MCQ
Let $A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right)$ and $(10)B = \left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right)$. If $B$ is the inverse of matrix $A$, then $\alpha $ is
  • $5$
  • B
    $-1$
  • C
    $2$
  • D
    $-2$

Answer

Correct option: A.
$5$
a
(a) Given, $\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right)\, = \,10{A^{ - 1}}$

$ \Rightarrow $ $\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right)\,\,\left( {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{10}&0&0\\0&{10}&0\\0&0&{10}\end{array}} \right)$

$ \Rightarrow $ $ - 5 + \alpha = 0 \Rightarrow \alpha = 5$

(Equating the element of  $2nd$  row and first column).

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