If A = 2 0 0 - 1 2 3 3 3 5 then find A ( adj A ) .

Given that, A = 2 0 0 - 1 2 3 3 3 5 We know that, A · adj ( A ) = | A | I ⇒ A · adj ( A ) = 2 0 0 - 1 2 3 3 3 5 1 0 0 0 1 0 0 0 1 ⇒ A · adj ( A ) = 2 1 0 0 0 1 0 0 0 1 ⇒ A · adj ( A ) = 2 I Thus the value of A · adj ( A ) is 2 I .

If A = 2 0 0 - 1 2 3 3 3 5 then find A ( adj A ) .

Given that, A = 2 0 0 - 1 2 3 3 3 5 We know that, A · adj ( A ) = | A | I ⇒ A · adj ( A ) = 2 0 0 - 1 2 3 3 3 5 1 0 0 0 1 0 0 0 1 ⇒ A · adj ( A ) = 2 1 0 0 0 1 0 0 0 1 ⇒ A · adj ( A ) = 2 I Thus the value of A · adj ( A ) is 2 I .

Using properties of determinants, prove that ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) .

We have to prove that ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) L . H . S . = ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b Operating C 1 → C 1 + C 2 - 2 C 3 , we have = b 2 + c 2 + 2 b c + a 2 - 2 b c a 2 b c c 2 + a 2 + a c + b 2 - 2 c a b 2 c a a 2 + b 2 + 2 a b + c 2 - 2 a b c 2 a b = b 2 + c 2 + a 2 a 2 b c c 2 + a 2 + b 2 b 2 c a a 2 + b 2 + c 2 c 2 a b Taking ( a 2 + b 2 + c 2 ) common from first column, we have = ( a 2 + b 2 + c 2 ) 1 a 2 b c 1 b 2 c a 1 c 2 a b Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , we have = ( a 2 + b 2 + c 2 ) 1 a 2 b c 0 b 2 - a 2 c a - b c 0 c 2 - a 2 a b - b c = ( a 2 + b 2 + c 2 ) 1 a 2 b c 0 ( b - a ) ( b + a ) c ( a - b ) 0 ( c - a ) ( c + a ) b ( a - c ) Taking out ( a - b ) common from R 2 and ( c - a ) from R 3 , we have = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) 1 a 2 b c 0 - 1 ( b + a ) c 0 ( c + a ) - b Expanding along first column, we have = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ ( a + b ) b - c ( a + c ) ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a b + b 2 - a c - c 2 ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a ( b - c ) + b 2 - c 2 ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a ( b - c ) + ( b - c ) ( b + c ) ] [ ∵ m 2 - n 2 = ( m - n ) ( m + n ) ] = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) = R . H . S . Hence, the given result is proved.

Using properties of determinants, prove that ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) .

We have to prove that ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) L . H . S . = ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b Operating C 1 → C 1 + C 2 - 2 C 3 , we have = b 2 + c 2 + 2 b c + a 2 - 2 b c a 2 b c c 2 + a 2 + a c + b 2 - 2 c a b 2 c a a 2 + b 2 + 2 a b + c 2 - 2 a b c 2 a b = b 2 + c 2 + a 2 a 2 b c c 2 + a 2 + b 2 b 2 c a a 2 + b 2 + c 2 c 2 a b Taking ( a 2 + b 2 + c 2 ) common from first column, we have = ( a 2 + b 2 + c 2 ) 1 a 2 b c 1 b 2 c a 1 c 2 a b Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , we have = ( a 2 + b 2 + c 2 ) 1 a 2 b c 0 b 2 - a 2 c a - b c 0 c 2 - a 2 a b - b c = ( a 2 + b 2 + c 2 ) 1 a 2 b c 0 ( b - a ) ( b + a ) c ( a - b ) 0 ( c - a ) ( c + a ) b ( a - c ) Taking out ( a - b ) common from R 2 and ( c - a ) from R 3 , we have = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) 1 a 2 b c 0 - 1 ( b + a ) c 0 ( c + a ) - b Expanding along first column, we have = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ ( a + b ) b - c ( a + c ) ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a b + b 2 - a c - c 2 ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a ( b - c ) + b 2 - c 2 ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a ( b - c ) + ( b - c ) ( b + c ) ] [ ∵ m 2 - n 2 = ( m - n ) ( m + n ) ] = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) = R . H . S . Hence, the given result is proved.

If A and B are square matrices of the same order 3 , such that | A | = 2 , and AB = I , write the value of B .

Since A and B are square matrices of the same order, by the property of determinants we get | A B | = | A | × | B | . . . . i It is given that | A | = 2 , A B = I ⇒ | A B | = I ⇒ A B = 1 ⇒ | A | × | B | = 1 [Using i ] ⇒ 2 × | B | = 1 Hence, | B | = 1 2

If A = 2 - 3 5 3 2 - 4 1 1 - 2 , find A - 1 . Use it to solve the system of equations 2 x - 3 y + 5 z = 11 3 x + 2 y - 4 z = - 5 x + y - 2 z = - 3

Given that A = 2 - 3 5 3 2 - 4 1 1 - 2 A = 2 - 4 + 4 + 3 - 6 + 4 + 5 3 - 2 = 0 - 6 + 5 = - 1 Let C i j denote the cofactors of a i j in A = a i j for all i , j C 11 = - 1 1 + 1 2 - 4 1 - 2 = - 4 + 4 = 0 C 12 = - 1 1 + 2 3 - 4 1 - 2 = - - 6 + 4 = 2 C 13 = - 1 1 + 3 3 2 1 1 = 3 - 2 = 1 C 21 = - 1 2 + 1 - 3 5 1 - 2 = - 6 - 5 = - 1 C 22 = - 1 2 + 2 2 5 1 - 2 = - 4 - 5 = - 9 C 23 = - 1 2 + 3 2 - 3 1 1 = - 2 + 3 = - 5 C 31 = - 1 3 + 1 - 3 5 2 - 4 = 12 - 10 = 2 C 32 = - 1 3 + 2 2 5 3 - 4 = - - 8 - 15 = 23 C 33 = - 1 3 + 3 2 - 3 3 2 = 4 + 9 = 13 adj A = C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 T = C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 = 0 - 1 2 2 - 9 23 1 - 5 13 Hence, A - 1 = adj A A = 0 1 - 2 - 2 9 - 23 - 1 5 - 13 Given system of equations 2 x - 3 y + 5 z = 11 3 x + 2 y - 4 z = - 5 x + y - 2 z = - 3 It can be written in matrix form as A X = B Where, A = 2 - 3 5 3 2 - 4 1 1 - 2 , X = x y z , B = 11 - 5 - 3 Now, X = A - 1 B Putting Value of A - 1 and B , X = 0 1 - 2 - 2 9 - 23 - 1 5 - 13 · 11 - 5 - 3 = 0 - 5 + 6 - 22 - 45 + 69 - 11 - 25 + 39 x y z = 1 2 3 We know that the two matrices are equal if and only if the corresponding elements of the two are equal Therefore, x = 1 , y = 2 , z = 3 .

Using properties of determinants, prove that: 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 = 9 3 x y z + x y + y z + z x .

To prove: ∆ = 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 = 9 3 x y z + x y + y z + z x Consider LHS as, ∆ = 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 ⇒ ∆ = 1 1 1 + 3 x 3 y 0 - 3 x 0 3 z - 3 x [Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ] Expanding along R 1 , we get: ∆ = 1 0 + 9 x z - 1 - 9 x y - 0 + 1 + 3 x 9 y z - 0 = 9 x z + 9 x y + 9 y z + 27 x y z = 9 3 x y z + x y + y z + z x = RHS Hence, proved.

Using the properties of determinants, prove that a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 = ( a - 1 ) 3 .

Consider a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 Applying R 1 → R 1 - R 2 and R 2 → R 2 - R 3 = a 2 - 1 a - 1 0 2 a - 2 a - 1 0 0 3 1 = ( a - 1 ) 2 a + 1 1 0 2 1 0 0 3 1 Expanding along C 3 = ( a - 1 ) 2 [ 0 + 0 + ( a + 1 - 2 ) ] = ( a - 1 ) 3 . Hence, proved

If A = 1 1 1 1 0 2 3 1 1 . Find A - 1 . Hence, solve the system of equations x + y + z = 6 , x + 2 z = 7 , 3 x + y + z = 12 .

Given A = 1 1 1 1 0 2 3 1 1 Expand along R 1 , we get ⇒ | A | = 1 × ( 0 - 2 ) - 1 × ( 1 - 6 ) + 1 × 1 - 0 ⇒ | A | = 4 ≠ 0 ⇒ A - 1 exists. We know that C i j = - 1 i + j M i j , M i j is minor of a matrix C 11 = - 1 1 + 1 0 2 1 1 = 0 - 2 = - 2 , C 12 = - 1 1 + 2 1 2 3 1 = - 1 - 6 = 5 C 13 = - 1 1 + 3 1 0 3 1 = 1 - 0 = 1 , C 21 = - 1 2 + 1 1 1 1 1 = 0 - 0 = 0 C 22 = - 1 2 + 2 1 1 3 1 = 1 - 3 = - 2 , C 23 = - 1 2 + 3 1 1 3 1 = - 1 - 3 = 2 C 31 = - 1 3 + 1 1 1 0 2 = 2 - 0 = 2 , C 32 = - 1 3 + 2 1 1 1 2 = - 2 - 1 = - 1 C 33 = - 1 3 + 3 1 1 1 0 = 0 - 1 = - 1 ∴ co factor matrix C = - 2 5 1 0 - 2 2 2 - 1 - 1 ∴ a d j A = C T = - 2 0 2 5 - 2 - 1 1 2 - 1 ∴ A - 1 = a d j A | A | = 1 4 - 2 0 2 5 - 2 - 1 1 2 - 1 Given system of equations: x + y + z = 6 ; x + 2 z = 7 ; 3 x + y + z = 12 We can write the above equations as 1 1 1 1 0 1 3 1 1 x y z = 6 7 12 ⇒ A X = B ⇒ X = A - 1 B (pre multiply by A - 1 on both sides) ⇒ X = 1 4 - 2 0 2 5 - 2 - 1 1 2 - 1 6 7 12 ⇒ X = 1 4 - 2 × 6 + 0 × 7 + 2 × 12 5 × 6 - 2 × 7 - 1 × 12 1 × 6 + 2 × 7 - 1 × 12 (Apply row into column multiplication) ⇒ X = 1 4 - 12 + 0 + 24 30 - 14 - 12 6 + 14 - 12 ⇒ X = 1 4 12 4 8 = 3 1 2 ⇒ x y z = 3 1 2 ∴ x = 3 , y = 1 , z = 2 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2020>Q 8

EASY

12th CBSE

IMPORTANT

Earn 100

If $A = [\begin{matrix} 2 & 0 & 0 \\ - 1 & 2 & 3 \\ 3 & 3 & 5 \end{matrix}]$ then find $A (adj A)$ .

Important Questions on Determinants

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2020>Q 9

If

A = [\begin{array}{ccc} 1 & 2 & - 3 \\ 3 & 2 & - 2 \\ 2 & - 1 & 1 \end{array}]

. Find

A^{- 1}

and use it to solve the following system of equations

x + 2 y - 3 z = 6, 3 x + 2 y - 2 z = 3, 2 x - y + z = 2

.

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2020>Q 10

Using properties of determinants, prove that $|\begin{matrix} (b + c)^{2} & a^{2} & b c \\ (c + a)^{2} & b^{2} & c a \\ (a + b)^{2} & c^{2} & a b \end{matrix}| = (a^{2} + b^{2} + c^{2}) (a + b + c) (b - c) (c - a) (a - b)$ .