Question

first and perfect answer will be given brainliest answer​

Answer 1

Given,

$\displaystyle\longrightarrow\sf {A=\left [\begin {array}{ccc}1&-1&1\\&&\\2&1&-3\\&&\\1&1&1\end {array}\right]}$

$\displaystyle\longrightarrow\sf {10B=\left [\begin {array}{ccc}4&2&2\\&&\\-5&0&\alpha\\&&\\1&-2&3\end {array}\right]}$

Given that B is the inverse of A, i.e.,

$\displaystyle\longrightarrow\sf {AB=I}$

On multiplying 10 on both sides,

$\displaystyle\longrightarrow\sf {10AB=10I}$

$\displaystyle\longrightarrow\sf {A\cdot10B=10I}$

$\displaystyle\longrightarrow\sf {\left [\begin {array}{ccc}1&-1&1\\&&\\2&1&-3\\&&\\1&1&1\end {array}\right]\cdot\left [\begin {array}{ccc}4&2&2\\&&\\-5&0&\alpha\\&&\\1&-2&3\end {array}\right]=\left [\begin {array}{ccc}10&0&0\\&&\\0&10&0\\&&\\0&0&10\end {array}\right]}$

So we can say that using first row of A and third column of B we obtain $\displaystyle\sf {I_{13}=0,}$ i.e.,

$\displaystyle\longrightarrow\sf {A_{11}\cdot(10B)_{13}+A_{12}\cdot(10B)_{23}+A_{13}\cdot (10B)_{33}=I_{13}}$

$\displaystyle\longrightarrow\sf {1\times2+(-1)\alpha+1\times3=0}$

$\displaystyle\longrightarrow\sf {2-\alpha+3=0}$

$\displaystyle\longrightarrow\sf {5-\alpha=0}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\alpha=5}}}$