QBManager

Matrix and Determinant

78963 If $A$ and $B$ are square matrices of the same order such that $(A + B) (A - B) = A^{2} - B^{2}$ , then ${({A B A}^{- 1})}^{2} =$

1

A^{2} B^{2}

2

A^{2}

3

B^{2}

4 1

Explanation:

(C) : According to given summation,
$(A + B) (A - B) = A^{2} - B^{2}$
$A A + B A - A B - B B = A^{2} - B^{2}$
$A^{2} + B A - A B - B^{2} = A^{2} - B^{2}$
$B A - A B = 0 \Rightarrow B A = A B$
Now, ${({ABA}^{- 1})}^{2} = {({BAA}^{- 1})}^{2}$
$= (B \cdot 1)^{2} [∵ {AA}^{- 1} = 1]$
$= B^{2}$

Karnataka CET-2009

Matrix and Determinant

78964 If $A (adj A) = 5 I$ where $I$ is the identity matrix of order 3 , then $| adj A |$ is equal to

1 5

2 10

3 125

4 25

Explanation:

(D) : According to given summation,
${AA}^{- 1} = I$
$A \cdot \frac{1}{| A |} (adj A) = I$
$A \cdot (adj A) = | A | I$
After comparing the above equation expression with given expression, we get -
$| A | = 5$
$adj A = | A | A^{- 1}$
$| adj A | = | | A | A^{- 1} | = | A^{2} | = 25$

Karnataka CET-2008

Matrix and Determinant

78965 If $A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$ , then $A \cdot adj (A)$ is equal to=

1

[\begin{array}{lll} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{array}]

2

[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}]

3

[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]

4

[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

Explanation:

(C) : We have,
$A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$
$= 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6$
$= 2 + 18 - 12 = 2 + 6 = 8$
$| A | = 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6)$
As we know that,
$A^{- 1} = \frac{Adj A}{| A |}$
${AA}^{- 1} = \frac{A (Adj A)}{| A |}$
$I | A | = A (Adj A)$
$(∵ {AA}^{- 1} = I)$
Now, $| A | = 8$
$A (Adj A) = 8 I = [\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]$

Karnataka CET-2008

Matrix and Determinant

78966 If $A = [\begin{array}{rrr} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}], B = (adj A)$ and $C = 5 A$ ,
then $\frac{| C |}{| adjB |}$ is equal to

1 125

2 -1

3 5

4 -5

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]$
So, $| A | = 3 (- 3 + 4) + 3 (2 - 0) + 4 (- 2 - 0)$
$| A | = 1$
Here, it is given that,
$B = adj A, C = 5 A$
$| C | = | 5 A | = 5^{3} | A | = 5^{3} = 125$
Also, $| adj B | = | adj (adj A) | = | A |^{(3 - 1)^{2}} = | A |^{4}$
$∴ \frac{| C |}{| adj B |} = \frac{125}{| adj (adj A) |} = \frac{125}{| A |^{4}} = 125 [∵ | A | = 1]$

Karnataka CET-2007

Matrix and Determinant

78963 If $A$ and $B$ are square matrices of the same order such that $(A + B) (A - B) = A^{2} - B^{2}$ , then ${({A B A}^{- 1})}^{2} =$

1

A^{2} B^{2}

2

A^{2}

3

B^{2}

4 1

Explanation:

(C) : According to given summation,
$(A + B) (A - B) = A^{2} - B^{2}$
$A A + B A - A B - B B = A^{2} - B^{2}$
$A^{2} + B A - A B - B^{2} = A^{2} - B^{2}$
$B A - A B = 0 \Rightarrow B A = A B$
Now, ${({ABA}^{- 1})}^{2} = {({BAA}^{- 1})}^{2}$
$= (B \cdot 1)^{2} [∵ {AA}^{- 1} = 1]$
$= B^{2}$

Karnataka CET-2009

Matrix and Determinant

78964 If $A (adj A) = 5 I$ where $I$ is the identity matrix of order 3 , then $| adj A |$ is equal to

1 5

2 10

3 125

4 25

Explanation:

(D) : According to given summation,
${AA}^{- 1} = I$
$A \cdot \frac{1}{| A |} (adj A) = I$
$A \cdot (adj A) = | A | I$
After comparing the above equation expression with given expression, we get -
$| A | = 5$
$adj A = | A | A^{- 1}$
$| adj A | = | | A | A^{- 1} | = | A^{2} | = 25$

Karnataka CET-2008

Matrix and Determinant

78965 If $A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$ , then $A \cdot adj (A)$ is equal to=

1

[\begin{array}{lll} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{array}]

2

[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}]

3

[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]

4

[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

Explanation:

(C) : We have,
$A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$
$= 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6$
$= 2 + 18 - 12 = 2 + 6 = 8$
$| A | = 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6)$
As we know that,
$A^{- 1} = \frac{Adj A}{| A |}$
${AA}^{- 1} = \frac{A (Adj A)}{| A |}$
$I | A | = A (Adj A)$
$(∵ {AA}^{- 1} = I)$
Now, $| A | = 8$
$A (Adj A) = 8 I = [\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]$

Karnataka CET-2008

Matrix and Determinant

78966 If $A = [\begin{array}{rrr} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}], B = (adj A)$ and $C = 5 A$ ,
then $\frac{| C |}{| adjB |}$ is equal to

1 125

2 -1

3 5

4 -5

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]$
So, $| A | = 3 (- 3 + 4) + 3 (2 - 0) + 4 (- 2 - 0)$
$| A | = 1$
Here, it is given that,
$B = adj A, C = 5 A$
$| C | = | 5 A | = 5^{3} | A | = 5^{3} = 125$
Also, $| adj B | = | adj (adj A) | = | A |^{(3 - 1)^{2}} = | A |^{4}$
$∴ \frac{| C |}{| adj B |} = \frac{125}{| adj (adj A) |} = \frac{125}{| A |^{4}} = 125 [∵ | A | = 1]$

Karnataka CET-2007

Matrix and Determinant

78963 If $A$ and $B$ are square matrices of the same order such that $(A + B) (A - B) = A^{2} - B^{2}$ , then ${({A B A}^{- 1})}^{2} =$

1

A^{2} B^{2}

2

A^{2}

3

B^{2}

4 1

Explanation:

(C) : According to given summation,
$(A + B) (A - B) = A^{2} - B^{2}$
$A A + B A - A B - B B = A^{2} - B^{2}$
$A^{2} + B A - A B - B^{2} = A^{2} - B^{2}$
$B A - A B = 0 \Rightarrow B A = A B$
Now, ${({ABA}^{- 1})}^{2} = {({BAA}^{- 1})}^{2}$
$= (B \cdot 1)^{2} [∵ {AA}^{- 1} = 1]$
$= B^{2}$

Karnataka CET-2009

Matrix and Determinant

78964 If $A (adj A) = 5 I$ where $I$ is the identity matrix of order 3 , then $| adj A |$ is equal to

1 5

2 10

3 125

4 25

Explanation:

(D) : According to given summation,
${AA}^{- 1} = I$
$A \cdot \frac{1}{| A |} (adj A) = I$
$A \cdot (adj A) = | A | I$
After comparing the above equation expression with given expression, we get -
$| A | = 5$
$adj A = | A | A^{- 1}$
$| adj A | = | | A | A^{- 1} | = | A^{2} | = 25$

Karnataka CET-2008

Matrix and Determinant

78965 If $A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$ , then $A \cdot adj (A)$ is equal to=

1

[\begin{array}{lll} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{array}]

2

[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}]

3

[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]

4

[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

Explanation:

(C) : We have,
$A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$
$= 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6$
$= 2 + 18 - 12 = 2 + 6 = 8$
$| A | = 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6)$
As we know that,
$A^{- 1} = \frac{Adj A}{| A |}$
${AA}^{- 1} = \frac{A (Adj A)}{| A |}$
$I | A | = A (Adj A)$
$(∵ {AA}^{- 1} = I)$
Now, $| A | = 8$
$A (Adj A) = 8 I = [\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]$

Karnataka CET-2008

Matrix and Determinant

78966 If $A = [\begin{array}{rrr} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}], B = (adj A)$ and $C = 5 A$ ,
then $\frac{| C |}{| adjB |}$ is equal to

1 125

2 -1

3 5

4 -5

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]$
So, $| A | = 3 (- 3 + 4) + 3 (2 - 0) + 4 (- 2 - 0)$
$| A | = 1$
Here, it is given that,
$B = adj A, C = 5 A$
$| C | = | 5 A | = 5^{3} | A | = 5^{3} = 125$
Also, $| adj B | = | adj (adj A) | = | A |^{(3 - 1)^{2}} = | A |^{4}$
$∴ \frac{| C |}{| adj B |} = \frac{125}{| adj (adj A) |} = \frac{125}{| A |^{4}} = 125 [∵ | A | = 1]$

Karnataka CET-2007

Matrix and Determinant

78963 If $A$ and $B$ are square matrices of the same order such that $(A + B) (A - B) = A^{2} - B^{2}$ , then ${({A B A}^{- 1})}^{2} =$

1

A^{2} B^{2}

2

A^{2}

3

B^{2}

4 1

Explanation:

(C) : According to given summation,
$(A + B) (A - B) = A^{2} - B^{2}$
$A A + B A - A B - B B = A^{2} - B^{2}$
$A^{2} + B A - A B - B^{2} = A^{2} - B^{2}$
$B A - A B = 0 \Rightarrow B A = A B$
Now, ${({ABA}^{- 1})}^{2} = {({BAA}^{- 1})}^{2}$
$= (B \cdot 1)^{2} [∵ {AA}^{- 1} = 1]$
$= B^{2}$

Karnataka CET-2009

Matrix and Determinant

78964 If $A (adj A) = 5 I$ where $I$ is the identity matrix of order 3 , then $| adj A |$ is equal to

1 5

2 10

3 125

4 25

Explanation:

(D) : According to given summation,
${AA}^{- 1} = I$
$A \cdot \frac{1}{| A |} (adj A) = I$
$A \cdot (adj A) = | A | I$
After comparing the above equation expression with given expression, we get -
$| A | = 5$
$adj A = | A | A^{- 1}$
$| adj A | = | | A | A^{- 1} | = | A^{2} | = 25$

Karnataka CET-2008

Matrix and Determinant

78965 If $A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$ , then $A \cdot adj (A)$ is equal to=

1

[\begin{array}{lll} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{array}]

2

[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}]

3

[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]

4

[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

Explanation:

(C) : We have,
$A = [\begin{array}{ccc} 1 & - 2 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array}]$
$= 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6$
$= 2 + 18 - 12 = 2 + 6 = 8$
$| A | = 1 (8 - 6) + 2 (0 + 9) + 2 (0 - 6)$
As we know that,
$A^{- 1} = \frac{Adj A}{| A |}$
${AA}^{- 1} = \frac{A (Adj A)}{| A |}$
$I | A | = A (Adj A)$
$(∵ {AA}^{- 1} = I)$
Now, $| A | = 8$
$A (Adj A) = 8 I = [\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}]$

Karnataka CET-2008

Matrix and Determinant

78966 If $A = [\begin{array}{rrr} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}], B = (adj A)$ and $C = 5 A$ ,
then $\frac{| C |}{| adjB |}$ is equal to

1 125

2 -1

3 5

4 -5

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]$
So, $| A | = 3 (- 3 + 4) + 3 (2 - 0) + 4 (- 2 - 0)$
$| A | = 1$
Here, it is given that,
$B = adj A, C = 5 A$
$| C | = | 5 A | = 5^{3} | A | = 5^{3} = 125$
Also, $| adj B | = | adj (adj A) | = | A |^{(3 - 1)^{2}} = | A |^{4}$
$∴ \frac{| C |}{| adj B |} = \frac{125}{| adj (adj A) |} = \frac{125}{| A |^{4}} = 125 [∵ | A | = 1]$

Karnataka CET-2007