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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
The element of second row and third column in the inverse of $\left[ {\begin{array}{*{20}{c}}1&2&1\\2&1&0\\{ - 1}&0&1\end{array}} \right]$ is
  • A
    $-2$
  • ✓
    $-1$
  • C
    $1$
  • D
    $2$
✓

Answer

Correct option: B.
$-1$
b
(b)In ${A^{ - 1}},$ the element of  $2^{nd}$  row and $ 3^{rd}$  column is the ${c_{32}}$ element of the matrix $({c_{ij}})$ of cofactors of element of $Adj\,(A) = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&4\\4&1&{ - 2}\\{ - 2}&4&1\end{array}} \right]$, (due to transposition) divided by $\Delta = \,|A|\, = - 2$.
$\therefore $ Required element = $\frac{{{{( - 1)}^{3 + 2}}{M_{32}}}}{{ - 2}}\,{\rm{ }} = \frac{{ - ( - 2)}}{{ - 2}} = - 1$,
where ${M_{32}} = $minor of ${c_{32}}$ in $A = \left[ {\begin{array}{*{20}{c}}
  1&1 \\ 
  2&0 
\end{array}} \right] = 0 - 2 =  - 2$

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