Properties of DeterminantDeterminant: To each square matrix A we can associate a expression or number (real or complex) known as its determinant denoted by det A or ∣A∣If A=[ a ] then ∣A∣=a .Determinant of a square matrix of order 2:

If A=[a bc d ] then ∣A∣=ad – bc

For example ∣2 3– 3 4∣=2×4– 3×–3=89=17

Determinant of a square matrix matrix of order 3:

Let A=[a b cx y zu v w ] then we can find its determinant in six possible ways

Expansion Along First Row:

∣A∣= a∣y zv w∣– b∣x z

u w∣ c∣x yu v∣

Expansion Along Second Row:

∣A∣=−x∣b cv w∣ y∣a c

u w∣− z∣a bu v∣

Expansion Along Third Row:

∣A∣=u∣b cy z∣−v∣a c

x z∣w∣a bx y∣

Expansion along first column:

∣A∣=a∣y zv w∣– x∣b c

v w∣u∣b cy z∣

Expansion along second column:

∣A∣=−b∣x zu w∣ y∣a c

u w∣– v∣a cx z∣

Expansion along third column:

∣A∣=c∣x yu v∣– z∣a b

u v∣w∣a bx y∣

Note: We expand the determinant along a row or column which has maximum number of zeros.

Example-1: Evaluate the determinant ∣2 −3 13 −2 41 2 5∣ by expanding it along first row.

Solution: ∣2 −3 13 −2 41 2 5∣=2∣−2 4

2 5∣−−3∣3 41 5∣1∣3 −2

1 2 ∣⇒∣2 −3 1

3 −2 41 2 5∣=2−10−83 15−462=−36338=5

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Example-2: Find the value of the determinant ∣ 2 3 −1−1 2 −42 −1 3 ∣ by expanding it along the second row.

Solution:

∣2 3 −1−1 2 −42 −1 3 ∣=−−1∣ 3 −1

−1 3 ∣2∣2 −12 3 ∣−−4∣2 3

2 −1∣=9−12 624 −2−6

⇒∣2 3 −1−1 2 −42 −1 3 ∣=816−32=−8

Example-3: Find the value of the determinant ∣3 2 −1−2 1 14 −2 3 ∣ by expanding along it third row.

Solution: ∣ 3 2 −1−2 1 14 −2 3 ∣=4∣2 −1

1 1 ∣−−2∣ 3 −1−2 1 ∣3∣ 3 2

−2 1∣⇒∣ 3 2 −1

−2 1 14 −2 3 ∣=42123−23 34=12221=35

Example-4: Find the value of the determinant ∣2 1 13 −2 31 2 1∣ by expanding along first column.

Solution: ∣2 1 13 −2 31 2 1∣=2∣−2 3

2 1∣−3∣1 12 1∣1∣ 1 1

−2 3∣

∣2 1 13 −2 31 2 1∣=2−2−6−31−232=−1635=−8

Example-5: Find the value of the determinant ∣2 −1 34 2 −15 2 1 ∣ by expanding it along second column.

Solution: ∣2 −1 34 2 −15 2 1 ∣=−−1∣4 −1

5 1 ∣2∣2 35 1∣−2∣2 3

4 −1∣⇒∣2 −1 3

4 2 −15 2 1 ∣=4522−15−2 −2−12=9−2628=11

Example-6: Find the value of the determinant ∣3 −1 −12 2 −12 3 −1∣ by expanding it along the third column.

Solution: ∣3 −1 −12 2 −12 3 −1∣=−1∣2 2

2 3∣−−1∣3 −12 3 ∣−1∣3 −1

2 2 ∣

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∣3 −1 −12 2 −12 3 −1∣=−6−492−62=−211−8=1

Exercise-4.1

1. Find the value of the determinant ∣3 1 22 −1 34 1 2∣ by expanding it along the first row. Also find

its value by expanding it along first column.

2. Find the value of the determinant ∣ 3 −1 1−2 1 14 2 1∣ by expanding it along the second row. Also

find its value by expanding it along second column.

3. Find the value of the determinant ∣ 3 4 0−1 2 32 4 5∣ by expanding it along the third row. Also find

its value by expanding it along third column.

4. Find the value of the determinant ∣ a 0 11 2 1−1 1 2∣

5. Find the value of the determinant ∣ 2 0 1−1 0 21 2 1∣

6. Find the value of the determinant ∣cos −sinsin cos ∣

7. Find the values of x for which ∣3 xx 1∣=∣3 2

4 1∣8. Find the value of thew determinant ∣x2−x1 x−1

x1 x1∣9. Find the value of the determinant ∣ 0 sin −cos

−sin 0 sincos −sin 0 ∣

Properties of Determinant

Property-1: Let A=[ aij ] be a square matrix of order n then ∣A∣=∣AT∣

For example ∣a b cx y zu v w∣=∣

a x ub y vc z w∣

Property-2: Let A=[ aij ] be a square matrix of order n n≥2 and B be matrix obtained from A by interchanging any two rows(columns) of A , then ∣A∣=–∣B∣

4

For example ∣a b cx y zu v w∣= – ∣x y z

a b cu v w∣ or ∣a b c

x y zu v w∣= – ∣b a c

y x zv u w∣

Property-3: If any two rows(columns) of a square matrix A=[ aij ] of order n are identical, then its determinant is zero i.e. ∣A∣=0

For example ∣a b cx y za b c∣= 0 or ∣a b a

x y xu v u∣= 0

Property-4: Let A=[ aij ] be a square matrix of order n , and let B be the matrix obtained from A by multiplying each element of a row(column) of A by a scalar then ∣B∣=k∣A∣

For example ∣ka kb kcx y zu v w ∣= k∣a b c

x y zu v w∣ or ∣ka b c

kx y zku v w∣= k∣a b c

x y zu v w∣

Property-5: Let A be a square matrix such that each element of a row(column) of A is expressed as the sum of two or more terms. Then, the determinant of A can be expressed as the sum of determinants of two or more matrices of of the same order.

For example ∣a a ' b b ' c c 'x y zu v w ∣=∣a b c

x y zu v w∣∣

a ' b ' c 'x y zu v w ∣

or ∣ a b cx x ' y zu u ' v w∣=∣

a b cx y zu v w∣∣

a ' b cx ' y zu ' v w∣

Property-6: Let A be any square matrix and B be a matrix obtained from A by adding to a row(column) of A a scalar multiple of another row(column) of A , then ∣B∣=∣A∣

For example ∣a b cx y zu v w∣=∣

a ku b kv c kwx y zu v w ∣ or

∣a b cx y zu v w∣=∣

a kc b cx kz y zu kw v w∣

Example1. Prove that ∣ 1 1 1a b c

bc ca ab∣=0

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Solution: Let =∣ 1 1 1a b c

bc ca ab∣Applying R2R2R3 we get

=∣ 1 1 1abc abc abcbc ca ab ∣

Taking abc common from R2 we get

=abc ∣ 1 1 11 1 1

bc ca ab∣

⇒=0 .̈ R1and R2are identical

Example-2: Prove that ∣a−b b−c c−ab−c c−a a−bc−a a−b b−c∣=0

Solution: Let =∣a−b b−c c−ab−c c−a a−bc−a a−b b−c∣

Applying R1R1C 2R3

=∣ 0 0 0b−c c−a a−bc−a a−b b−c∣

⇒ =0 ( .̈ all elements in R1 are zero)

Example-3 Prove that ∣ 1 1 1ab bc ca

c ab a bc b ca ∣=0

Solution: Let =∣ 1 1 1ab bc ca

c ab abc bca∣Applying R2R2R3 we get

=∣ 1 1 1abbcca abbcca abbccac ab a bc bca ∣

⇒ =abbcca∣ 1 1 11 1 1

c ab a bc bca∣⇒=0 (As R1 and R2 are identical)

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Example-4: Prove that ∣1 1 1a b ca2 b2 c2∣=a−bb−c c−a

Solution: Let =∣1 1 1a b ca2 b2 c2∣

Applying C1C1– C2 and C2C2−C3 we get

=∣ 0 0 1a−b b−c ca2−b2 b2−c2 c2∣

Taking a – b and b – c common from C1 and C2 respectively we get

=a−bb−c ∣ 0 0 11 1 c

ab bc c2∣

Expanding along R1 we get =a – bb– c {bc – ab}=a−b b−cc−a

Example-5 Prove that ∣1 1 1a b ca3 b3 c3∣=a – bb – cc – aabc

Solution: Let =∣1 1 1a b ca3 b3 c3∣

Applying C1C1−C2 andC 2C2−C3 we get

=∣ 0 0 1a−b b−c ca3−b3 b3−c3 c3∣

⇒=a – bb – c∣ 0 0 11 1 c

a2abb2 b2bcc2 c3∣ Applying C1C1– C2 we get

=a−bb−c ∣ 0 0 10 1 c

a2ab−bc−c2 b2bcc2 c2∣=a−bb−c ∣ 0 0 1

0 1 ca−cabc b2bcc2 c2∣

=a−bb−ca−cabc ∣0 0 10 1 c1 b2bcc2 c2∣ (Taking out a−c abc from C1 )

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=a−bb−ca−cabc 1⋅∣0 11 b2bcc2∣ (Expanding along R3 )

=a – bb– c a−cabc×−1=a−bb−cc−aabc

Example 6: Prove that ∣1a 1 11 1b 11 1 1c∣=abc 11

a1b1c =abcbccaab

Solution: Let =∣1a 1 11 1b 11 1 1c∣

Taking a ,b , c common from R1, R2 and R3 we get

=abc∣1a1 1

a1a

1b

1b1 1

b1c

1c

1c1∣

Applying R1 R2R3 we get

=abc∣11a1b1c

1 1a1b1c

11a 1b1c

1b

11b

1b

1c

1c

11c

∣Taking 11

a1b1c common from R1 we get

=abc11a1b1c ∣1 1 1

1b

11b

1b

1c

1c

11c∣

Applying C1 C1– C2 and C2 C2 –C3 we get

=abc11a1b1c ∣ 0 0 1

– 1 1 1b

0 – 1 11c∣

Expanding along R1 we get

=abc11a1b1c 1⋅∣–1 1

0 –1∣=abc11a1b1c =abcbccaab

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Example-7 Prove that ∣1a2– b2 2ab – 2b2ab 1– a2b2 2a2b – 2 a 1 – a2 – b2∣=1a2b23

Solution: Let =∣1a2 – b2 2 ab – 2b2ab 1 – a2b2 2a2b – 2a 1– a2– b2∣

Applying C1C1−bC3 and C2C2aC3 we get

=∣ 1a2b2 0 – 2b0 1a2b2 2 a

b 1a2b2 – a1a2b2 1– a2– b2∣Taking 1a2b2 common from both C1 and C2 we get

=1a2b22∣1 0 – 2b0 1 2ab – a 1– a2– b2∣

Applying R3 R3– bR1aR2 we get

=1a2b22∣1 0 – 2b0 1 2a0 0 1a2b2∣

Expanding along R3 we get

=1a2b221a2b2⋅∣1 00 1∣

⇒ =1a2b23

Example 8: Prove ∣bc 2 a2 a2

b2 ca2 b2

c2 c2 ab2∣=2abc abc3

Solution: Let =∣bc2 a2 a2

b2 ca2 b2

c2 c2 ab2∣Applying C1C1– C3 and C2C2 –C 3 we get

=∣bc2 –a2 0 a2

0 ca2– b2 b2

c2– ab2 c2– ab2 ab2∣Taking abc common from C 1 and C3 we get

=abc2∣bc – a 0 a2

0 ca – b b2

c – a – b c – a – b ab2∣Applying R3R3 – R1R2 we get

=abc2∣bc – a 0 a2

0 ca – b b2

– 2 b – 2a 2ab∣ 9

⇒ =abc 2

ab ∣abac – a2 0 a2

0 bcab – b2 b2

– 2ab – 2ab 2 ab∣ C1aC1 andC 2 bC2

Applying C1C1C3 and C2C 2C3

=abc 2

ab ∣abac a2 a2

b2 bcab b2

0 0 2ab∣=

abc 2

ab2ab⋅∣abbc a2

b2 bcab∣ (By Expanding along R3 )

Taking a common from R1 and b common from R2 we get

=2ab abc 2∣bc ab ca∣

⇒ =2ab abc2 {bcbac2ac−ab }⇒=2abc abc3

Example 9: Prove that ∣1 1 1a b cbc ca ab∣=a – bb – cc – a

Solution: Let =∣1 1 1a b cbc ca ab∣=a – bb – cc – a

Applying C1C1– C2 and C2C2 –C 3 we get

=∣ 0 0 1a – b b –c c

– c a – b – ab– c ab∣⇒ =a – bb– c ∣ 0 0 1

1 1 c– c – a ab∣ (Taking a – b common from C1 and b – c from C2 )

Expanding along R1 we get

=a – bb– c c – a1⋅∣ 1 1– c – a∣

⇒=a – bb – cc – a

Example-10 Prove that ∣y z x xy zx yz z x y∣=4 xyz

Solution: Let =∣yz x xy zx yz z x y∣

Applying C1C1−C3 and C2C2−C3 we get

=∣yz−x 0 x0 z x− y y

z−x− y z− x− y x y∣ 10

Applying R3R3 – R1R2 we get

=∣yz−x 0 x0 z x− y y

−2y −2x 0∣Expanding along R1 we get= yz – x2 xyx 2 y z x – y ⇒=2 xy yz – xzx – y =4 xyz

Example-11 Prove that ∣bc ca abqr r p pqy z z x x y∣=2∣a b c

p q rx y z∣

Solution: Let =∣bc ca abqr rp pqyz zx x y∣

Applying C1C2C3 we get

=∣2 abc ca ab2 pqr r p pq2 x yz z x x y∣

⇒=2∣abc ca abpqr rp pqx yz zx x y∣

Applying C2C2 –C 1 and C3C3−C1 we get

=2∣abc −b −cpqr −q −rx yz − y −z∣

Applying C1C1C2C3 we get

=2∣a −b −cp −q −rx − y −z∣

⇒=∣a b cp q rx y z∣

Example-12: Prove that ∣a21 ab acab b21 bcca cb c21∣=1a2b2c2

Solution: Let =∣a21 ab acab b21 bcca cb c21∣

⇒= 1abc∣aa

21 a2b a2 cab2 bb21 b2 cc2a c2b c c21∣ ( R1

1aR1 , R2

1bR2 and R3

1cR3 )

11

⇒=abcabc∣a21 a2 a2

b2 b21 b2

c2 c2 c21∣⇒=∣a21 a2 a2

b2 b21 b2

c2 c2 c21∣Applying R1R1R2R3 we get

=∣1a2b2 1a2b2 1a2b2

b2 b21 b2

c2 c2 c21 ∣=1a2b2∣1 1 1

b2 b21 b2

c2 c2 c21∣ (By taking 1a2b2 coomon from R1 )

Applying C1C1– C2 and C2C 2–C 3 we get

=1a2b2∣ 0 0 1–1 1 b2

0 –1 c21∣Expanding along R1 we get

=1a2b2∣–1 10 –1∣=1a2b2

Example-13: Prove that ∣a – b – c 2a 2 a2b b – c – a 2 b2c 2c c – a – b∣=abc3

Solution: Let =∣a – b – c 2a 2a2b b – c – a 2b2c 2c c –a – b∣

Applying R1R1R2R3 we get

=∣abc abc abc2b b – c – a 2b2c 2c c – a – b∣

⇒=abc∣1 1 12 b b – c – a 2b2c 2c c – a – b∣

Applying C1C1– C2 and C2C 2–C 3 we get

=abc∣ 0 0 1abc – abc 2b

0 abc c−a−b∣=abc3∣0 0 1

1 – 1 2b0 1 c – a – b∣ (Taking abc common from C1 and C2 )

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=abc3∣1 – 10 1 ∣ (Expandibg along R1 )

⇒=abc3

Example 14: Prove that ∣sin cos cos sin cos cos sin cos cos ∣=0

Solution: Let =∣sin cos cos sin cos cos sin cos cos ∣

⇒=∣sin cos cos cos – sin sinsin cos coscos – sin sinsin cos cos cos – sin sin∣

By applying C3C3sinC1– cosC 2

⇒=∣sin cos 0sin cos 0sin cos 0∣

⇒=0

Example-15: Prove that ∣ a2 bc acc2

a2ab b2 acab b2bc c2 ∣=4a2b2 c2

Solution: Let =∣ a2 bc acc2

a2ab b2 acab b2bc c2 ∣

⇒=a bc∣ a c acab b ab bc c ∣

Applying R1R1R2R3 we get

=abc∣2ab 2 bc 2 caab b ab bc c ∣

=2abc∣ab bc caab b ab bc c ∣

Applying R1R1– R2 and we get

=2abc∣ 0 c cab b ab bc c∣

Applying C2C 2–C 3 we get

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=2abc∣ 0 0 cab b – a ab b c∣

=2abc {c abb2– b2ab }=4 a2b2 c2

Example-16 If a ,b , c are positive and unequal, show that the value of the determinant

=∣a b cb c ac a b∣

is negative.Solution: Applying C1C1C2C3 we get

=∣abc b cabc c aabc a b∣=abc ∣1 b c

1 c a1 a b∣

⇒=abc∣0 b – c c – a0 c – a a –b1 a b ∣ ApplyingR1R1−R2 and R2 R2−R3

⇒=abc {b−ca−b−c−a2}⇒=abcba– b2– cabc – c2 – a22ca⇒=abc– a2– b2– c2abbcca=– abca2b2c2– ab – bc – ca

⇒=−12abc2a22b22c2– 2 ab– 2bc – 2ca

⇒=−12abc {a – b2b – c 2c – a2}

As a ,b , c are positive and unequal therefore abc0 and a – b2b – c2c – a20⇒0

Example 17: Solve ∣a x a – x a – xa – x ax a – xa – x a – x ax∣=0

Solution: Let =∣ax a – x a – xa – x ax a – xa – x a – x a x∣

Applying C1C1C2C3 we get =∣3a – x a – x a – x3a – x a x a – x3a – x a – x ax∣

⇒=3a– x ∣1 a– x a – x1 a x a – x1 a– x ax∣

Applying R2R2– R1 and R3R3 – R1 we get

=3a – x ∣1 a– x a – x0 2 x 00 0 2 x ∣

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⇒=3a– x ×1×∣2 x 00 2 x∣ (Expanding along C1 )

⇒=3a– x 4 x2

Thus =0⇒3a – x4 x2=0⇒ x=0,3a

Example 18: Prove that =∣x x2 1 px3

y y 2 1 py3

z z 2 1 pz3∣Solution: We have =∣x x2 1 px3

y y 2 1 py3

z z 2 1 pz3∣⇒=∣x x 2 1

y y2 1z z2 1∣ pxyz∣1 x x2

1 y y2

1 z z2∣ {As each element of C3 is sum of two elements}

⇒ = – ∣1 x2 x1 y2 y1 z2 z ∣ pxyz∣1 x x2

1 y y2

1 z z 2∣⇒=∣1 x x 2

1 y y2

1 z z 2∣ pxyz∣1 x x2

1 y y2

1 z z 2∣⇒= 1 pxyz ∣1 x x2

1 y y2

1 z z 2∣⇒= 1 pxyz ∣1 x x2

0 y – x y2 – x2

0 z – x z2 – x2∣ (Applying R2R2– R3 and R3R3 – R1 )

⇒= 1 pxyz y – x z – x ∣1 x x2

0 1 y x0 1 z x∣

⇒=1pxyz y – x z – x∣1 yx1 zx∣ (Expanding along C1 )

⇒=1pxyz y – z z – x z x – y – x=1 pxyz x – y y – z z – x

Example-19: Show that ∣a b ca2 b2 c2

bc ca ab∣=∣1 1 1a2 b2 c2

a3 b3 c3∣=a – bb – c c – aabbcca

Solution: Let =∣a b ca2 b2 c2

bc ca ab∣Applying C1

1aC1 , C2

1bC2 and C3

1cC3 we get

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= 1abc∣ a2 b2 c2

a3 b3 c3

abc abc abc∣⇒= abc

abc∣a2 b2 c2

a3 b3 c3

1 1 1 ∣=∣a2 b2 c2

a3 b3 c3

1 1 1 ∣Applying R1⇔ R3 we get

= – ∣1 1 1a3 b3 c3

a2 b2 c2∣Applying R2⇔R3 we get

=∣1 1 1a2 b2 c2

a3 b3 c3∣Applying C1C1– C2 and C2C 2–C 3 we get

=∣ 0 0 1a2 – b2 b2 – c2 c2

a3 – b3 b3 – c3 c3∣=∣ 0 0 1a – ba b b – cb c c2

a – ba2 ab b2 b – cb2 bc c2 c3∣⇒= a – bb – c∣ 0 0 1

a b b c c2

a2 ab b2 b2 bc c2 c3∣Applying C2C 2–C 1 we get

= a – bb – c ∣ 0 0 1a b c – a c2

a2ab b2 c – aa b c c3∣⇒= a – bb – cc – a∣ 0 0 1

a b 1 c2

a2 ab b2 a b c c3∣Expanding along R1 we get=a – bb– c c – a {ababc – a2abb2}⇒=a – bb – cc – a a2abacabb2bc – a2– ab−b2⇒=a – bb – cc – a abbcca

Example-20: Prove that ∣ a a b a b c2a 3a 2 b 4a 3b 2c3a 6a 3b 10a 6b 3c∣=a3

Solution: Let =∣a a b a b c2 a 3a 2b 4a 3b 2c3a 6a 3b 10 a 6b 3c∣

Since each element of the second column is sum of two elements we get

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=∣ a a a b c2a 3a 4a 3b 2c3a 6a 10a 6b 3c∣∣

a b a b c2 a 2b 4a 3b 2c3a 3b 10a 6 b 3c∣

⇒ =∣a a a b c2 a 3a 4 a 3b 2c3a 6a 10a 6b 3c∣ ab⋅∣1 1 a b c

2 2 4a 3b 2 c3 3 10a 6b 3c∣

⇒=∣a a a b c2a 3a 4a 3b 2c3a 6a 10 a 6b 3c∣ab⋅0 [ .̈ C2 and C3 are identical in second determinant)

As each element of C3 is sum of three elements therefore

=∣a a a2 a 3a 4 a3a 6a 10a∣∣

a a b2a 3a 3b3a 6 a 6b∣∣

a a c2a 3a 2c3a 6a 3c∣

⇒=a3∣1 1 12 3 43 6 10∣ a2b∣1 1 1

2 3 33 6 6∣ a2 c∣1 1 1

2 3 23 6 3∣

As C2 and C3 are identical in second determinant and C1 and C3 are identical in third determinant thus we get

=a3∣1 1 12 3 43 6 10∣ a2 b ⋅0 a2 c ⋅0

⇒=a3∣1 0 02 1 23 3 7∣ [Applying C2C 2–C 1 and C3C3– C1 ]

⇒=a37– 6 [Expanding along R1 ]⇒ =a3

Exercise 4.2Using Properties of determinants, prove the following

1. ∣x a x ay b y bz c z c∣= 0

2. ∣1 bc a b c1 ca b c a1 ab c a b∣= 0

3. ∣0 a – b– a 0 – cb c 0 ∣= 0

4. ∣– a2 ab acba – b2 bcca cb – c2∣=0

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5. ∣b2 c2 bc b cc2a2 ca c aa2b2 ab a b∣= 0

6. ∣x x2 yzy y2 zxz z2 xy∣= x – y y – z z – xxy yz zx

7. ∣x 4 2 x 2 x2 x x 4 2 x2 x 2 x 2 x 4∣=5 x 4 4 – x 2

8. ∣y k y yy y k yy y y k∣= k23y k

9. ∣1 a a2 – bc1 b b2 – ca1 c c2 – ab∣= 0

10. ∣b c a ab c a bc c a b∣= 4abc

11. ∣a – b – c 2a 2 a2b b – c – a 2 b2c 2c c – a – b∣= a b c 3

12. ∣x y 2 z x yz y z 2 x yz x z x 2 y∣= 2 x y z 3

13. ∣1 x x2

x 2 1 xx x2 1 ∣= 1 – x23

14. ∣x y x x5 x 4 y 4 x 2 x10 x 8 y 8 x 3 x ∣= x3

15. ∣b c a – b ac a b – c ba b c – a c ∣= 3abc – a3 – b3 – c3

16. ∣ y z2 xy zxxy x z 2 yzxz yz x y 2∣= 2 xyz x y z 3

17. ∣b2 c2 ab acba c2 a2 bcca cb a2 b2∣= 4 a2b2c2

18

18. ∣3a – a b – a c– b a 3b – b c– c a – c b 3c ∣= 3a b c ab bc ca

19. ∣1 1 p 1 p q2 3 2 p 4 3 p 2q3 6 3 p 10 6 p 3q∣= 1

20. Prove that ∣x sin cos– sin – x 1cos 1 x ∣ is independent of

21. Evaluate ∣cos cos cos sin – sin – sin cos 0sin cos sinsin cos ∣

22. Solve the equation ∣x a x xx x a xx x x a∣= 0 , a≠0

23. If x≠ y≠z and ∣x x2 1 x3

y y2 1 y3

z z2 1 z3 ∣= 0 , then prove that xyz=–1

24. Solve ∣x – 2 2 x – 3 3 x – 4x – 4 2 x – 9 3 x – 16x – 8 2 x – 27 3 x – 64∣=0

25. If a , b , c are in A.P, find value of ∣2 y 4 5 y 7 8 y a3 y 5 6 y 8 9 y b4 y 6 7 y 9 10 y c∣

Area of TriangleResult : Area of triangle whose vertices are x1, y1, z 1 , x2, y2, z2; x3, y3, z 3 is given by

=±12∣x1 y1 1x2 y 2 1x3 y3 1∣

Collinearity of three points: Three points x1, y1, z 1 , x2, y2, z2; x3, y3, z 3 will be collinear if and only

if ∣x1 y1 1x 2 y2 1x3 y3 1∣= 0

Example-1: Find the area of the triangle whose vertices are 1,2 ,2 ;3 , 3; – 1

Solution: =±12∣1 2 1

2 3 13 −1 1∣=±1

2 {131−22−31−2−9}

⇒=±1242 – 11=5

2 square units

Example-2: Show that the points 1,2; 4,−1; 5.−2 are collinear.

19

Solution: ∣1 2 14 – 1 15 – 2 1∣=1 –12– 2 4 –51 – 85=12 – 3=0

Example-3: If points a ,0 ,0,b ,x , y are collinear then show that xa yb=1

Solution: As points a ,0 ,0,b ,x , y are collinear ∣a 0 10 b 1x y 1∣=0

⇒ab – y – bx=0 ⇒ab– ay – bx=0⇒aybx=abDivide both sides by ab we getxa yb=1

Exercise 4.31. Find the area of the triangle whose vertices are

(i) 2,3 ,4,– 1 ,0,2 (ii) 3,– 1 ,2,4 ,9,2 (iii) 4, –1 , 4,1 ,7,32. Show that the points 3,1 ,4,0 ,6, – 2 are collinear.3. Show that the points 3,1 ,2, – 1 ,1,– 3 are collinear.4. If the points a ,0 ,0,b ,1,1 are collinear then show ab=ab5. Using determinants prove that the points a ,b ,a ' , b ' and a – a ' , b – b ' are collinear if

ab '=ba '6. Find the value of such that 2,3 , ,1 , – 1,2 are collinear.7. Using determinants find the equation of the line joining the points 2,3 and –3,1

Minors and Co-FactorsMinors: Let A=[ aij ] be a square matrix of order n . Then we define minor of a ij as a determinant of order n – 1 obtained from A be deleting its ith row and jth column. Minor of a ij is denoted by M ij

Co-Factor: Let A=[ aij ] be a square matrix of order n . Then we define co-factor of a ij as –i j M ij

. We denote the co-factor of a ij as Aij or C ij

Thus Aij=–1i j M ij

or Aij={M ij if i j is even−M ij if i j is odd

Result: If A=[ aij ] be a square matrix of order n

∣A∣=∑j=1

n

aij Aij (Expansion along ith row)

∣A∣=∑j=1

n

aijC ij (Expanding along jth column)

To be more specific let A=[ aij ] be a square matrix of order 3.Then ∣A∣=a11 A11a12 A12a13 A13 (Expansion along first column)∣A∣=a21 A21a22 A22a23 A23 (Expanding along second row)∣A∣=a31 A31a32 A32a33 A33 (Expanding along third row)∣A∣=a11 A11a21A21a31C31 (Expansion along first column)∣A∣=a12 A12a22 A22a32 A32 (Expanding along second column)∣A∣=a13 A13a23 A23a33 A33 (Expanding along third column)Note: If we multiply the elements of a row(column) by the co-factors of elements of different row(column) and then add the value of such expression is zero.

20

For example for a square matrix of order 3 a11 A21a12 A22a13 A23=0

Example-1: Find the minor and co-factor of each element of the matrix [ 2 3 1– 1 4 22 5 1]

Solution: M 11=∣4 25 1∣=4−10=−6 ; A11=– 6

M 12=∣−1 22 1∣=−1−4=−5, A12=5

M 13=∣−1 42 5∣=−5−8=−13, A13=−13

M 21=∣3 15 1∣=3−5=−2, A21=2

M 22=∣2 12 1∣=2−2=0, A22=0

M 23=∣2 33 5∣=10−9=1,A23=−1

M 31=∣3 14 2∣=6−4=2, A31=2

M 32=∣ 2 1−1 2∣=41=5, A32=−5

M 33=∣ 2 3−1 4∣=83=11, A33=11

Example-2: Find the determinant of the matrix [ 2 3 4– 1 2 33 2 5] and verify the relation

∣A∣=a11 A11a12 A12a13 A13

Solution: ∣A∣=2∣2 32 5∣– 3∣–1 3

3 5∣4∣– 1 23 2∣=210 – 6–3 – 5 –94 – 2 – 6=842– 32=18

A11=−111∣2 32 5∣=10−6=4 , A12=−112∣−1 3

3 5∣=−−5−9=14

A13=−113∣−1 23 2∣=−2−6=−8

Thus a11 A11a12 A12a13 A13=243144−8=842−32=18Thus ∣A∣=a11 A11a12 A12a13 A13

Example-3: For the matrix [ 2 3 11 2 4– 1 2 1] , verify that a21A11a22 A12a23 A13=0

Solution: A11=−111∣2 42 1∣=2– 8=– 6

A12=−112∣1 4−1 2∣=−14=−5

21

A13=−113∣ 1 2−1 2∣=22=4

Thus a21A11a22 A12a23 A13=1−62−544=−6−1016=0

Exercise-4.4

1. Write the minors and co-factors of each element the determinant ∣2 31 4∣

2. Write the minors and cofactors of each element of the determinant ∣ 2 −1 3−2 4 −15 1 2 ∣

3. Using co-factors of the elements of the 2nd column find the value fo the determinant

∣2 – 1 03 2 –12 1 4 ∣

4. Using co-factors of elements of 3rd row find the value of the determinant ∣3 – 1 1– 1 0 23 5 1∣

5. For the determinant ∣2 −1 34 1 51 0 2∣ show that a11 A21a12A22a13 A23=0

6. For the determinant ∣– 1 2 51 2 – 12 5 7 ∣ , show that a13A12a23A22a33 A23=0

Adjoint of a MatrixDefinition: : The adjoint of a square matrix A=[ aij ]n×n is defined as the transpose of the matrix [ Aij ]n× n , where Aij is the co-factor of the element a ij . Adjoint of the matrix A is denoted by adjA

Example-1: Find the adjoint of the matrix [2 34 1]

Solution: A11=1, AA12=−4,A21=−3, A22=2

Hence adj A= transposeof [ 1 – 4– 3 2 ]=[ 1 – 3

– 4 2 ]Note: For a 2×2 matrix A=[a b

c d ] adjA=[ d – b– c a ]

Thus the rule to write the adjoint of 2×2 matrix is that (i) interchange the position of diagonal elements and(ii) change the sign of the non-diagonal elements Result: ∣adj A∣=∣A∣n−1

Result: A adjA=∣A∣I

22

Example-2: Find the adjoint of the matrix [ 2 1 3– 1 4 12 5 1]

Solution: A11=−1, A12=3, A13=−13 ; A21=14, A22=−4, A23=−8 ; A31=−11, A32=−5, A33=9

adjA=[ – 1 14 –113 – 4 –5

– 13 – 8 9 ]Example 3: Find the adjoint of the matrix A=[2 3 –1

2 1 –14 1 2 ] and verify that A adjA =∣A∣I . Also

verify that ∣adj A∣=∣A∣n−1

Solution: A11=3, A12=−8, A13=−2 ; A21=−7, A22=8, A23=10 ; A31=−2, A32=0, A33=−4

Thus adjA=[ 3 –7 – 2– 8 8 0– 2 10 – 4]

and ∣A∣=233−8−−2=6−242=−16

Now A adj A=[2 3 −12 1 −14 1 2 ][ 3 −7 −2

−8 8 0−2 10 −4]=[– 16 0 0

0 – 16 00 0 – 16]=–16 I=∣A∣I

Now

∣adj A∣=∣3 −7 −2−8 8 0−2 10 −4∣=3−32−0−−732−0−2 −8016=−96224128=256=−162

Thus ∣adjA∣=∣A∣2=∣A∣3−1=∣A∣n−1 (Here n=3 )

Exercise 4.5

1. Find the adjoint of the matrix [ 2 3– 1 5] and verify the result A adjA=∣A∣I

2. Find the adjoint of the matrix [ 3 –1 2– 3 1 21 3 1] and verify the result A adjA=∣A∣I

3. Find the adjoint of the matrix [2 31 –3] and verify the result ∣adj A∣=∣A∣n−1

4. Find the adjoint of the matrix [3 –1 02 3 14 1 2] and verify the result ∣adj A∣=∣A∣n−1

5. For the matrix A=[2 3 11 2 43 5 5 ] show that A adj A=0

Inverse of a MatrixSingular Matrix: A square matrix A is said to be singular if ∣A∣=0Non-Singular Matrix: A square matrix A is said to be non-singular if ∣A∣≠0Result: If A and B are nonsingular matrices of the same order, then AB and BA are also non

23

singular matrices of the same order.Result: ∣AB∣=∣A∣∣B∣where A and B are square matrices of the same order.Invertible Matrix: A square matrix A is invertible if and only if A is nonsingular.

and A–1= 1∣A∣

adj A

Example-1 Show that the matrix [2 31 3] is invertible and find its inverse.

Solution: Let A=[2 31 3] . Then

∣A∣=6−3=3≠0 ⇒ A is nonsingular. Thus A is invertible.

A11=3, A12=–1, A21=– 3, A22=2 ⇒adj A=[ 3 −3−1 2 ]

Thus A−1= 1∣A∣

adj A=13 [ 3 −3−1 2 ]=[ 1 −1

−13

23 ]

Example-2: Show that A=[2 3 –11 4 22 –1 3 ] is invertible and find A–1

Solution: ∣A∣=∣2 3 −11 4 22 −1 3 ∣=2 122– 33– 4– 1– 1– 8=2839=40

A11=14, A12=1, A13=– 9, A21=– 8, A22=8, A23=8, A31=10, A32=−5, A33=5

Thus adj A=[14 −8 101 8 −5−9 8 5 ]

.̇. A−1= 1∣A∣

adj A

⇒ A−1= 140 [14 −8 10

1 8 −5−9 8 5 ]=[

720

– 15

14

140

15

−18

– 940

15

18]

Example-3:: If A=[2 31 – 4] and B=[ 1 – 2

– 1 3 ] , then verify that AB −1=B−1 A−1

Solution: We have AB=[2 31 – 4 ][ 1 – 2

–1 3 ]=[–1 55 –14]

Since ∣AB∣=14−25=−11≠0 , AB −1 exists and is given by

AB−1= 1∣AB∣

adjAB=− 111 [– 14 – 5

– 5 – 1]= 111 [14 5

5 1]Further ∣A∣=−11≠0 and ∣B∣=1≠0 . Therefore, A−1 and B–1 both exist and are given by

A−1=− 111 [−4 −3

−1 2 ], B−1=[3 21 1]

24

Therefore B−1 A−1=−111 [3 2

1 1][−4 −3−1 2 ]= 1

11 [14 55 1]

Hence AB−1=B−1 A−1

Example-4: If A=[2 34 5] then show that A2 – 7 A– 2 I=0 and hence find A–1 .

Solution: A2=[2 34 5][2 3

4 5]=[16 2128 37] , 7A=7[2 3

4 5]=[14 2128 35] , 2I=[2 0

0 2]Thus A2−7A−2I=[16 21

28 37 ]−[14 2128 35]−[2 0

0 2]=[0 00 0]

Thus A2 – 7 A– 2 I=0 Pre-Multiply by A−1 we getA–1 A2– 7 A–1 A– 2 A–1 A=0 ⇒ A –7 I – 2 A–1=0

⇒2A−1=A−7I=[2 34 5]−[7 0

0 7]=[– 5 34 – 2]

⇒ A−1=12 [– 5 3

4 – 2]⇒ A−1=[– 5

232

2 –1]Example-5: If A=[2 1 1

1 2 11 1 2] and B=[ 3 –1 –1

– 1 3 –1– 1 –1 3 ] then find AB and hence find A−1

Solution: AB=[2 1 11 2 11 1 2][ 3 – 1 – 1

– 1 3 – 1– 1 – 1 3 ]=[4 0 0

0 4 00 0 4]

Thus AB=4 I

Pre-Multiply by A–1 AB=4 A–1 I ⇒ IB=4 A–1 ⇒B=4 A– 1 ⇒ A–1= 14B

Thus A−1=[34

– 14

– 14

– 14

34

– 14

– 14

– 14

34]

Example:6 Find the inverse of the matrix [cos −sin 0sin cos 0

0 0 1]Solution: Let A=[cos −sin 0

sin cos 00 0 1]

∣A∣=cos2sin2=1 (By expanding along third row)A11=cos , A12=−sin , A13=0 ; A21=sin , A22=cos , A23=0 ; A31=0, A32=0, A33=1

25

adj A=[ cos sin 0−sin cos 0

0 0 1]Thus A−1= 1

∣A∣adj A

⇒ A−1=[ cos sin 0−sin cos 0

0 0 1]Example: -7: If A= [2 4

3 – 2 ] then show that A−1= 116

A

Solution: We have ∣A∣=−4−12=−16≠0 Thus A−1 exists.

A−1= 1∣A∣

adj A=− 116 [−2 −4

−3 2 ]= 116 [2 4

3 −2]= 116

A

Example -8: If A=[ 1 tan x−tan x 1 ] then show that AT A−1=[cos 2 x −sin 2 x

sin 2 x cos 2 x ]Solution: ∣A∣=1tan 2 x≠0 , thus A−1 exists and is given by

A−1= 1∣A∣

adj A= 11tan2 x [ 1 −tan x

tan x 1 ]AT=[ 1 −tan x

tan x 1 ]Thus AT A= 1

1tan2 x [ 1 −tan xtan x 1 ][ 1 −tan x

tan x 1 ]= 11tan2 x∣1−tan2 x −2tan x

2tan x 1−tan2 x∣⇒ AT A−1=[1−tan 2 x

1tan 2 x−

2tan x1tan 2 x

2 tan x1tan 2 x

1−tan2 x1tan2 x

]=[cos 2 x −sin 2 xsin 2 x cos 2 x ]

Example-9: For the matrix A=[2 34 –1] , find the values of a ad b such that A2aAbI=0 .

Hence find A– 1

Solution: We have A=[2 34 −1]

Therefore A2=[2 34 −1][2 3

4 −1]=[16 34 13]

As A2aAbI=0

Therefore [16 34 13][2a 3a

4a −a][b 00 b]=[0 0

0 0]⇒[162ab 33a

44a 13−ab]=[0 00 0]

⇒162ab=0, 33a=0, 44a=0,13−ab=0⇒a=−1 and b=−14

Thus A2a Ab I=0⇒ A2−A−14I=0

26

Pre-Multiply by A−1 we getA−1 A2−A−1 A−14 A−1 I=0

⇒ A−I−14A−1=0 ⇒14A−1=A−I ⇒ A−1= 114 {[2 3

4 −1]−[1 00 1]}

⇒ A−1=[ 114

314

414

− 214 ]=[

114

314

27

−17 ]

Example-10: Find a 2×2 matrix B such that B[1 −21 4 ]=[6 0

0 6]Solution: Let A=[1 −2

1 4 ] and C=[6 00 6 ]

Then, ∣A∣=42=6≠0 . Thus A−1 exits and is given by

A−1= 1∣A∣

adj A =16 [ 4 2−1 1]

Now B[1 −21 4 ]=[6 0

0 6] ⇒B A=C

Post-Multiply both side by A−1 we get B A A−1=C A−1 or B I=C A−1 or

B=C A−1=16 [6 0

0 6 ][ 4 2−1 1]=1

6 [24 12−6 6 ]

Thus B=[ 4 2−1 1]

Example:11 Find a matrix A such that [2 31 2]A[1 – 2

3 1 ]=[2 11 3]

Solution: Let B = [2 31 2 ],C = [1 – 2

3 1 ],D = [2 11 3 ]

Thus we have BAC=DPre-Multiply by B–1 and post multiply by C–1 we get B– 1 BAC C– 1=B– 1 DC– 1 ⇒ IAI=B– 1 DC–1

⇒A=B– 1D C –1

∣B∣=1,∣C∣=7

Thus B–1 = 1∣B∣

adjB = [ 2 – 3– 1 2 ]= [ 2 – 3

– 1 2 ]and C–1= 1

∣C∣adjC=1

7 [ 1 2– 3 1]

Thus A = 17 [ 2 – 3

– 1 2 ][2 11 3 ][ 1 2

– 3 1]Thus A=1

7 [1 – 70 5 ] [ 1 2

– 3 1]

27

Thus A=17 [ 22 – 5

– 15 5 ]= [ 227

– 57

– 157

57 ]

Exercise 4.6

1. Show that [2 3 41 2 11 1 4] is non singular.

2. Show that the matrix [3 1 22 1 35 2 5] is singular

3. Show that the matrix [a – b b – c c – ab– c c – a a – bc – a a – b b – c] is singular.

4. Find the value of such that the matrix [2 31 2 30 1 2 ] is singular.

5. Find the inverse of the matrix [2 31 – 2 ]

6. Find the inverse of the matrix [ 3 –1– 2 4 ]

7. Find the inverse of the matrix [cos – sinsin cos ]

8. Find the inverse of the matrix [a bc d ]

9. Find the inverse of the matrix [2 3 – 11 2 52 0 – 1]

10. Find the inverse of the matrix [3 –1 – 22 1 34 1 2 ]

11. If A=[2 34 – 1] and B=[4 1

2 – 1] then verify that AB– 1=B– 1 A– 1

12. If A = [ 1 2 22 1 3

– 1 2 1 ] and B = [0 – 1 22 1 12 – 1 1 ] then verify that AB–1=B– 1 A– 1

13. If A = [2 34 – 1] then prove that A2 – A – 14I=0 and hence find A^{-1}

28

14. If A = [2 13 2 ] then find a and b such that A2aAbI=0 and hence find A– 1

15. Find the inverse of the matrix [1 0 00 cos sin 0 sin – cos]

16. For the matrix A=[ 2 – 1 1– 1 2 – 11 – 1 2 ] , show that A3 – 6A29A – 4 I=0 . Hence find A– 1

17. For the matrix A=[1 1 11 2 – 32 – 1 3 ] show that A3 – 6A25A11 I=0

18. Show that the matrix A=[ 2 3– 1 4] is zero of the polynomial f x=x2 – 6x11 and hence find

A– 1

19. Find the matrix X for which [3 27 5]X=[1 −2

1 3 ]20. Find the matrix X for which X[3 – 1

2 4 ]= [1 23 4 ]

21. Find the matrix A for which [3 12 2 ]A [ – 1 2

3 4 ]=[ 1 3−1 0]

22. Find the inverse of the matrix A = [a b

c 1 bca ]and show that aA–1=A2bc1I –aA

23. Given A = [ 2 – 3– 4 7 ] , compute A– 1 and show that 2A –1=9I – A

24. If A = [4 52 1 ] , then show that A−3I=2I3A–1

25. Given A = [5 0 42 3 21 2 1 ], B– 1 = [1 3 3

1 4 31 3 4 ] . Compute AB–1

26. Show that [ 1 – tan 2

tan 2

1 ][ 1 tan 2

– tan 2

1 ]– 1

= [cos – sinsin cos ]

27. If A = 19 [ – 8 1 4

4 4 71 – 8 4 ] , prove that A– 1=AT

28. If A = [3 – 3 42 – 3 40 – 1 1 ] , prove that A– 1=A3

29. If A = [ – 1 2 0– 1 1 10 1 0 ] , show that A2=A–1

29

System of Linear Equations

Consider a system of linear equations {a1 xb1 y=c1

a2 xb2 y=c2.

Using matrix notations this can be written as [a1 b1

a2 b2][xy]=[c1

c2] or AX=B where A=[a1 b1

a2 b2] ,

X=[xy ] and B=[c1

c2]If ∣A∣≠0 then A– 1 existsPre-Multiplying AX=B by A– 1 we get A– 1 AX=A – 1 B ⇒ I X=A – 1 B⇒X=A–1 B

Consider the system of equations {a1 xb1 yc1 z=d1

a2 xb2 yc2 z=d 2

a3 xb3 yc3 z=d3

This can be written as [a1 b1 c1

a2 b2 c2

a3 b3 c3][xyz]=[d1

d2

d3]

or A X=B where A=[a1 b1 c1

a2 b2 c2

a3 b3 c3] , X=[xyz ] and B=[d1

d 2

d 3]

If ∣A∣≠0 then A−1 exists.Pre-Multiplying A X=B by A−1 we get A−1 A X=A−1B⇒ I X=A−1B⇒ X=A−1 B

Note: The system of equations A X=B will have(i) unique solution if ∣A∣≠0 and the solution is given by X=A−1B(ii) infinitely many solutions if adj AB=0(iii) no solution if adj AB≠0

Example-1: Solve the system of linear equations {2 x4 y=103 x− y=8

Solution: The given system of equations can be written as [2 43 −1][xy]=[10

8 ]or A X=B where A=[2 4

3 −1], X=[xy ], B=[108 ]

Now ∣A∣=∣2 43 −1∣=−2−12=−14≠0 , thus A−1 exists.

A11=−1, A12=−3, A21=−4 , A22=2

Thus adj A=[−1 −4−3 2 ]

.̇. A−1= 1∣A∣

adj A=− 114 [−1 −4

−3 2 ]

30

Now X=A−1B ⇒ X=− 114 [−1 −4

−3 2 ][108 ]

Thus X=−114 [−42

−14]=[31] ⇒[xy]=[31]Hence solution of the system of equations is x=3, y=1

Example:2 Solve the system of equations {3 x−2 y z=3x2 y−z=5x− y z=1

Solution: The above system of equations can be written as A X=B where

A=[3 −2 11 2 −11 −1 1 ], X=[

xyz ], B=[

351]

∣A∣=∣3 −2 11 2 −11 −1 1 ∣=32−1−−2111−1−2=4≠0 . Thus A−1 exists.

Now A11=1, A12=−2, A13=−3, A21=1, A22=2, A23=1, A31=0, A32=4, A33=8

Thus adj A=[ 1 1 0−2 2 4−3 1 8]

Using A−1= 1∣A∣

adj A we get

A−1=14 [ 1 1 0

−2 2 4−3 1 8]

Using X=A−1B we get

X=14 [ 1 1 0

−2 2 4−3 1 8][351 ]=1

4 [884]=[221] ⇒[xyz]=[221]

Thus solution of the given system of equations is x=2, y=2, z=1

Example-3: If A=[1 2 11 0 32 −3 0] then find A−1 and hence solve the system of equations

x2 yz=7, x3 z=11, 2x−3y=1

Solution: ∣A∣=∣1 2 11 0 32 −3 0∣=109−2 0−61−3−0=912−3=18≠0

Thus A−1 exists. A11=9, A12=6, A13=−3 ; A21=−3, A22=−2, A23=7, A31=6, A32=−2, A33=−2

Thus adj A=[ 9 −3 66 −2 −2−3 7 −2]

31

⇒ A−1= 1∣A∣

adj A= 118 [ 9 −3 6

6 −2 −2−3 7 −2]

Now given system of equations can be expressed as [1 2 11 0 32 −3 0][

xyz ]=[

7111 ] or

A X=B where A=[1 2 11 0 32 −3 0] , X=[xyz ] and B=[ 7

111 ] .

.̇. X=A−1 B= 118 [ 9 −3 6

6 −2 −2−3 7 −2][

7112 ]= 1

18 [361854]=[

213]

.̇.[xyz ]=[213]Hence solution of the system of equations is x=2, y=1, z=3

Example-5: If A=[1 −1 12 1 −31 1 1 ] , find A−1 and hence solve the system of equations

x2 y z=4,−x yz=0, x−3yz=2

Solution: A=[1 −1 12 1 −31 1 1 ]

.̇.∣A∣=∣1 −1 12 1 −31 1 1 ∣=1 1312312−1=10≠0

So, A is invertible.A11=5, A12=−5, A13=1, A21=2, A22=0, A23=−2, A31=2, A32=5, A33=3

.̇. adj A=[ 4 2 2−5 0 51 −2 3]

⇒ A−1= 1∣A∣

adj A= 110 [ 4 2 2

−5 0 51 −2 3 ]

Now the given system of equations is expressible as [ 1 2 1−1 1 11 −3 1][

xyz ]=[

402]

or, AT X=B where X=[xyz ], B=[402 ]

Now, ∣AT∣=∣A∣=10≠0 . So, the system of equations is consistent with a unique solution.X= AT −1 B=A−1T B

32

[xyz ]= 110 [ 4 2 2

−5 0 51 −2 3 ]

T

[402]⇒[xyz]= 1

10 [4 −5 12 0 −22 5 3 ][402]= 1

10 [18414]=[

952575]

Hence x= 95,

y= 25,

z=75 is the solution of the given system of linear equations.

Example-6: Find the product of the matrices [2 1 11 2 11 1 2] and [ 3 – 1 – 1

– 1 3 – 1– 1 – 1 3 ] and hence solve the

system of equations 2 x y z=4, x2 yz=4, x y2 z=4

Solution:Let A=[2 1 11 2 11 1 2] and B=[ 3 −1 −1

−1 −3 −1−1 −1 3 ]

A B=[2 1 11 2 11 1 2][ 3 −1 −1

−1 −3 −1−1 −1 3 ]=[4 0 0

0 4 00 0 4]=4 I

Pre-Multiplying by A−1 we get A−1 AB=4 A−1 I⇒ I B=4 A−1⇒4A−1=B⇒ A−1= 14B

Thus A−1=14 [ 3 −1 −1−1 3 −1−1 −1 3 ]

Given system of equations can be written as [2 1 11 2 11 1 2][

xyz ]=[

444 ] or A X=C where

X=[xyz ],C=[444].̇.X=A−1C=1

4 [ 3 −1 −1−1 3 −1−1 −1 3 ][4444]=1

4 [444]⇒[xyz]=[

111]

Hence the solution of the given system of equations is x=1, y=1, z=1Example-7: Solve the system of equations 2 x− y z=4, x2 y−z=3,3 x y=7 if consistent.

Solution: The given system of equations can be expressed as [2 −1 11 2 −13 1 0 ][xyz ]=[

437 ]

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or A X=B where A=[2 −1 11 2 −13 1 0 ], X=[xyz], B=[

437]

∣A∣=2011 031 1−6=23−5 =0 ⇒ A−1 does not exist.Thus system of equations does not have unique solution.A11=1, A12=−3, A13=−5, A21=1, A22=−3, A23=−5, A31=−1, A32=3, A33=5

Thus adj A=[ 1 1 −1−3 −3 3−5 −5 5 ]

⇒adj A B=[ 1 1 −1−3 −3 3−5 −5 5 ][437 ]=[000]

Thus system of equations has infinitely many solutions.To find the solutions let us put z=k in the first two equations. Thus we get

2 x− yk=4, x2 y−k=3or 2 x− y=4−k , x2y=3k This can be written as

[2 −11 2 ][xy]=[4−k

3k ]⇒[xy]=[2 −1

1 2 ]−1

[4−k3k ]

⇒[xy]=15 [ 2 1−1 2][4−k

3k ]⇒[xy]=1

5[11−k23k ]

⇒[xy]=[11−k5

23k5

]Thus solution of the given system of equations is x=11−k

5, y=23 k

5,z=k where k is any real

number.Example-8: Show that the system of equations 2 x y z=4 , x y z=1,3 x2 y2 z=4 is inconsistent.

Solution: The given system of equations can be written as [2 1 11 1 13 2 2][

xyz ]=[

414 ]

or A X=B where A=[2 1 11 1 13 2 2], X=[xyz ], B=[

414]

∣A∣=22−2−12−312−3=0 . Thus A is singular matrix. Thus A−1 does not exist.Now A11=0, A12=1, A13=−1, A21=0, A22=1, A23=−1, A31=0, A32=−1, A33=1

Thus adj A=[ 0 0 01 1 −1−1 −1 1 ]

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⇒adj A B=[ 0 0 01 1 −1−1 −1 1 ][414]=[ 0

1−1]

As ∣A∣=0and adj AB≠0 , therefore the system of equations do not have any solution.Thus the given system of equations is inconsistent.Homogeneous system of equations: A system of equations is said to be homogeneous if it can be written as A X=BTrivial Solution: A solution in which each variable is zero is known as trivial solution. Every system homogeneous linear equations always has trivial solution. A homogeneous system of linear equations will have trivial solution only if ∣A∣=0Non-trivial Solution: A homogeneous system of linear equations AX=B will have non-trivial solution if ∣A∣≠0Example-9: Solve the following homogeneous system of linear equations

2x− y z=0, x y− z=0, x− y=0Solution:The given system of equations can be written as A X=0 where

A=[2 −1 11 1 −11 −1 0 ], X=[xyz]

Now ∣A∣=20−11 011−1−1=−21−2=−3≠0Thus the given system of equations has trivial solution.Thus solution is x=0, y=0, x=0

Exercise 4.7Solve the following system of linear equations by matrix method

1. 5 x7 y24 x6 y3=0

2. 3 x4 y−5=0x− y3=0

3.x yz=3

2 x− y z=−12 x y−3z=−9

4.6 x−12 y25 z=44 x15 y−20 z=32 x18 y15 z=10

5.3 x4 y7 z=162 x− y3 z=19x2 y−3z=25

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6.

2x−3

y3

z=10

1x1

y1

z=10

3x−

1y

2z=13

7.3 x4 y2 z=8

2 y−3 z=3x−2 y6 z=−2

8.2 x6 y=23 x−z=−8

2 x− yz=−3

9.8 x4 y3 z=18

2 x y z=5x2 y z=5

10.2 x yz=4x2 yz=4x y2 z=4

11. If A=[1 −1 02 3 40 1 2] and B=[ 2 2 −4

−4 2 −42 −1 5 ] are two square matrices, find AB and hence

solve the system of equations x− y=3, 2 x3 y4 z=17, y2 z=7

12. If A=[2 −3 53 2 −41 1 −2 ] find A−1 and hence solve the system of equations

2 x−3 y5 z=11,3 x2 y−4 z=−5, x y2 z=−3

13. Find A−1 if A=[1 2 51 −1 −12 3 −1] . Hence solve the system of equations

x2 y5 z=10, x− y−z=−2,2 x3 y− z=−11

14. Find A−1 if A=[ 4 5 2−5 −4 2−2 2 8 ] and hence solve the system of equations

4 x−5 y−2 z=2,5 x−4 y2 z=−2, 2 x2 y8 z=−115. Solve the following system of equations if consistent

2 x y z=4, x yz=1,3 x2 y2 z=516. Show that the system of equations 3 x yz=1, x y−z=2,4 x2 y=217. Solve the following homogeneous\system of linear equations

(i) 2 x− yz=0

3 x2 y− z=0x4 y3 z=0

(ii) 2 x− y− z=0x y z=0

3 x2 y− z=0(iii)

x yz=0x− y−5 z=0x2 y4 z=0

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(iv) x y−z=0x−2 yz=0

3 x6 y−5 z=0(v)

2 x yz=0xx2 y z=0x y2z=0

(vi) 2 x− y2 z=05 x3 y−z=0x5 y−5 z=0