leqalign Inverse - of - the - matrix begin bmatrix 3 -2 -1 1-4 - 1 -1 - 12 - 0 - 1- endbmatrix is -MP PET 1990- -

The correct option is D Let A = nbsp;[ 3 amp; nbsp; 2 amp; 1 nbsp;; nbsp; 4 amp; 1 amp; 1; 2 amp; 0 amp; 1 ] nbsp; ...

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Standard XII

Mathematics

Singular and Non Singualar Matrices

leqalign Inve...

Question

$\begin{matrix} I n v e r s e o f t h e m a t r i x ⎡ ⎢ ⎣ \begin{matrix} 3 & - 2 & - 1 - 4 & 1 & - 1 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}$ is [MP PET 1990]

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Solution

The correct option is C

$\begin{matrix} L e t A = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 2 & - 1 - 4 & 1 & - 1 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}$

then |A| = $\begin{matrix} ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 1 - 4 & 1 & - 1 2 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 1 \end{matrix}$

The matrix of cofactors of A
$\begin{matrix} ⎡ ⎢ ⎣ \begin{matrix} c_{11} & c_{12} & c_{13} c_{21} & c_{22} & c_{23} c_{31} & c_{32} & c_{33} \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & - 2 2 & 5 & - 4 3 & 7 & - 5 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}$
Therefore, $\begin{matrix} a d j (A) = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 5 & 7 - 2 & - 4 & - 5 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}$
$\begin{matrix} ∴ A^{- 1} = \frac{1}{| A |} . a d j A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 5 & 7 - 2 & - 4 & - 5 \end{matrix} ⎤ ⎥ ⎦ (∵ | A | = 1) \end{matrix}$

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Inverse of the matrix⎡⎢⎣3−2−1−41−1201⎤⎥⎦ is [MP PET 1990]

$\begin{matrix} I n v e r s e o f t h e m a t r i x ⎡ ⎢ ⎣ \begin{matrix} 3 & - 2 & - 1 - 4 & 1 & - 1 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}$ is [MP PET 1990]