1. ⇒ (MHT CET 2023 12th May Morning Shift )

If the matrix $A = [\begin{array}{cc} 1 & 2 \\ - 5 & 1 \end{array}]$ and $A^{- 1} = x A + y I$ , when $I$ is a unit matrix of order 2 , then the value of $2 x + 3 y$ is

A. $\frac{8}{11}$

B. $\frac{4}{11}$

C. $\frac{- 8}{11}$

D. $\frac{- 4}{11}$

Correct answer option is (B)

$\begin{aligned} A = [\begin{array}{cc} 1 & 2 \\ - 5 & 1 \end{array}] \\ ∴ | A | = 11 \\ A^{- 1} = \frac{1}{| A |} [\begin{array}{cc} 1 & - 2 \\ 5 & 1 \end{array}] = \frac{1}{11} [\begin{array}{cc} 1 & - 2 \\ 5 & 1 \end{array}] \\ A^{- 1} = x A + y I, we get \\ [\begin{array}{cc} \frac{1}{11} & \frac{- 2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{array}] = [\begin{array}{cc} x & 2 x \\ - 5 x & x \end{array}] + [\begin{array}{ll} y & 0 \\ 0 & y \end{array}] \\ \begin{array}{ll} ∴ & [\begin{array}{cc} \frac{1}{11} & \frac{- 2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{array}] = [\begin{array}{cc} x + y & 2 x \\ - 5 x & x + y \end{array}] \\ \Rightarrow & x = \frac{- 1}{11} and y = \frac{2}{11} \\ ∴ & 2 x + 3 y = 2 (\frac{- 1}{11}) + 3 (\frac{2}{11}) = \frac{4}{11} \end{array} \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Evening Shift )

If $A = [\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}]$ where $i = \sqrt{- 1}$ and $B = A^{2029}$ , then $B^{- 1} =$

A. $- A$

B. $adj A$

C. $- I$

D. $- adj A$

Correct answer option is (D)

$\begin{aligned} A = [\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}] \\ ∴ A^{2} = [\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}] \times [\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}] = [\begin{array}{ll} 0 & i \\ i & 1 \end{array}] \\ ∴ A^{3} = [\begin{array}{ll} 0 & i \\ i & 1 \end{array}] [\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}] = [\begin{array}{ll} i & 0 \\ 0 & i \end{array}] \\ ∴ A^{6} = A^{3} \times A^{3} \\ = [\begin{array}{ll} i & 0 \\ 0 & i \end{array}] \times [\begin{array}{ll} i & 0 \\ 0 & i \end{array}] \\ = [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] = (- 1) I_{2} \\ Now, B = A^{2029} = A^{(6 \times 338 + 1)} \\ ∴ B = {(A^{6})}^{338} \times A \\ = {((- 1) I_{2})}^{338} \times A \\ = I_{2} \times A \\ ∴ B = A \\ Now, | A | = - 1 \\ ∴ B^{- 1} = A^{- 1} = \frac{1}{| A |} adj A \\ ∴ B^{- 1} = - adj A \end{aligned}$

3. ⇒ (MHT CET 2023 9th May Evening Shift )

Let $A = [\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & - 2 & 1 \end{array}], B = [\begin{matrix} 6 \\ 11 \\ 0 \end{matrix}]$ and $X = [\begin{array}{l} a \\ b \\ c \end{array}]$ , if $AX = B$ , then the value of $2 a + b + 2 c$ is

A. 10

B. 8

C. 6

D. 12

Correct answer option is (A)

$\begin{array}{l} A X = B \\ [\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & - 2 & 1 \end{array}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{array}{c} 6 \\ 11 \\ 0 \end{array}] \\ ∴ a + b + c = 6 .... (i) \\ b + 3 c = 11 .... (ii) \\ a - 2 b + c = 0 \\ i.e., a + c = 2 b .... (iii) \\ From (i) and (ii), we get b = 2 \\ From (ii), c = 3 \\ From (i), a = 1 \\ ∴ 2 a + b + 2 c = 2 (1) + 2 + 2 (3) = 10 \end{array}$

4. ⇒ (MHT CET 2023 9th May Morning Shift )

If $A = [\begin{array}{cc} 2 & - 1 \\ - 1 & 3 \end{array}]$ , then the inverse of $(2 A^{2} + 5 A)$ is

A. $\frac{1}{95} [\begin{array}{ll} 7 & 3 \\ 3 & 4 \end{array}]$

B. $\frac{1}{95} [\begin{array}{cc} - 7 & 3 \\ 3 & - 4 \end{array}]$

C. $\frac{1}{95} [\begin{array}{cc} - 7 & - 3 \\ 3 & 4 \end{array}]$

D. $\frac{1}{95} [\begin{array}{ll} 4 & 3 \\ 3 & 7 \end{array}]$

Correct answer option is (A)

$\begin{aligned} A = [\begin{array}{cc} 2 & - 1 \\ - 1 & 3 \end{array}] \\ A^{2} = [\begin{array}{cc} 2 & - 1 \\ - 1 & 3 \end{array}] [\begin{array}{cc} 2 & - 1 \\ - 1 & 3 \end{array}] \\ ∴ A^{2} = [\begin{array}{cc} 5 & - 5 \\ - 5 & 10 \end{array}] \\ Now, 2 A^{2} + 5 A = 2 [\begin{array}{cc} 5 & - 5 \\ - 5 & 10 \end{array}] + 5 [\begin{array}{cc} 2 & - 1 \\ - 1 & 3 \end{array}] \\ = [\begin{array}{cc} 20 & - 15 \\ - 15 & 35 \end{array}] \\ If A = [\begin{array}{ll} a & b \\ c & d \end{array}], then A^{- 1} = \frac{1}{a d - b c} [\begin{array}{cc} d & - b \\ - c & a \end{array}] \\ ∴ {(2 A^{2} + 5 A)}^{- 1} = \frac{1}{475} [\begin{array}{ll} 35 & 15 \\ 15 & 20 \end{array}] \\ = \frac{1}{95} [\begin{array}{ll} 7 & 3 \\ 3 & 4 \end{array}] \end{aligned}$

5. ⇒ (MHT CET 2021 21th September Evening Shift )

If $A = [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{array}]$ and $A^{- 1} = \frac{1}{2} [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 6 & 2 c \\ 5 & - 3 & 1 \end{array}]$ , then values of a and c are respectively

A. $\frac{1}{2}, \frac{1}{2}$

B. $- 1, 1$

C. $2, \frac{- 1}{2}$

D. $1, - 1$

Correct answer option is (D)

We know that ${AA}^{- 1} = I$

$\begin{aligned} ∴ [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{array}] [\begin{array}{ccc} \frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ - 4 & 3 & c \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ ∴ [\begin{array}{ccc} 1 & 0 & c + 1 \\ 0 & 1 & 2 + 2 c \\ 4 - 4 a & 3 a - 3 & 2 + a c \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{aligned}$

Thus $c + 1 = 0 \Rightarrow c = - 1$ and $4 - 4 a = 0 \Rightarrow a = 1$

⇒Matrices