If A - lbegin bmatrix 3 - 411-1 - 2 0 - 1 e n d bmatrix and B - begin bmatrix -1 - 2 - 1182 - 3 1lend bmatrix - then verify that i A - B - A - B -ii A - B - A - B -

∴ A=(A')'=[ 3 amp; 4; 1 amp; 2; 0 amp; 1 ]=[ 3 amp; 1 amp; 0; 4 amp; 2 amp; 1; ] Here, =A+B =[ 3 amp; 1 amp; 0; ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Matrix

If A '=. lbeg...

Question

If $A^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 - 1 & 2 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ and $B = [\begin{matrix} - 1 & 2 & 1 1 & 2 & 3 \end{matrix}]$ , then verify that
(i)(A+B)'=A'+B'

(ii)(A-B)'=A'-B'

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Solution

$∴ A = (A^{'})^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 - 1 & 2 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = [\begin{matrix} 3 & - 1 & 0 4 & 2 & 1 \end{matrix}]$
Here, $= A + B = [\begin{matrix} 3 & - 1 & 0 4 & 2 & 1 \end{matrix}] + [\begin{matrix} - 1 & 2 & 1 1 & 2 & 3 \end{matrix}] = [\begin{matrix} 2 & 1 & 1 5 & 4 & 4 \end{matrix}]$
$∴ L H S = (A + B)^{'} = [\begin{matrix} 2 & 1 & 1 5 & 4 & 4 \end{matrix}] = ⎡ ⎢ ⎣ \begin{matrix} 2 & 5 1 & 4 1 & 4 \end{matrix} ⎤ ⎥ ⎦ B^{'} = [\begin{matrix} - 1 & 2 & 1 1 & 2 & 3 \end{matrix}] = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 2 & 2 1 & 3 \end{matrix} ⎤ ⎥ ⎦, a l s o A^{'} ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 - 1 & 2 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
Here, $R H S = A^{'} + B^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 - 1 & 2 0 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 2 & 2 1 & 3 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 2 & 5 1 & 4 1 & 4 \end{matrix} ⎤ ⎥ ⎦$
So, verfied that (A+B)'=A'+B'.

$∴ A = (A^{'})^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 - 1 & 2 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = [\begin{matrix} 3 & - 1 & 0 4 & 2 & 1 \end{matrix}]$
Here, $R H S = A^{'} - B^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 - 1 & 2 0 & 1 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} - 1 & 1 2 & 2 1 & 3 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 3 + 1 & 4 - 1 - 1 - 2 & 2 - 2 0 - 1 & 1 - 3 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 4 & 3 - 3 & 0 - 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$
Also $(A - B)^{'} = {([\begin{matrix} 3 & - 1 & 0 4 & 2 & 1 \end{matrix}] - [\begin{matrix} - 1 & 2 & 1 1 & 2 & 3 \end{matrix}])}^{'} (∵ A = (A^{'})^{'}) = [\begin{matrix} 3 + 1 & - 1 - 2 & 0 - 1 4 - 1 & 2 - 2 & 1 - 3 \end{matrix}] = {[\begin{matrix} 4 & - 3 & - 1 3 & 0 & - 2 \end{matrix}]}^{'} = ⎡ ⎢ ⎣ \begin{matrix} 4 & 3 - 3 & 0 - 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$
So, verified that (A-B)'=A'-B'

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