Find the reciprocal or inverse of the matrix M - - 0 1 1- 1 0 1- 1 1 0 - and the transform of the matrix A - 1-2- b-c c-a b-a- c-b c-a a-b- b-c a-c a-b - by MI-e- MAM-1 is a -

The correct option is C [ 1/2 1/2 1/2; 1/2 1/2 1/2; 1/2 1/2 1/2 ] , Diagonal Matrix M = [ 0 amp; 1 amp; 1; 1 amp; 0 amp; 1; 1 ...

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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

Find the reci...

Question

Find the reciprocal (or inverse) of the matrix
$M = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$ and the transform of the matrix
$A = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} b + c & c - a & b - a c - b & c + a & a - b b - c & a - c & a + b \end{matrix} ⎤ ⎥ ⎦$ by $M$
I.e. $M A M^{- 1}$ is a ___________.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, Diagonal Matrix

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⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, Assymetric Matrix

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⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, Diagonal Matrix

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None of these.

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Solution

The correct option is C $⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$ , Diagonal Matrix
$M = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$
$| M | = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ = 2$
$A d j M = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 & 1 1 & - 1 & 1 1 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow M^{- 1} = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 & 1 1 & - 1 & 1 1 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎦$
$M A = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} b + c & c - a & b - a c - b & c + a & a - b b - c & a - c & a + b \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 0 & a & a b & 0 & b c & c & 0 \end{matrix} ⎤ ⎥ ⎦$
$M A M^{- 1} = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} 0 & a & a b & 0 & b c & c & 0 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 & 1 1 & - 1 & 1 1 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} a & 0 & 0 0 & b & 0 0 & 0 & c \end{matrix} ⎤ ⎥ ⎦$
$∴ M A M^{- 1}$ is a digonal matrix.
Hence, option C.

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Similar questions

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Find the reciprocal (or inverse) of the matrixM=⎡⎢⎣011101110⎤⎥⎦ and the transform of the matrix A=12⎡⎢⎣b+cc−ab−ac−bc+aa−bb−ca−ca+b⎤⎥⎦ by MI.e. MAM−1 is a ___________.