1Exam 1: Solutions

1. Given the linear system x1 + 4x2 2x3 = 3x1 + 7x2 5x3 = 13x1 + x2 + x3 = 55x1 + 9x2 3x3 = 112x1 3x2 + 3x3 = 2

(1)

use Gauss elimination to obtain an equivalent system whose coefficient matrix is in rowechelon form. If the system is consistent and involves no free variables, use back substitutionto find the unique solution. If the system is consistent and there are free variables, transformit to reduced row echelon form and find all solutions.

Solution. First, we write the augmented matrix for this system1 4 2

1 7 53 1 15 9 32 3 3

315112

.Performing the elementary row operations [2]+[1], [3]3[1], [4]5[1], and [5]2[1] (where[i] denotes the ith row) one arrives with the equivalent augmented matrix

1 4 20 11 70 11 70 11 70 11 7

34

444

which, in its turn, is equivalent to

1 4 20 1 7110 0 00 0 00 0 0

3411

000

.The last matrix is already in row echelon form which implies that the linear system underconsideration is consistent. The unknown x3 is free variable and x1 and x2 are lead variables.Performing the elementary row operation [1] 4[2] we finally transform the matrix to thefollowing reduced row echelon form

1 0 6110 1 7110 0 00 0 00 0 0

1711411

000

.Assume that x3 = where is an arbitrary real number. Then x1 = 1711 611 and x2 =411 +

711. Therefore, the solutions of the system (1) are(

1711 611, 411 + 711,

), R.

22. Let B be an n n matrix and let C = B BT . Show that C is skew-symmetric1.Solution. The property (A D)T = AT DT which holds for arbitrary matrices A and Dof the same dimensions, and the property (AT )T = A imply

CT = (B BT )T = BT (BT )T = BT B = (B BT ) = C.Therefore, CT = C. Then by definition of skew-symmetricity (see footnote on this page)one concludes that the matrix C is skew-symmetric.

3. Let I denote the n n identity matrix. Find a block form for the inverse of the 2n 2nmatrix (

B II 0

).

Solution. SinceA =

(B II 0

)is a 2 2 block matrix, we look for its inverse X = A1 (if it exists) also in the 2 2 blockform

X =(

X11 X12X21 X22

),

where Xij , i, j = 1, 2 are nn matrices. Being the inverse of A the matrix X should satisfyin particular the equation AX = I2n where I2n =

(I 00 I

)is the 2n 2n identity matrix,

that is, (B II 0

)(X11 X12X21 X22

)=(

I 00 I

)which is equivalent to(

BX11 +X21 BX12 +X22X11 X12

)=(

I 00 I

). (2)

Matrix equation (2) splits into the two following systems of equations for the entries of thefirst and second columns of X , respectively,{

BX11 + X21 = IX11 = 0

and{

BX12 + X22 = 0X12 = I

. (3)

From the first system in (3) one infers that X11 = 0 and then X21 = 0, and from the secondthat X12 = I and then X22 = B. Thus A is invertible and

A1 = X =(

0 II B

).

1Recall that a matrix A is said to be skew-symmetric if AT = A.

34. Given

A =

2 1 14 6 22 6 4

(a) Find elementary matrices E1, E2, E3 such that

E3E2E1A = U

where U is an upper triangular matrix.(b) Determine the inverses of E1, E2, E3 and compute L = E11 E12 E13 . What type of

matrix is L?

Solution. (a) The elementary row operations [2] 2[1] and [3] [1] transform the matrix Ato the form 2 1 10 4 0

0 5 3

. (4)Performing the operations [2] 2[1] and [3] [1] is equivalent to premultiplication by theelementary matrices

E1 =

1 0 02 1 00 0 1

and E2 = 1 0 00 1 01 0 1 ,

respectively. Further, the elementary row operation [3] 54 [2] transforms the matrix (4) tothe upper triangular form

U =

2 1 10 4 00 0 3

. (5)The elementary matrix E3 corresponding to this operation is given by

E3 =

1 0 00 1 00 54 1

.Therefore, we have found the elementary matrices E1, E2, E3 such that E3E2E1A = Uwhere U is the upper triangular matrix (5).(b) The inverses of E1, E2, E3 read as follows

E11 =

1 0 02 1 00 0 1

, E12 = 1 0 00 1 0

1 0 1

, E13 = 1 0 00 1 0

0 54 1

.

4Computing the product L = E11 E12 E13 one obtains

L =

1 0 02 1 01 54 1

.The matrix L has lower triangular form.

5. Find the inverse of the matrix

A =

1 3 212

12 0

12 12 12

.Use A1 to solve Ax = b for b = (2 1 1)T .

Solution. We write the augmented matrix (A|I) and first perform the type II elementary rowoperations (1)[1], 2[2] and (2)[3] transforming this matrix to a more convenient form.Later on we perform the operations [2] [1], [3] [1], (12)[2], and [3]+4[1]. In this processwe deal with the following chain of augmented matrices 1 3 21

212 0

12 12 12

1 0 00 1 00 0 1

1 3 21 1 01 1 1

1 0 00 2 00 0 2

1 3 20 2 2

0 4 3

1 0 01 2 01 0 2 1 3 20 1 1

0 0 1

1 0 012 1 01 4 2

.Performing the row operations [2] [3], [1] 2[3], and then [1] 3[2] we arrive with anaugmented matrix in reduced row echelon form: 1 3 00 1 0

0 0 1

1 8 412 3 2

1 4 2

1 0 00 1 0

0 0 1

12 1 2

12 3 2

1 4 2

.This yields

A1 =

12 1 212 3 21 4 2

.The remaining step consists in computing x = A1b,

x =

12 1 212 3 21 4 2

211

= 468

.

Problem 1Problem 2Problem 3Problem 4Problem 5