)


This gives

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Recall

1 −2 10 −1 20 0 1

∣∣∣∣∣∣5−1331

(∗)


This gives

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⇒

1 0 00 1 00 0 1

∣∣∣∣∣∣∣1247531

⇒ This simple augmented matrix quickly gives thedesired solution

x = 124, y = 75, z = 31.

It is useful to have a shorthand notation for the threetypes of elementary row operations.Notation: Let Ri denote the i th row of a given matrix.

Operation NotationInterchange Ri and Rj Ri ↔ Rj

Multiply Ri by a (nonzero) scalar c cRi

Multiply Rj by a scalar c and add to Ri Ri + cRj

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⇒

1 0 00 1 00 0 1

∣∣∣∣∣∣∣1247531

⇒ This simple augmented matrix quickly gives the

desired solution

x = 124, y = 75, z = 31.





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⇒

1 0 00 1 00 0 1

∣∣∣∣∣∣∣1247531

⇒ This simple augmented matrix quickly gives the

desired solution

x = 124, y = 75, z = 31.





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Example 2 (A system with infinitely many solutions):

x − 2y + z − u + v = 52x − 5y + 4z + u − v = −3x − 4y + 6z + 2u − v = 10

⇒

1 −2 1 −1 12 −5 4 1 −11 −4 6 2 −1

∣∣∣∣∣∣5−310

We shall use the notation introduced above for therow operations

.R2 − 2R1

R3 − R1

−→

1 −2 1 −1 10 −1 2 3 −30 −2 5 3 −2

∣∣∣∣∣∣5−13

5

. R3 − 2R2

−→

1 −2 1 −1 10 −1 2 3 −30 0 1 −3 4

∣∣∣∣∣∣5−1331

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⇒

1 −2 1 −1 12 −5 4 1 −11 −4 6 2 −1

∣∣∣∣∣∣5−310


.R2 − 2R1

R3 − R1

−→

1 −2 1 −1 10 −1 2 3 −30 −2 5 3 −2

∣∣∣∣∣∣5−13

5

. R3 − 2R2

−→

1 −2 1 −1 10 −1 2 3 −30 0 1 −3 4

∣∣∣∣∣∣5−1331

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⇒

1 −2 1 −1 12 −5 4 1 −11 −4 6 2 −1

∣∣∣∣∣∣5−310


.R2 − 2R1

R3 − R1

−→

1 −2 1 −1 10 −1 2 3 −30 −2 5 3 −2

∣∣∣∣∣∣5−13

5

. R3 − 2R2

−→

1 −2 1 −1 10 −1 2 3 −30 0 1 −3 4

∣∣∣∣∣∣5−1331

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⇒

1 −2 1 −1 12 −5 4 1 −11 −4 6 2 −1

∣∣∣∣∣∣5−310


.R2 − 2R1

R3 − R1

−→

1 −2 1 −1 10 −1 2 3 −30 −2 5 3 −2

∣∣∣∣∣∣5−13

5

. R3 − 2R2

−→

1 −2 1 −1 10 −1 2 3 −30 0 1 −3 4

∣∣∣∣∣∣5−1331

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⇒

1 −2 1 −1 12 −5 4 1 −11 −4 6 2 −1

∣∣∣∣∣∣5−310


.R2 − 2R1

R3 − R1

−→

1 −2 1 −1 10 −1 2 3 −30 −2 5 3 −2

∣∣∣∣∣∣5−13

5

. R3 − 2R2

−→

1 −2 1 −1 10 −1 2 3 −30 0 1 −3 4

∣∣∣∣∣∣5−1331

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. −R2

−→

1 −2 1 −1 10 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣51331

. R1 + 2R2

−→

1 0 −3 −7 70 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣311331

.

R2 + 2R3

R1 + 3R3

−→

1 0 0 −16 190 1 0 −9 110 0 1 −3 4

∣∣∣∣∣∣1247531

The system of linear equations corresponding to thelast augmented matrix is:

x = 124 + 16u − 19vy = 75 + 9u − 11vz = 31 + 3u − 4v .

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. −R2

−→

1 −2 1 −1 10 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣51331

. R1 + 2R2

−→

1 0 −3 −7 70 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣311331

.R2 + 2R3

R1 + 3R3

−→

1 0 0 −16 190 1 0 −9 110 0 1 −3 4

∣∣∣∣∣∣1247531


x = 124 + 16u − 19vy = 75 + 9u − 11vz = 31 + 3u − 4v .

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. −R2

−→

1 −2 1 −1 10 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣51331

. R1 + 2R2

−→

1 0 −3 −7 70 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣311331

.

R2 + 2R3

R1 + 3R3

−→

1 0 0 −16 190 1 0 −9 110 0 1 −3 4

∣∣∣∣∣∣1247531


x = 124 + 16u − 19vy = 75 + 9u − 11vz = 31 + 3u − 4v .

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. −R2

−→

1 −2 1 −1 10 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣51331

. R1 + 2R2

−→

1 0 −3 −7 70 1 −2 −3 30 0 1 −3 4

∣∣∣∣∣∣311331

.

R2 + 2R3

R1 + 3R3

−→

1 0 0 −16 190 1 0 −9 110 0 1 −3 4

∣∣∣∣∣∣1247531


x = 124 + 16u − 19vy = 75 + 9u − 11vz = 31 + 3u − 4v .

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We say that u and v are independent variables and x,y, z are dependent variables

(x , y , z,u, v)T

= (124 + 16t1 − 19t2,75 + 9t1 − 11t2,31 + 3t1 − 4t2, t1, t2)T

= (124,75,31,0,0)T + t1(16,9,3,1,0)T

+t2(−19,−11,−4,0,1)T .

The above equation gives the general solution to thesystem. (124,75,31,0,0) is a particular solution ofthe inhomogeneous system.v1 = (16,9,3,1,0) and v2 = (−19,−11,−4,0,1) aresolutions of the corresponding homogeneous system.(These two solutions are linearly independent and)every other solution of the homogeneous system is alinear combination of these two solutions.

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(x , y , z,u, v)T

= (124 + 16t1 − 19t2,75 + 9t1 − 11t2,31 + 3t1 − 4t2, t1, t2)T

= (124,75,31,0,0)T + t1(16,9,3,1,0)T

+t2(−19,−11,−4,0,1)T .


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(x , y , z,u, v)T

= (124 + 16t1 − 19t2,75 + 9t1 − 11t2,31 + 3t1 − 4t2, t1, t2)T

= (124,75,31,0,0)T + t1(16,9,3,1,0)T

+t2(−19,−11,−4,0,1)T .

The above equation gives the general solution to thesystem. (124,75,31,0,0) is a particular solution ofthe inhomogeneous system.

v1 = (16,9,3,1,0) and v2 = (−19,−11,−4,0,1) aresolutions of the corresponding homogeneous system.(These two solutions are linearly independent and)every other solution of the homogeneous system is alinear combination of these two solutions.

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(x , y , z,u, v)T

= (124 + 16t1 − 19t2,75 + 9t1 − 11t2,31 + 3t1 − 4t2, t1, t2)T

= (124,75,31,0,0)T + t1(16,9,3,1,0)T

+t2(−19,−11,−4,0,1)T .

The above equation gives the general solution to thesystem. (124,75,31,0,0) is a particular solution ofthe inhomogeneous system.v1 = (16,9,3,1,0) and v2 = (−19,−11,−4,0,1) aresolutions of the corresponding homogeneous system.

(These two solutions are linearly independent and)every other solution of the homogeneous system is alinear combination of these two solutions.

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(x , y , z,u, v)T

= (124 + 16t1 − 19t2,75 + 9t1 − 11t2,31 + 3t1 − 4t2, t1, t2)T

= (124,75,31,0,0)T + t1(16,9,3,1,0)T

+t2(−19,−11,−4,0,1)T .


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TheoremSuppose Ax = b is a system of linear equations whereA = ((aij)) is a m × n matrix and x = (x1, x2, . . . , xn)

T ,b = (b1,b2, . . . ,bm)

T .

Suppose c = (c1, c2, . . . , cn)T is a

solution of Ax = b and S is the set of all solutions to theassociated homogeneous system Ax = 0.

Then the set of all solutions to Ax = b isc + S := {c + v|v ∈ S}.

Proof: Let r ∈ Rn be a solution of Ax = b. Then

A(r− c) = Ar− Ac = b− b = 0.

Hence r− c ∈ S. Thus r ∈ c + S.

Conversely, let v ∈ S. Then

A(c + v) = Ac + Av = b + 0 = b.Hence c + v is a solution to Ax = b. 2

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T ,b = (b1,b2, . . . ,bm)

T . Suppose c = (c1, c2, . . . , cn)T is a




A(r− c) = Ar− Ac = b− b = 0.




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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.




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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.




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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.




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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.

Hence r− c ∈ S.

Thus r ∈ c + S.



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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.




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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.




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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.



A(c + v) = Ac + Av = b + 0 = b.

Hence c + v is a solution to Ax = b. 2

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T ,b = (b1,b2, . . . ,bm)





A(r− c) = Ar− Ac = b− b = 0.



A(c + v) = Ac + Av = b + 0 = b.Hence c + v is a solution to Ax = b. 2 36/45

Example 3 (A system with no solution):

x − 5y + 4z = 3x − 5y + 3z = 6

2x − 10y + 13z = 5

⇒

1 −5 41 −5 32 −10 13

∣∣∣∣∣∣365

→ Apply Gauss Elimination Method to get 1 −5 4

0 0 −10 0 0

∣∣∣∣∣∣3314

→ The bottom row corresponds to the equation

0.z = 14.→ Hence the system has no solutions.

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x − 5y + 4z = 3x − 5y + 3z = 6

2x − 10y + 13z = 5

⇒

1 −5 41 −5 32 −10 13

∣∣∣∣∣∣365

→ Apply Gauss Elimination Method to get 1 −5 40 0 −10 0 0

∣∣∣∣∣∣3314



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x − 5y + 4z = 3x − 5y + 3z = 6

2x − 10y + 13z = 5

⇒

1 −5 41 −5 32 −10 13

∣∣∣∣∣∣365


0 0 −10 0 0

∣∣∣∣∣∣3314

→ The bottom row corresponds to the equation0.z = 14.

→ Hence the system has no solutions.

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x − 5y + 4z = 3x − 5y + 3z = 6

2x − 10y + 13z = 5

⇒

1 −5 41 −5 32 −10 13

∣∣∣∣∣∣365


0 0 −10 0 0

∣∣∣∣∣∣3314


0.z = 14.

→ Hence the system has no solutions.

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x − 5y + 4z = 3x − 5y + 3z = 6

2x − 10y + 13z = 5

⇒

1 −5 41 −5 32 −10 13

∣∣∣∣∣∣365


0 0 −10 0 0

∣∣∣∣∣∣3314



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Stage 1: Forward Elimination Phase

The steps in Gaussian elimination can be summarized as follows:

1. Search the first column of [A|b] from the top to the bottom for the firstnon-zero entry, and then if necessary, the second column (the case whereall the coefficients corresponding to the first variable are zero), and thenthe third column, and so on. The entry thus found is called the currentpivot.

2. Interchange, if necessary, the row containing the current pivot with thefirst row.

3. Keeping the row containing the pivot (that is, the first row) untouched,subtract appropriate multiples of the first row from all the other rows toobtain all zeroes below the current pivot in its column.4. Repeat the preceding steps on the submatrix consisting of all thoseelements which are below and to the right of the current pivot.

5. Stop when no further pivot can be found.

Neela Nataraj Lecture 3

REF

The m × n coefficient matrix A of the linear system Ax = b is thusreduced to an (m × n) row echelon matrix U and the augmentedmatrix [A|b] is reduced to[U|c] =

0 . . . p1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ . . . ∗ c1

0 . . . 0 . . . p2 ∗ ∗ ∗ ∗ ∗ . . . ∗ c2

0 . . . 0 0 0 . . . p3 ∗ ∗ ∗ . . . ∗ c3...

......

... 0 0 . . ....

......

...0 . . . 0 0 0 0 0 0 pr ∗ . . . ∗ cr

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...

......

......

......

......

......

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm

.

The entries denoted by ∗ and the ci ’s are real numbers; they may or may

not be zero. The pi ’s denote the pivots; they are non-zero. Note that

there is exactly one pivot in each of the first r rows of U and that any

column of U has at most one pivot. Hence r ≤ m and r ≤ n.


REF


0 . . . p1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ . . . ∗ c1

0 . . . 0 . . . p2 ∗ ∗ ∗ ∗ ∗ . . . ∗ c2

0 . . . 0 0 0 . . . p3 ∗ ∗ ∗ . . . ∗ c3...

......

... 0 0 . . ....

......

...0 . . . 0 0 0 0 0 0 pr ∗ . . . ∗ cr

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...

......

......

......

......

......

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm

.


not be zero.

The pi ’s denote the pivots; they are non-zero. Note that




REF


0 . . . p1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ . . . ∗ c1

0 . . . 0 . . . p2 ∗ ∗ ∗ ∗ ∗ . . . ∗ c2

0 . . . 0 0 0 . . . p3 ∗ ∗ ∗ . . . ∗ c3...

......

... 0 0 . . ....

......

...0 . . . 0 0 0 0 0 0 pr ∗ . . . ∗ cr

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...

......

......

......

......

......

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm

.


not be zero. The pi ’s denote the pivots; they are non-zero.

Note that




REF


0 . . . p1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ . . . ∗ c1

0 . . . 0 . . . p2 ∗ ∗ ∗ ∗ ∗ . . . ∗ c2

0 . . . 0 0 0 . . . p3 ∗ ∗ ∗ . . . ∗ c3...

......

... 0 0 . . ....

......

...0 . . . 0 0 0 0 0 0 pr ∗ . . . ∗ cr

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...

......

......

......

......

......

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm

.




column of U has at most one pivot.

Hence r ≤ m and r ≤ n.


REF


0 . . . p1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ . . . ∗ c1

0 . . . 0 . . . p2 ∗ ∗ ∗ ∗ ∗ . . . ∗ c2

0 . . . 0 0 0 . . . p3 ∗ ∗ ∗ . . . ∗ c3...

......

... 0 0 . . ....

......

...0 . . . 0 0 0 0 0 0 pr ∗ . . . ∗ cr

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cr+1...

......

......

......

......

......

0 . . . 0 0 0 0 0 0 0 0 . . . 0 cm

.






Consistent & Inconsistent Systems

If r < m (the number of non-zero rows is less than the number ofequations)

and cr+k 6= 0 for some k ≥ 1, then the (r + k)th rowcorresponds to the self-contradictory equation 0 = cr+k and so the

system has no solutions (inconsistent system) .

If (i) r = m or (ii) r < m and cr+k = 0 for all k ≥ 1, then there

exists a solution of the system (consistent system) .

(Basic & Free Variables)

If the jth column of U contains a pivot, then xj is called a basicvariable; otherwise xj is called a free variable.

In fact, there are n − r free variables , where n is the number ofcolumns (unknowns) of A (and hence of U).



If r < m (the number of non-zero rows is less than the number ofequations) and cr+k 6= 0 for some k ≥ 1,

then the (r + k)th rowcorresponds to the self-contradictory equation 0 = cr+k and so the









If r < m (the number of non-zero rows is less than the number ofequations) and cr+k 6= 0 for some k ≥ 1, then the (r + k)th rowcorresponds to the self-contradictory equation 0 = cr+k

and so the









If r < m (the number of non-zero rows is less than the number ofequations) and cr+k 6= 0 for some k ≥ 1, then the (r + k)th rowcorresponds to the self-contradictory equation 0 = cr+k and so the























If the jth column of U contains a pivot, then xj is called a basicvariable;

otherwise xj is called a free variable.





















Stage 2: (Back Substitution Phase)

In the case of a consistent system, if xj is a free variable, then itcan be set equal to a parameter sj which can assume arbitraryvalues.

If xj is a basic variable, then we solve for xj in terms ofxj+1, . . . , xm, starting from the last basic variable and working ourway up row by row.


Stage 2: (Back Substitution Phase)

In the case of a consistent system, if xj is a free variable, then itcan be set equal to a parameter sj which can assume arbitraryvalues.If xj is a basic variable, then we solve for xj in terms ofxj+1, . . . , xm, starting from the last basic variable and working ourway up row by row.


REF - A revisit

The forward elimination phase of the Gauss elimination methodleads to the “row echelon form” of a matrix which can be definedas follows:

(Row-Echelon Form (REF))

A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the followingproperties:(a) The all-zero rows (if any) are at the bottom.(b) If we call the left most non-zero entry of a non-zero row its leadingentry, then the leading entry of each non-zero row is to the right of theleading entry of the preceding row.(c) All entries in a column below a leading entry is zero.

( Row Reduced Echelon Form (RREF))

If a matrix in echelon form satisfies the following additional conditions,then it is in row reduced echelon form:(d) The leading entry in each nonzero row is 1.(e) Each leading 1 is the only nonzero entry in its column.


REF - A revisit



A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the followingproperties:

(a) The all-zero rows (if any) are at the bottom.(b) If we call the left most non-zero entry of a non-zero row its leadingentry, then the leading entry of each non-zero row is to the right of theleading entry of the preceding row.(c) All entries in a column below a leading entry is zero.




REF - A revisit



A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the followingproperties:(a) The all-zero rows (if any) are at the bottom.

(b) If we call the left most non-zero entry of a non-zero row its leadingentry, then the leading entry of each non-zero row is to the right of theleading entry of the preceding row.(c) All entries in a column below a leading entry is zero.




REF - A revisit



A matrix is said to be in a row echelon form (or to be a row echelonmatrix) if it has a staircase-like pattern characterized by the followingproperties:(a) The all-zero rows (if any) are at the bottom.(b) If we call the left most non-zero entry of a non-zero row its leadingentry, then the leading entry of each non-zero row is to the right of theleading entry of the preceding row.

(c) All entries in a column below a leading entry is zero.




REF - A revisit







REF - A revisit





If a matrix in echelon form satisfies the following additional conditions,then it is in row reduced echelon form:

(d) The leading entry in each nonzero row is 1.(e) Each leading 1 is the only nonzero entry in its column.


REF - A revisit





If a matrix in echelon form satisfies the following additional conditions,then it is in row reduced echelon form:(d) The leading entry in each nonzero row is 1.

(e) Each leading 1 is the only nonzero entry in its column.


REF - A revisit







Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.

Question : How to obtain RREF?Remarks :

1 Any row reduced echelon matrix is also a row-echelon matrix.2 Any nonzero matrix may be row reduced into more than one

matrix in echelon form. But the row reduced echelon formthat one obtains from a matrix is unique.

3 A pivot is a nonzero number in a pivot position that is used asneeded to create zeros with the help of row operations.

4 Different sequences of row operations might involve adifferent set of pivots.


Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.







Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.







Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.







Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.

Question : How to obtain RREF?

Remarks :1 Any row reduced echelon matrix is also a row-echelon matrix.2 Any nonzero matrix may be row reduced into more than one





Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.

Question : How to obtain RREF?

Remarks :1 Any row reduced echelon matrix is also a row-echelon matrix.2 Any nonzero matrix may be row reduced into more than one





Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.


1 Any row reduced echelon matrix is also a row-echelon matrix.

2 Any nonzero matrix may be row reduced into more than onematrix in echelon form. But the row reduced echelon formthat one obtains from a matrix is unique.




Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.



matrix in echelon form.

But the row reduced echelon formthat one obtains from a matrix is unique.




Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.







Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.







Examples

1

2 3 −2 40 −9 7 −80 0 0 200 0 0 0

is in REF.

2

1 0 0 40 1 0 −80 0 1 20

is in RREF.







EXAMPLES :

Determine whether the following statements are true or false.1 If a matrix is in row echelon form, then the leading entry of

each nonzero row must be 1.

F2 If a matrix is in reduced row echelon form, then the leading

entry of each nonzero row is 1. T3 Every matrix can be transformed into one in reduced row

echelon form by a sequence of elementary row operations.4 If the reduced row echelon form of the augmented matrix of a

system of linear equations contains a zero row, then thesystem is consistent. F

5 If the only nonzero entry in some row of an augmented matrixof a system of linear equations lies in the last column, thenthe system is inconsistent. T

6 If the reduced row echelon form of the augmented matrix of aconsistent system of m linear equations in n variables containsr nonzero rows, then its general solution contains r basicvariables. T (number of free variables = n − r).


EXAMPLES :


each nonzero row must be 1. F2 If a matrix is in reduced row echelon form, then the leading

entry of each nonzero row is 1.

T3 Every matrix can be transformed into one in reduced row






EXAMPLES :




echelon form by a sequence of elementary row operations.

4 If the reduced row echelon form of the augmented matrix of asystem of linear equations contains a zero row, then thesystem is consistent. F




EXAMPLES :





system of linear equations contains a zero row, then thesystem is consistent.

F5 If the only nonzero entry in some row of an augmented matrix

of a system of linear equations lies in the last column, thenthe system is inconsistent. T



EXAMPLES :






5 If the only nonzero entry in some row of an augmented matrixof a system of linear equations lies in the last column, thenthe system is inconsistent.

T6 If the reduced row echelon form of the augmented matrix of a

consistent system of m linear equations in n variables containsr nonzero rows, then its general solution contains r basicvariables. T (number of free variables = n − r).


EXAMPLES :







6 If the reduced row echelon form of the augmented matrix of aconsistent system of m linear equations in n variables containsr nonzero rows, then its general solution contains r basicvariables.

T (number of free variables = n − r).


EXAMPLES :







6 If the reduced row echelon form of the augmented matrix of aconsistent system of m linear equations in n variables containsr nonzero rows, then its general solution contains r basicvariables. T

(number of free variables = n − r).


EXAMPLES :









More Examples - REF, RREF

Determine whether the following matrices in REF, RREF :

1

1 2 3 4 50 0 1 −1 20 0 0 1 5

is an example of a matrix in REF, not

RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3

is in RREF. Is it in REF?? YES.

3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3

is in RREF and hence in REF.

Note that the left side of a matrix in RREF need not be identitymatrix.




1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5

is an example of a matrix in REF,

not

RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3

is in RREF. Is it in REF??

YES.

3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT

in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3

is in RREF

and hence in REF.





1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3






1

1 2 3 4 50 0 1 −1 20 0 0 1 5


RREF.

2

1 0 0 0 10 1 0 0 20 0 0 1 3


3 The matrix

2 −1 2 1 50 1 1 −3 30 2 0 0 50 0 0 3 2

is NOT in REF.

4

1 0 0 1/2 0 10 0 1 −1/3 0 20 0 0 0 1 3




Solution given REF

Example

Consider the REF form of the augmented matrix given by1 1 2 3 20 1 5 0 10 0 0 1 20 0 0 0 0

.

In this case r = 3 (number of pivots) and n = 4 (number ofvariables).

Note that x3 is a free variable and x1, x2, x4 are basic variables.Back substitution yields x4 = 2,

x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.

We get infinitely many solutions (since x3 is a free parameter).


Solution given REF

Example

Consider the REF form of the augmented matrix given by

1 1 2 3 20 1 5 0 10 0 0 1 20 0 0 0 0

.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example

Consider the REF form of the augmented matrix given by1 1 2 3 2

0 1 5 0 10 0 0 1 20 0 0 0 0

.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example

Consider the REF form of the augmented matrix given by1 1 2 3 20 1 5 0 1

0 0 0 1 20 0 0 0 0

.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example

Consider the REF form of the augmented matrix given by1 1 2 3 20 1 5 0 10 0 0 1 2

0 0 0 0 0

.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.

In this case r = 3 (number of pivots)

and n = 4 (number ofvariables).


x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.


Note that x3 is a free variable

and x1, x2, x4 are basic variables.Back substitution yields x4 = 2,

x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.


Note that x3 is a free variable and x1, x2, x4 are basic variables.

Back substitution yields x4 = 2,

x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.


Note that x3 is a free variable and x1, x2, x4 are basic variables.Back substitution yields

x4 = 2,

x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1

=⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3

= 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2

=⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.

We get infinitely many solutions

(since x3 is a free parameter).


Solution given REF

Example


.



x2 + 5x3 = 1 =⇒ x2 = 1− 5x3 = 1− 5s

x1 + x2 + 2x3 + 3x4 = 2 =⇒ x1 = 2− (1− 5s)− 2s − 3× 2

=⇒ x1 = −5 + 3s.



Elementary Matrices

An m ×m elementary matrix is a matrix obtained from the m ×midentity matrix Im by one of the elementary operations, namely,

1 interchange of two rows,

2 multiplying a row by a non-zero constant,

3 adding a constant multiple of a row to anotherrow.

That is, an elementary matrix is a matrix which differs from theidentity matrix by one single elementary row operation.


Elementary Matrices







Elementary Matrices







Elementary Matrices







Elementary Matrices







Elementary Matrices







Elementary Matrices







1. Row Switching Operation

Interchange of two rows - Ri ↔ Rj .

The elementary matrix Ei ,j corresponding to this operation on Im isobtained by swapping row i and row j of the identity matrix.

Ei ,j =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 0 . . . 1...

. . .

Rj 1 . . . 0...

. . .

Rm 1

Example : Consider I3. R1 ↔ R2 gives E1,2 =

0 1 01 0 00 0 1

.





Ei ,j =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 0 . . . 1...

. . .

Rj 1 . . . 0...

. . .

Rm 1


0 1 01 0 00 0 1

.





Ei ,j =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 0 . . . 1...

. . .

Rj 1 . . . 0...

. . .

Rm 1


0 1 01 0 00 0 1

.





Ei ,j =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 0 . . . 1...

. . .

Rj 1 . . . 0...

. . .

Rm 1

Example : Consider I3.

R1 ↔ R2 gives E1,2 =

0 1 01 0 00 0 1

.





Ei ,j =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 0 . . . 1...

. . .

Rj 1 . . . 0...

. . .

Rm 1


0 1 01 0 00 0 1

.





Ei ,j =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 0 . . . 1...

. . .

Rj 1 . . . 0...

. . .

Rm 1


0 1 01 0 00 0 1

.


2. Row Multiplying Transformation

Multiplying a row by a non-zero constant - Ri −→ kRi .

Theelementary matrix Ei (k) corresponding to this operation on Im isobtained by multiplying row i of the identity matrix by a non-zeroconstant k.

Ei (k) =

C1 C2 . . . Ci . . . Cm

R1 1R2 1...

. . .

Ri k...

. . .

Rm 1

Example : Consider I3. R3 → 7R3 gives E3(7) =

1 0 00 1 00 0 7

.



Multiplying a row by a non-zero constant - Ri −→ kRi . Theelementary matrix Ei (k) corresponding to this operation on Im isobtained by multiplying row i of the identity matrix by a non-zeroconstant k.

Ei (k) =

C1 C2 . . . Ci . . . Cm

R1 1R2 1...

. . .

Ri k...

. . .

Rm 1


1 0 00 1 00 0 7

.




Ei (k) =

C1 C2 . . . Ci . . . Cm

R1 1R2 1...

. . .

Ri k...

. . .

Rm 1


1 0 00 1 00 0 7

.




Ei (k) =

C1 C2 . . . Ci . . . Cm

R1 1R2 1...

. . .

Ri k...

. . .

Rm 1

Example : Consider I3.

R3 → 7R3 gives E3(7) =

1 0 00 1 00 0 7

.




Ei (k) =

C1 C2 . . . Ci . . . Cm

R1 1R2 1...

. . .

Ri k...

. . .

Rm 1


1 0 00 1 00 0 7

.




Ei (k) =

C1 C2 . . . Ci . . . Cm

R1 1R2 1...

. . .

Ri k...

. . .

Rm 1


1 0 00 1 00 0 7

.


3. Row Addition Transformation

Ri −→ Ri + kRj :

The elementary matrix Ei ,j(k) corresponding

to this operation on Im is obtained by multiplying row j of theidentity matrix by a non-zero constant k and adding with row i .

Ei ,j(k) =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 1 k...

. . .

Rj 1...

. . .

Rm 1

Example : R2 → R2 + (−3)R1 for I3 gives E2,1(−3) =

1 0 0−3 1 00 0 1

.



Ri −→ Ri + kRj : The elementary matrix Ei ,j(k) corresponding


Ei ,j(k) =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 1 k...

. . .

Rj 1...

. . .

Rm 1


1 0 0−3 1 00 0 1

.





Ei ,j(k) =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 1 k...

. . .

Rj 1...

. . .

Rm 1


1 0 0−3 1 00 0 1

.





Ei ,j(k) =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 1 k...

. . .

Rj 1...

. . .

Rm 1

Example :

R2 → R2 + (−3)R1 for I3 gives E2,1(−3) =

1 0 0−3 1 00 0 1

.





Ei ,j(k) =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 1 k...

. . .

Rj 1...

. . .

Rm 1


1 0 0−3 1 00 0 1

.





Ei ,j(k) =

C1 . . . Ci . . . Cj . . . Cm

R1 1...

. . .

Ri 1 k...

. . .

Rj 1...

. . .

Rm 1


1 0 0−3 1 00 0 1

.


Proposition 1

Let A be an m×n matrix. If A is obtained from A by an elementaryrow operation, and E is the corresponding m×m elementary matrix,then EA = A.

Proof :

Let Im =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

=

eT1

eT2...

eTm

where ei =

0...1...0

, with 1 is in the i th position, 1 ≤ i ≤ m.

Also, eTi A = A(i), the i th row of A.


Proposition 1


Proof :

Let Im =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

=

eT1

eT2...

eTm

where ei =

0...1...0




Proposition 1


Proof :

Let Im =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

=

eT1

eT2...

eTm

where ei =

0...1...0




Proposition 1


Proof :

Let Im =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

=

eT1

eT2...

eTm

where ei =

0...1...0




Proposition 1


Proof :

Let Im =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

=

eT1

eT2...

eTm

where ei =

0...1...0


Also, eTi A =

A(i), the i th row of A.


Proposition 1


Proof :

Let Im =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

=

eT1

eT2...

eTm

where ei =

0...1...0




Proposition 1


Proof :

Let Im =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

=

eT1

eT2...

eTm

where ei =

0...1...0




Ri ↔ Rj(i < j)

Ei ,jA =

...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.

Ri −→ kRi Ei (k)A =

...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.


Ri ↔ Rj(i < j) Ei ,jA =

...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A

=

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.

Ri −→ kRi

Ei (k)A =

...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A

=

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A

=

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.



...eT

j...

eTi...

A =

...eT

j A...

eTi A...

=

...A(j)

...A(i)

...

= A.


...

keTi

...

A =

...

keTi A...

=

...

kA(i)...

= A.

Ri −→ Ri + kRj

Ei ,j(k)A =

...

eTi + keT

j...

A =

...

(eTi + keT

j )A...

=

...

A(i) + kA(j)...

= A.


Proposition 2 :

Elementary matrices are invertible and the inverses are also elemen-tary matrices.

In fact,• If E corresponds to Ri ↔ Rj , then E−1

i ,j = Ei ,j ;• if Ei (k) corresponds to Ri −→ kRi , then Ei (k)−1 is the

elementary matrix corresponding to Ri →1

kRi ; that is, Ei (1/k).

• if Ei ,j(k) corresponds to Ri −→ Ri + kRj , then Ei ,j(k)−1 is theelementary matrix corresponding to Ri → Ri − kRj , that is,Ei ,j(−k).

Exercise : In each case, check EE−1 = I .


Proposition 2 :


In fact,

• If E corresponds to Ri ↔ Rj , then E−1i ,j = Ei ,j ;

• if Ei (k) corresponds to Ri −→ kRi , then Ei (k)−1 is the






Proposition 2 :



i ,j

= Ei ,j ;• if Ei (k) corresponds to Ri −→ kRi , then Ei (k)−1 is the






Proposition 2 :



i ,j = Ei ,j ;

• if Ei (k) corresponds to Ri −→ kRi , then Ei (k)−1 is the






Proposition 2 :









Proposition 2 :









Proposition 2 :









Invertibility of elementary matrices

Since row operations are reversible, elementary matrices areinvertible,

for if E is produced by a row operation on Im, there isanother row operation of the same type, that changes E back toIm. Hence, there is an elementary operation F such that FE = Im.Since E and F correspond to reverse operations, EF = Im too. �

Example

Find the inverses of

E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 = E3,1(4) =

1 0 00 1 04 0 1

, E2(7)−1 =

1 0 00 1/7 00 0 1

& E−1

1,3 =

0 0 10 1 01 0 0

.



Since row operations are reversible, elementary matrices areinvertible, for if E is produced by a row operation on Im, there isanother row operation of the same type, that changes E back toIm. Hence, there is an elementary operation F such that FE = Im.

Since E and F correspond to reverse operations, EF = Im too. �

Example


E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 = E3,1(4) =

1 0 00 1 04 0 1

, E2(7)−1 =

1 0 00 1/7 00 0 1

& E−1

1,3 =

0 0 10 1 01 0 0

.



Since row operations are reversible, elementary matrices areinvertible, for if E is produced by a row operation on Im, there isanother row operation of the same type, that changes E back toIm. Hence, there is an elementary operation F such that FE = Im.Since E and F correspond to reverse operations, EF = Im too. �

Example


E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 = E3,1(4) =

1 0 00 1 04 0 1

, E2(7)−1 =

1 0 00 1/7 00 0 1

& E−1

1,3 =

0 0 10 1 01 0 0

.




Example


E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 =

E3,1(4) =

1 0 00 1 04 0 1

, E2(7)−1 =

1 0 00 1/7 00 0 1

& E−1

1,3 =

0 0 10 1 01 0 0

.




Example


E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 = E3,1(4)

=

1 0 00 1 04 0 1

, E2(7)−1 =

1 0 00 1/7 00 0 1

& E−1

1,3 =

0 0 10 1 01 0 0

.




Example


E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 = E3,1(4) =

1 0 00 1 04 0 1

,

E2(7)−1 =

1 0 00 1/7 00 0 1

& E−1

1,3 =

0 0 10 1 01 0 0

.




Example


E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 = E3,1(4) =

1 0 00 1 04 0 1

, E2(7)−1 =

1 0 00 1/7 00 0 1

& E−11,3 =

0 0 10 1 01 0 0

.




Example


E3,1(−4) =

1 0 00 1 0−4 0 1

, E2(7) =

1 0 00 7 00 0 1

& E1,3 =

0 0 10 1 01 0 0

.

Soln: E3,1(−4)−1 = E3,1(4) =

1 0 00 1 04 0 1

, E2(7)−1 =

1 0 00 1/7 00 0 1

& E−1

1,3 =

0 0 10 1 01 0 0

.


Corollary

Corollary

If A is an invertible matrix,

then A can be written as a product ofelementary matrices.

Proof : If A is invertible, then A is row equivalent to identitymatrix, that is, there exists elementary matrices E1, . . . , Ek suchthat Ek . . . E1A = I .This gives A = (Ek . . . E1)−1 = E−1

1 . . . E−1k .

Gauss-Jordan method for finding A−1

Let A be invertible, and Ek . . . E1A = I .Then, Ek . . . E1I = A−1.That is, the elementary row operations performed on A to reduceit to identity matrix, performed in the same order on I reduces I toA−1.Exercises : Tut. Sheet 2 - Qns 2, 6.


Corollary

Corollary

If A is an invertible matrix, then A can be written as a product ofelementary matrices.


1 . . . E−1k .




Corollary

Corollary


Proof :

If A is invertible, then A is row equivalent to identitymatrix, that is, there exists elementary matrices E1, . . . , Ek suchthat Ek . . . E1A = I .This gives A = (Ek . . . E1)−1 = E−1

1 . . . E−1k .




Corollary

Corollary


Proof : If A is invertible,

then A is row equivalent to identitymatrix, that is, there exists elementary matrices E1, . . . , Ek suchthat Ek . . . E1A = I .This gives A = (Ek . . . E1)−1 = E−1

1 . . . E−1k .




Corollary

Corollary


Proof : If A is invertible, then A is row equivalent to identitymatrix,

that is, there exists elementary matrices E1, . . . , Ek suchthat Ek . . . E1A = I .This gives A = (Ek . . . E1)−1 = E−1

1 . . . E−1k .




Corollary

Corollary


Proof : If A is invertible, then A is row equivalent to identitymatrix, that is, there exists elementary matrices E1, . . . , Ek

suchthat Ek . . . E1A = I .This gives A = (Ek . . . E1)−1 = E−1

1 . . . E−1k .




Corollary

Corollary


Proof : If A is invertible, then A is row equivalent to identitymatrix, that is, there exists elementary matrices E1, . . . , Ek suchthat Ek . . . E1A = I .

This gives A = (Ek . . . E1)−1 = E−11 . . . E−1

k .




Corollary

Corollary



1 . . . E−1k .




Corollary

Corollary



1 . . . E−1k .


Let A be invertible, and Ek . . . E1A = I .

Then, Ek . . . E1I = A−1.That is, the elementary row operations performed on A to reduceit to identity matrix, performed in the same order on I reduces I toA−1.Exercises : Tut. Sheet 2 - Qns 2, 6.


Corollary

Corollary



1 . . . E−1k .


Let A be invertible, and Ek . . . E1A = I .Then, Ek . . . E1I = A−1.

That is, the elementary row operations performed on A to reduceit to identity matrix, performed in the same order on I reduces I toA−1.Exercises : Tut. Sheet 2 - Qns 2, 6.


Corollary

Corollary



1 . . . E−1k .


Let A be invertible, and Ek . . . E1A = I .Then, Ek . . . E1I = A−1.That is, the elementary row operations performed on A to reduceit to identity matrix, performed in the same order on I reduces I toA−1.

Exercises : Tut. Sheet 2 - Qns 2, 6.


Corollary

Corollary



1 . . . E−1k .




Inverse of A : an example

[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]

This completes the forward elimination. The first half of elimination has

taken A to echelon form B, and now the second half will take B to I .

That is, we create 0′s above the pivots in the last matrix.



[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]

This completes the forward elimination.

The first half of elimination has





[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]






[A|I ] =

2 1 1

... 1 0 0

4 −6 0... 0 1 0

−2 7 2... 0 0 1

R2→R2+(−2)R1−−−−−−−−−−→R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 8 3... 1 0 1

R3→R3+(1)R1−−−−−−−−−→

2 1 1

... 1 0 0

0 −8 −2... −2 1 0

0 0 1... −1 1 1

= [B|L]



That is, we create 0′s above the pivots in the last matrix.Neela Nataraj Lecture 4

Example continued...

[B|L]

R2→R2+(2)R3−−−−−−−−−→R1→R1+(−1)R3−−−−−−−−−−→

2 1 0

... 2 −1 −1

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1+(1/8)R3−−−−−−−−−−→

2 0 0

... 128 −5

8 −68

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1/2−−−−−−→

R2→R2/−8−−−−−−−→

1 0 0

... 1216 − 5

16 − 616

0 1 0... 4

8 −38 −2

8

0 0 1... −1 1 1

= [I |A−1]

Exercises : Tut. Sheet 2 : Qn. 4, Tut. Sheet 3 : Qns. 1 & 2



[B|L]

R2→R2+(2)R3−−−−−−−−−→R1→R1+(−1)R3−−−−−−−−−−→

2 1 0

... 2 −1 −1

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1+(1/8)R3−−−−−−−−−−→

2 0 0

... 128 −5

8 −68

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1/2−−−−−−→

R2→R2/−8−−−−−−−→

1 0 0

... 1216 − 5

16 − 616

0 1 0... 4

8 −38 −2

8

0 0 1... −1 1 1

= [I |A−1]




[B|L]

R2→R2+(2)R3−−−−−−−−−→R1→R1+(−1)R3−−−−−−−−−−→

2 1 0

... 2 −1 −1

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1+(1/8)R3−−−−−−−−−−→

2 0 0

... 128 −5

8 −68

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1/2−−−−−−→

R2→R2/−8−−−−−−−→

1 0 0

... 1216 − 5

16 − 616

0 1 0... 4

8 −38 −2

8

0 0 1... −1 1 1

= [I |A−1]




[B|L]

R2→R2+(2)R3−−−−−−−−−→R1→R1+(−1)R3−−−−−−−−−−→

2 1 0

... 2 −1 −1

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1+(1/8)R3−−−−−−−−−−→

2 0 0

... 128 −5

8 −68

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1/2−−−−−−→

R2→R2/−8−−−−−−−→

1 0 0

... 1216 − 5

16 − 616

0 1 0... 4

8 −38 −2

8

0 0 1... −1 1 1

= [I |A−1]




[B|L]

R2→R2+(2)R3−−−−−−−−−→R1→R1+(−1)R3−−−−−−−−−−→

2 1 0

... 2 −1 −1

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1+(1/8)R3−−−−−−−−−−→

2 0 0

... 128 −5

8 −68

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1/2−−−−−−→

R2→R2/−8−−−−−−−→

1 0 0

... 1216 − 5

16 − 616

0 1 0... 4

8 −38 −2

8

0 0 1... −1 1 1

= [I |A−1]




[B|L]

R2→R2+(2)R3−−−−−−−−−→R1→R1+(−1)R3−−−−−−−−−−→

2 1 0

... 2 −1 −1

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1+(1/8)R3−−−−−−−−−−→

2 0 0

... 128 −5

8 −68

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1/2−−−−−−→R2→R2/−8−−−−−−−→

1 0 0

... 1216 − 5

16 − 616

0 1 0... 4

8 −38 −2

8

0 0 1... −1 1 1

= [I |A−1]




[B|L]

R2→R2+(2)R3−−−−−−−−−→R1→R1+(−1)R3−−−−−−−−−−→

2 1 0

... 2 −1 −1

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1+(1/8)R3−−−−−−−−−−→

2 0 0

... 128 −5

8 −68

0 −8 0... −4 3 2

0 0 1... −1 1 1

R1→R1/2−−−−−−→

R2→R2/−8−−−−−−−→

1 0 0

... 1216 − 5

16 − 616

0 1 0... 4

8 −38 −2

8

0 0 1... −1 1 1

= [I |A−1]



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Gaussian Elimination - IIT Bombay