QBManager

Matrix and Determinant

78967 If the product of the matrix $B=$
$\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$ with a matrix $A$ has the inverse
$C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$, then $A^{-1}$ equals

1 $\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 9\end{array}\right]$

2 $\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16\end{array}\right]$

3 $\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 0 & 2 \\ 2 & 14 & 6\end{array}\right]$

4 $\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$

COMEDK-2014

Matrix and Determinant

78968 If $A=\left[\begin{array}{ll}2 & -3 \\ 5 & -7\end{array}\right]$, then $A+A^{-1}=$

1 $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

2 $\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]$

3 $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$

4 $\left[\begin{array}{cc}4 & 0 \\ 0 & -5\end{array}\right]$

COMEDK-2017

Matrix and Determinant

78969 If $A^{2}-A+I=0$, then the inverse of $A$ is

1 $\mathrm{I}-\mathrm{A}$

2 $\mathrm{A}-\mathrm{I}$

3 $\mathrm{A}$

4 $\mathrm{A}+\mathrm{I}$

COMEDK-2019

Matrix and Determinant

78970 If $A=\left[\begin{array}{ccc}2 & 3 & -5 \\ 1 & 4 & 9 \\ 0 & 7 & -2\end{array}\right]$, then apply $C_{3} \rightarrow C_{3}+3 C_{2}$ to form a matrix $B=\left[b_{i j}\right]$ and then find the value of $a_{22}+b_{21}$ and $a_{11} b_{11} + a_{22} b_{22}$ respectively.

1 20 and 5

2 5 and 20

3 5 and -4

4 10 and 15

Explanation:

(B) : We have,
$A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$
Now, applying $C_{3} \to C_{3} + 3 C_{2}$
$A = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$ i.e., $B = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$
Here, $a_{22} + b_{21} = 4 + 1 = 5$
$a_{11} b_{11} + a_{22} b_{22} = 2 \times 2 + 4 \times 4 = 4 + 16 = 20$

COMEDK-2020

Matrix and Determinant

78971 If $A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \boldsymbol t a n θ & \boldsymbol s e c θ & 0 \\ 0 & 0 & 1 \end{array}]$ , then

1

A^{- 1}

exists

2

A^{- 1}

does not exist

3

| A | = 2

4 none of these

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \tan θ & \sec θ & 0 \\ 0 & 0 & 1 \end{array}]$
$| A | = \sec θ (\sec θ - 0) - \tan θ (\tan θ - 0) + 0$
$| A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$= \sec^{2} θ - \tan^{2} θ = 1$
$∴ | A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$∵ | A | \neq 0$
$∴ A^{- 1}$ exists

COMEDK-2020

Matrix and Determinant

78967 If the product of the matrix $B =$
$[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$ with a matrix $A$ has the inverse
$C = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$ , then $A^{- 1}$ equals

1

[\begin{array}{ccc} - 3 & - 5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 9 \end{array}]

2

[\begin{array}{ccc} - 3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16 \end{array}]

3

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 0 & 2 \\ 2 & 14 & 6 \end{array}]

4

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]

Explanation:

(D) : It is given that,
$(BA)^{- 1} = C$
Above expression can be written as,
$A^{- 1} B^{- 1} = C$
$A^{- 1} {[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]}^{- 1} = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$
Multiply by B on both sides, we get -
$A^{- 1} (B^{- 1} B) = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}] [\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$
$A^{- 1} = [\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]$

COMEDK-2014

Matrix and Determinant

78968 If $A = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}]$ , then $A + A^{- 1} =$

1

[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]

2

[\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]

3

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

4

[\begin{array}{cc} 4 & 0 \\ 0 & - 5 \end{array}]

Explanation:

(B) : Given matrix,
$A = [\begin{array}{rr} 2 & - 3 \\ 5 & - 7 \end{array}]$
And, $| A | = - 14 + 15 = 1 \neq 0$
$∵ | A | \neq 0$ ,
$∴ A^{- 1}$ exists.
$∴$ adj $A = [\begin{array}{cc} - 7 & 3 \\ 5 & 2 \end{array}]$
$\Rightarrow A^{- 1} = \frac{1}{| A |} (adj A) = [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}]$
Here,
$A + A^{- 1} = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}] + [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}] = [\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]$

COMEDK-2017

Matrix and Determinant

78969 If $A^{2} - A + I = 0$ , then the inverse of $A$ is

1

I - A

2

A - I

3

A

4

A + I

Explanation:

(A) : According to given summation,
If $A$ is any square matrix then,
$\begin{array}{ll} {AA}^{- 1} = I and A \\ Here, & A^{2} - A + I = 0 \end{array}$
$A^{- 1} A^{2} - A^{- 1} A + A^{- 1} I = 0$
$(A^{- 1} A) A - (A^{- 1} A) + A^{- 1} = 0$
$I A - I + A^{- 1} = 0$
$A - I + A^{- 1} = 0$
$A^{- 1} = I - A$

COMEDK-2019

Matrix and Determinant

78970 If $A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$ , then apply $C_{3} \to C_{3} + 3 C_{2}$ to form a matrix $B = [b_{i j}]$ and then find the value of $a_{22} + b_{21}$ and $a_{11} b_{11} + a_{22} b_{22}$ respectively.

1 20 and 5

2 5 and 20

3 5 and -4

4 10 and 15

Explanation:

(B) : We have,
$A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$
Now, applying $C_{3} \to C_{3} + 3 C_{2}$
$A = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$ i.e., $B = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$
Here, $a_{22} + b_{21} = 4 + 1 = 5$
$a_{11} b_{11} + a_{22} b_{22} = 2 \times 2 + 4 \times 4 = 4 + 16 = 20$

COMEDK-2020

Matrix and Determinant

78971 If $A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \boldsymbol t a n θ & \boldsymbol s e c θ & 0 \\ 0 & 0 & 1 \end{array}]$ , then

1

A^{- 1}

exists

2

A^{- 1}

does not exist

3

| A | = 2

4 none of these

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \tan θ & \sec θ & 0 \\ 0 & 0 & 1 \end{array}]$
$| A | = \sec θ (\sec θ - 0) - \tan θ (\tan θ - 0) + 0$
$| A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$= \sec^{2} θ - \tan^{2} θ = 1$
$∴ | A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$∵ | A | \neq 0$
$∴ A^{- 1}$ exists

COMEDK-2020

Matrix and Determinant

78967 If the product of the matrix $B =$
$[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$ with a matrix $A$ has the inverse
$C = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$ , then $A^{- 1}$ equals

1

[\begin{array}{ccc} - 3 & - 5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 9 \end{array}]

2

[\begin{array}{ccc} - 3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16 \end{array}]

3

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 0 & 2 \\ 2 & 14 & 6 \end{array}]

4

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]

Explanation:

(D) : It is given that,
$(BA)^{- 1} = C$
Above expression can be written as,
$A^{- 1} B^{- 1} = C$
$A^{- 1} {[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]}^{- 1} = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$
Multiply by B on both sides, we get -
$A^{- 1} (B^{- 1} B) = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}] [\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$
$A^{- 1} = [\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]$

COMEDK-2014

Matrix and Determinant

78968 If $A = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}]$ , then $A + A^{- 1} =$

1

[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]

2

[\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]

3

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

4

[\begin{array}{cc} 4 & 0 \\ 0 & - 5 \end{array}]

Explanation:

(B) : Given matrix,
$A = [\begin{array}{rr} 2 & - 3 \\ 5 & - 7 \end{array}]$
And, $| A | = - 14 + 15 = 1 \neq 0$
$∵ | A | \neq 0$ ,
$∴ A^{- 1}$ exists.
$∴$ adj $A = [\begin{array}{cc} - 7 & 3 \\ 5 & 2 \end{array}]$
$\Rightarrow A^{- 1} = \frac{1}{| A |} (adj A) = [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}]$
Here,
$A + A^{- 1} = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}] + [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}] = [\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]$

COMEDK-2017

Matrix and Determinant

78969 If $A^{2} - A + I = 0$ , then the inverse of $A$ is

1

I - A

2

A - I

3

A

4

A + I

Explanation:

(A) : According to given summation,
If $A$ is any square matrix then,
$\begin{array}{ll} {AA}^{- 1} = I and A \\ Here, & A^{2} - A + I = 0 \end{array}$
$A^{- 1} A^{2} - A^{- 1} A + A^{- 1} I = 0$
$(A^{- 1} A) A - (A^{- 1} A) + A^{- 1} = 0$
$I A - I + A^{- 1} = 0$
$A - I + A^{- 1} = 0$
$A^{- 1} = I - A$

COMEDK-2019

Matrix and Determinant

78970 If $A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$ , then apply $C_{3} \to C_{3} + 3 C_{2}$ to form a matrix $B = [b_{i j}]$ and then find the value of $a_{22} + b_{21}$ and $a_{11} b_{11} + a_{22} b_{22}$ respectively.

1 20 and 5

2 5 and 20

3 5 and -4

4 10 and 15

Explanation:

(B) : We have,
$A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$
Now, applying $C_{3} \to C_{3} + 3 C_{2}$
$A = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$ i.e., $B = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$
Here, $a_{22} + b_{21} = 4 + 1 = 5$
$a_{11} b_{11} + a_{22} b_{22} = 2 \times 2 + 4 \times 4 = 4 + 16 = 20$

COMEDK-2020

Matrix and Determinant

78971 If $A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \boldsymbol t a n θ & \boldsymbol s e c θ & 0 \\ 0 & 0 & 1 \end{array}]$ , then

1

A^{- 1}

exists

2

A^{- 1}

does not exist

3

| A | = 2

4 none of these

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \tan θ & \sec θ & 0 \\ 0 & 0 & 1 \end{array}]$
$| A | = \sec θ (\sec θ - 0) - \tan θ (\tan θ - 0) + 0$
$| A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$= \sec^{2} θ - \tan^{2} θ = 1$
$∴ | A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$∵ | A | \neq 0$
$∴ A^{- 1}$ exists

COMEDK-2020

Matrix and Determinant

78967 If the product of the matrix $B =$
$[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$ with a matrix $A$ has the inverse
$C = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$ , then $A^{- 1}$ equals

1

[\begin{array}{ccc} - 3 & - 5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 9 \end{array}]

2

[\begin{array}{ccc} - 3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16 \end{array}]

3

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 0 & 2 \\ 2 & 14 & 6 \end{array}]

4

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]

Explanation:

(D) : It is given that,
$(BA)^{- 1} = C$
Above expression can be written as,
$A^{- 1} B^{- 1} = C$
$A^{- 1} {[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]}^{- 1} = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$
Multiply by B on both sides, we get -
$A^{- 1} (B^{- 1} B) = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}] [\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$
$A^{- 1} = [\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]$

COMEDK-2014

Matrix and Determinant

78968 If $A = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}]$ , then $A + A^{- 1} =$

1

[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]

2

[\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]

3

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

4

[\begin{array}{cc} 4 & 0 \\ 0 & - 5 \end{array}]

Explanation:

(B) : Given matrix,
$A = [\begin{array}{rr} 2 & - 3 \\ 5 & - 7 \end{array}]$
And, $| A | = - 14 + 15 = 1 \neq 0$
$∵ | A | \neq 0$ ,
$∴ A^{- 1}$ exists.
$∴$ adj $A = [\begin{array}{cc} - 7 & 3 \\ 5 & 2 \end{array}]$
$\Rightarrow A^{- 1} = \frac{1}{| A |} (adj A) = [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}]$
Here,
$A + A^{- 1} = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}] + [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}] = [\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]$

COMEDK-2017

Matrix and Determinant

78969 If $A^{2} - A + I = 0$ , then the inverse of $A$ is

1

I - A

2

A - I

3

A

4

A + I

Explanation:

(A) : According to given summation,
If $A$ is any square matrix then,
$\begin{array}{ll} {AA}^{- 1} = I and A \\ Here, & A^{2} - A + I = 0 \end{array}$
$A^{- 1} A^{2} - A^{- 1} A + A^{- 1} I = 0$
$(A^{- 1} A) A - (A^{- 1} A) + A^{- 1} = 0$
$I A - I + A^{- 1} = 0$
$A - I + A^{- 1} = 0$
$A^{- 1} = I - A$

COMEDK-2019

Matrix and Determinant

78970 If $A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$ , then apply $C_{3} \to C_{3} + 3 C_{2}$ to form a matrix $B = [b_{i j}]$ and then find the value of $a_{22} + b_{21}$ and $a_{11} b_{11} + a_{22} b_{22}$ respectively.

1 20 and 5

2 5 and 20

3 5 and -4

4 10 and 15

Explanation:

(B) : We have,
$A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$
Now, applying $C_{3} \to C_{3} + 3 C_{2}$
$A = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$ i.e., $B = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$
Here, $a_{22} + b_{21} = 4 + 1 = 5$
$a_{11} b_{11} + a_{22} b_{22} = 2 \times 2 + 4 \times 4 = 4 + 16 = 20$

COMEDK-2020

Matrix and Determinant

78971 If $A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \boldsymbol t a n θ & \boldsymbol s e c θ & 0 \\ 0 & 0 & 1 \end{array}]$ , then

1

A^{- 1}

exists

2

A^{- 1}

does not exist

3

| A | = 2

4 none of these

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \tan θ & \sec θ & 0 \\ 0 & 0 & 1 \end{array}]$
$| A | = \sec θ (\sec θ - 0) - \tan θ (\tan θ - 0) + 0$
$| A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$= \sec^{2} θ - \tan^{2} θ = 1$
$∴ | A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$∵ | A | \neq 0$
$∴ A^{- 1}$ exists

COMEDK-2020

Matrix and Determinant

78967 If the product of the matrix $B =$
$[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$ with a matrix $A$ has the inverse
$C = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$ , then $A^{- 1}$ equals

1

[\begin{array}{ccc} - 3 & - 5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 9 \end{array}]

2

[\begin{array}{ccc} - 3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16 \end{array}]

3

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 0 & 2 \\ 2 & 14 & 6 \end{array}]

4

[\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]

Explanation:

(D) : It is given that,
$(BA)^{- 1} = C$
Above expression can be written as,
$A^{- 1} B^{- 1} = C$
$A^{- 1} {[\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]}^{- 1} = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}]$
Multiply by B on both sides, we get -
$A^{- 1} (B^{- 1} B) = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}] [\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 1 \end{array}]$
$A^{- 1} = [\begin{array}{ccc} - 3 & - 5 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{array}]$

COMEDK-2014

Matrix and Determinant

78968 If $A = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}]$ , then $A + A^{- 1} =$

1

[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]

2

[\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]

3

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

4

[\begin{array}{cc} 4 & 0 \\ 0 & - 5 \end{array}]

Explanation:

(B) : Given matrix,
$A = [\begin{array}{rr} 2 & - 3 \\ 5 & - 7 \end{array}]$
And, $| A | = - 14 + 15 = 1 \neq 0$
$∵ | A | \neq 0$ ,
$∴ A^{- 1}$ exists.
$∴$ adj $A = [\begin{array}{cc} - 7 & 3 \\ 5 & 2 \end{array}]$
$\Rightarrow A^{- 1} = \frac{1}{| A |} (adj A) = [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}]$
Here,
$A + A^{- 1} = [\begin{array}{ll} 2 & - 3 \\ 5 & - 7 \end{array}] + [\begin{array}{ll} - 7 & 3 \\ - 5 & 2 \end{array}] = [\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}]$

COMEDK-2017

Matrix and Determinant

78969 If $A^{2} - A + I = 0$ , then the inverse of $A$ is

1

I - A

2

A - I

3

A

4

A + I

Explanation:

(A) : According to given summation,
If $A$ is any square matrix then,
$\begin{array}{ll} {AA}^{- 1} = I and A \\ Here, & A^{2} - A + I = 0 \end{array}$
$A^{- 1} A^{2} - A^{- 1} A + A^{- 1} I = 0$
$(A^{- 1} A) A - (A^{- 1} A) + A^{- 1} = 0$
$I A - I + A^{- 1} = 0$
$A - I + A^{- 1} = 0$
$A^{- 1} = I - A$

COMEDK-2019

Matrix and Determinant

78970 If $A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$ , then apply $C_{3} \to C_{3} + 3 C_{2}$ to form a matrix $B = [b_{i j}]$ and then find the value of $a_{22} + b_{21}$ and $a_{11} b_{11} + a_{22} b_{22}$ respectively.

1 20 and 5

2 5 and 20

3 5 and -4

4 10 and 15

Explanation:

(B) : We have,
$A = [\begin{array}{ccc} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{array}]$
Now, applying $C_{3} \to C_{3} + 3 C_{2}$
$A = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$ i.e., $B = [\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 4 & 21 \\ 0 & 7 & 19 \end{array}]$
Here, $a_{22} + b_{21} = 4 + 1 = 5$
$a_{11} b_{11} + a_{22} b_{22} = 2 \times 2 + 4 \times 4 = 4 + 16 = 20$

COMEDK-2020

Matrix and Determinant

78971 If $A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \boldsymbol t a n θ & \boldsymbol s e c θ & 0 \\ 0 & 0 & 1 \end{array}]$ , then

1

A^{- 1}

exists

2

A^{- 1}

does not exist

3

| A | = 2

4 none of these

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} \sec θ & \tan θ & 0 \\ \tan θ & \sec θ & 0 \\ 0 & 0 & 1 \end{array}]$
$| A | = \sec θ (\sec θ - 0) - \tan θ (\tan θ - 0) + 0$
$| A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$= \sec^{2} θ - \tan^{2} θ = 1$
$∴ | A | = \sec θ (\sec θ) - \tan θ (\tan θ)$
$∵ | A | \neq 0$
$∴ A^{- 1}$ exists

COMEDK-2020