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Inverse of a Matrix

Let M =[[ 0 ...

Question

Let $M = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & a 1 & 2 & 3 3 & b & 1 \end{matrix} ⎤ ⎥ ⎦$ and $a d j M = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 & - 1 8 & - 6 & 2 - 5 & 3 & - 1 \end{matrix} ⎤ ⎥ ⎦$ where $a, b$ are real numbers. Which of the following option is/are correct?

a + b = 3

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(a d j M)^{- 1} + a d j M^{- 1} = - M

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d e t (a d j M^{2}) = 81

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M ⎡ ⎢ ⎣ \begin{matrix} α β γ \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 123 \end{matrix} ⎤ ⎥ ⎦,

then

α - β + γ = 3

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Solution

The correct options are
A $a + b = 3$
B
$(a d j M)^{- 1} + a d j M^{- 1} = - M$
D If $M ⎡ ⎢ ⎣ \begin{matrix} α β γ \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 123 \end{matrix} ⎤ ⎥ ⎦,$ then $α - β + γ = 3$
$(a d j M)_{11} = ∣ ∣ ∣ \begin{matrix} 2 & 3 b & 1 \end{matrix} ∣ ∣ ∣ = - 1$
$\Rightarrow 2 - 3 b = - 1 \Rightarrow b = 1$
$(a d j M)_{22} = ∣ ∣ ∣ \begin{matrix} 0 & a 3 & 1 \end{matrix} ∣ ∣ ∣ = - 6$
$\Rightarrow - 3 a = - 6 \Rightarrow a = 2$
so, $a + b = 2 + 1 = 3$

Therefore, $M = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 2 1 & 2 & 3 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$
$d e t M = 0 + - 1 (1 - 9) + 2 (1 - 6) = - 2$
$| a d j (M^{2}) | = | M^{2} |^{2} = | M |^{4} = (- 2)^{4} = 16$

$∵ M^{- 1} = \frac{a d j M}{d e t M} = \frac{a d j M}{- 2}$
so, $(a d j M)^{- 1} + a d j M^{- 1}$
$(M^{- 1} | M |)^{- 1} + (M^{- 1})^{- 1} | M^{- 1} |$
$\frac{(M^{- 1})^{- 1}}{| M |} + \frac{M}{| M |}$
$\frac{M}{- 2} + \frac{M}{- 2} = - M$
$∴ (a d j M)^{- 1} + a d j M^{- 1} = - M$

If $M ⎡ ⎢ ⎣ \begin{matrix} α β γ \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 123 \end{matrix} ⎤ ⎥ ⎦$
As, $M X = B \Rightarrow X = M^{- 1} B = \frac{a d j M}{| M |} . B$
$= - \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 & - 1 8 & - 6 & 2 - 5 & 3 & - 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 123 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦$

$X = ⎡ ⎢ ⎣ \begin{matrix} 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} α β γ \end{matrix} ⎤ ⎥ ⎦$
$∴ (α, β, γ) = (1, - 1, 1)$
Hence, $(α - β + γ) = 3$

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