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

If $A = [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}]$ , then $A^{- 1} =$

A. $(\frac{1}{2}) [\begin{array}{lll} 0 & 1 & 2 \\ 3 & 2 & 1 \\ 4 & 2 & 3 \end{array}]$

B. $[\begin{array}{ccc} \frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ - 4 & 3 & - 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{array}]$

C. $[\begin{array}{ccc} \frac{1}{2} & - 1 & \frac{5}{2} \\ 1 & - 6 & 3 \\ 1 & 2 & - 1 \end{array}]$

D. $(\frac{1}{2}) [\begin{array}{ccc} 1 & - 1 & - 1 \\ - 8 & 6 & - 2 \\ 5 & - 3 & 1 \end{array}]$

Correct answer option is (B)

$\begin{aligned} [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}] A^{- 1} = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ R_{1} \to R_{1} + R_{2}, R_{3} \to R_{3} - 3 R_{2} \\ [\begin{array}{ccc} 1 & 3 & 5 \\ 1 & 2 & 3 \\ 0 & - 5 & - 8 \end{array}] A^{- 1} = [\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & - 3 & 1 \end{array}] \\ R_{2} \to R_{2} - R_{1} \\ [\begin{array}{ccc} 1 & 3 & 5 \\ 0 & - 1 & - 2 \\ 0 & - 5 & - 8 \end{array}] A^{- 1} = [\begin{array}{ccc} 1 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & - 3 & 1 \end{array}] \\ R_{1} \to R_{1} + 3 R_{2} and R_{3} \to R_{3} - 5 R_{2} \\ [\begin{array}{ccc} 1 & 0 & - 1 \\ 0 & - 1 & - 2 \\ 0 & 0 & 2 \end{array}] A^{- 1} = [\begin{array}{ccc} - 2 & 1 & 0 \\ - 1 & 0 & 0 \\ 5 & - 3 & 1 \end{array}] \\ R_{2} \to R_{2} + R_{3} and R_{1} \to 2 R_{1} + R_{3} \\ [\begin{array}{lll} 2 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 2 \end{array}] A^{- 1} = [\begin{array}{ccc} 1 & - 1 & 1 \\ 4 & - 3 & 1 \\ 5 & - 3 & 1 \end{array}] \\ R_{1} \to \frac{1}{2} R_{1}, R_{2} \to - R_{2}, R_{3} \to \frac{1}{2} R_{3} \\ [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A^{- 1} = [\begin{array}{ccc} \frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ - 4 & 3 & - 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{array}] \end{aligned}$

7. ⇒ (MHT CET 2021 21th September Morning Shift )

If $F (\propto) = [\begin{array}{ccc} \cos \propto & - \sin \propto & 0 \\ \sin \propto & \cos \propto & 0 \\ 0 & 0 & 1 \end{array}]$ , where $\propto\in R$ , then $[F (\propto)]^{- 1} =$

A. $F (- \propto)$

B. $F (2 \propto)$

C. $F (\propto)$

D. $F (3 \propto)$

Correct answer option is (A)

$\begin{aligned} F (α) = [\begin{array}{ccc} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}] \\ ∴ | F (α) | = \cos α (\cos α) + \sin α (\sin α) = \cos^{2} α + \sin^{2} α = 1 \\ ∴ adj [F (α)] = [\begin{array}{ccc} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{ccc} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{array}] = F (- α) \\ [f (α)]^{- 1} = \frac{adj [F (α)]}{| F (α) |} = F (- α) \end{aligned}$

8. ⇒ (MHT CET 2021 21th September Morning Shift )

$A^{- 1} = \frac{- 1}{2} [\begin{array}{cc} 1 & - 4 \\ - 1 & 2 \end{array}]$ , then $2 A + I_{2} =$

where $I_{2}$ is a unit matrix of order 2

A. $[\begin{array}{ll} 5 & 8 \\ 1 & 2 \end{array}]$

B. $[\begin{array}{ll} 5 & 8 \\ 2 & 2 \end{array}]$

C. $[\begin{array}{ll} 2 & 4 \\ 1 & 1 \end{array}]$

D. $[\begin{array}{ll} 5 & 8 \\ 2 & 3 \end{array}]$

Correct answer option is (D)

$\begin{aligned} A^{- 1} = \frac{- 1}{2} [\begin{array}{cc} 1 & - 4 \\ - 1 & 2 \end{array}] = [\begin{array}{cc} \frac{- 1}{2} & 2 \\ \frac{1}{2} & - 1 \end{array}] \\ A^{- 1} = I \Rightarrow A [\begin{array}{cc} \frac{- 1}{2} & 2 \\ \frac{1}{2} & - 1 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \end{aligned} \begin{aligned} R_{2} \to R_{1} + R_{2} \\ A [\begin{array}{cc} \frac{- 1}{2} & 2 \\ 0 & 1 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}] \\ R_{1} \to R_{1} - 2 R_{2} \\ A [\begin{array}{cc} \frac{- 1}{2} & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} - 1 & - 2 \\ 1 & 1 \end{array}] \\ R_{1} \to - 2 R_{1} \\ A [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{ll} 2 & 4 \\ 1 & 1 \end{array}] \Rightarrow A = [\begin{array}{ll} 2 & 4 \\ 1 & 1 \end{array}] \\ ∴ 2 A + I_{2} = [\begin{array}{ll} 4 & 8 \\ 2 & 2 \end{array}] + [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{ll} 5 & 8 \\ 2 & 3 \end{array}] \end{aligned}$

9. ⇒ (MHT CET 2021 20th September Evening Shift )

If $A^{- 1} = [\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 5 & 5 \end{array}]$ , then $A =$

A. $[\begin{array}{ccc} - 5 & 20 & - 2 \\ - 1 & 3 & 0 \\ 3 & - 11 & 1 \end{array}]$

B. $[\begin{array}{ccc} - 5 & 20 & 2 \\ - 1 & 3 & 0 \\ 3 & 11 & 1 \end{array}]$

C. $[\begin{array}{ccc} - 5 & 20 & 2 \\ 1 & 3 & 0 \\ 3 & 11 & - 1 \end{array}]$

D. $[\begin{array}{ccc} - 5 & 20 & - 2 \\ 1 & 3 & 0 \\ 3 & 11 & 1 \end{array}]$

Correct answer option is (A)

$\begin{aligned} Let A^{- 1} A = I \\ ∴ [\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 5 & 5 \end{array}] A = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ R_{2} \to 3 R_{2} - R_{1} and R_{3} \to 3 R_{3} - 2 R_{1} \\ [\begin{array}{ccc} 3 & 2 & 6 \\ 0 & 1 & 0 \\ 0 & 11 & 3 \end{array}] A = [\begin{array}{ccc} 1 & 0 & 0 \\ - 1 & 3 & 0 \\ - 2 & 0 & 3 \end{array}] \\ R_{1} \to R_{1} - 2 R_{2} and R_{3} \to R_{3} - 11 R_{2} \\ [\begin{array}{lll} 3 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array}] A^{-} = [\begin{array}{ccc} 3 & - 6 & 0 \\ - 1 & 3 & 0 \\ 9 & - 33 & 3 \end{array}] \\ R_{1} \to R_{1} - 2 R_{3} \\ [\begin{array}{lll} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array}] A = [\begin{array}{ccc} - 15 & 60 & - 6 \\ - 1 & 3 & 0 \\ 9 & - 33 & 3 \end{array}] \\ R_{1} \to \frac{1}{3} R_{1} and R_{3} \to \frac{1}{3} R_{3} \\ [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A = [\begin{array}{ccc} - 5 & 20 & - 2 \\ - 1 & 3 & 0 \\ 3 & - 11 & 1 \end{array}] \end{aligned}$

10. ⇒ (MHT CET 2021 20th September Evening Shift )

If $A^{- 1} = [\begin{array}{cc} 2 & - 3 \\ - 1 & 2 \end{array}]$ and $B^{- 1} = [\begin{array}{cc} 1 & 0 \\ - 3 & 1 \end{array}]$ , then $(A B)^{- 1} =$

A. $[\begin{array}{cc} 2 & 7 \\ 3 & - 1 \end{array}]$

B. $[\begin{array}{cc} 2 & - 7 \\ - 3 & 11 \end{array}]$

C. $[\begin{array}{cc} 2 & - 3 \\ - 7 & 11 \end{array}]$

D. $[\begin{array}{cc} 2 & 3 \\ 7 & - 11 \end{array}]$

Correct answer option is (C)

$(AB)^{- 1} = B^{- 1} A^{- 1}$

$\begin{aligned} = [\begin{array}{cc} 1 & 0 \\ - 3 & 1 \end{array}] [\begin{array}{cc} 2 & - 3 \\ - 1 & 2 \end{array}] \\ = [\begin{array}{cc} 2 + 0 & - 3 + 0 \\ - 6 - 1 & 9 + 2 \end{array}] = [\begin{array}{cc} 2 & - 3 \\ - 7 & 11 \end{array}] \end{aligned}$

⇒Matrices