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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 in the first row and third column of the inverse of the matrix $\left[ {\begin{array}{*{20}{c}}1&2&{ - 3}\\0&1&2\\0&0&1\end{array}} \right]$ is
  • A
    $-2$
  • B
    $0$
  • C
    $1$
  • ✓
    $7$
✓

Answer

Correct option: D.
$7$
d
(d) Let $A = \left[ {\,\begin{array}{*{20}{c}}1&2&{ - 3}\\0&1&2\\0&0&1\end{array}\,} \right]$ ==>$ |A| = 1$

$adj\,(A) = {\left[ {\begin{array}{*{20}{c}}
1{2\,\,\,}{ - 1}\\
{ - 2}11\\
{7\,\,}{ - 2}1
\end{array}} \right]^T}$

Hence, ${A^{ - 1}} = \frac{{adj\,(A)}}{{|A|}}$

==> ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&7\\2&1&{ - 2}\\{ - 1}&1&1\end{array}} \right]$ .

Hence, element $A_{13} = 7$.

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