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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
If $A =$ $\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&a&1\end{array}} \right]$ ,$A^{-1} =$$\left[ {\begin{array}{*{20}{c}}{1/2}&{ - 1/2}&{1/2}\\{ - 4}&3&c\\{5/2}&{ - 3/2}&{1/2}\end{array}} \right]$, then
  • ✓
    $a = 1, c = - 1$
  • B
    $a = 2, c = - \frac{1}{2}$
  • C
    $a = - 1, c = 1$
  • D
    $a = \frac{1}{2}$, $c =\frac{1}{2}$
✓

Answer

Correct option: A.
$a = 1, c = - 1$
a
$| A | = 2 (a - 2)$ ==> $a \ne 2$

cofactor of $0$ in $| A |$ is $2 - 3a$. According to value of $A^{-1}$,

$\frac{{2 - 3a}}{{|A|}}$ $=$ $\frac{1}{2}$

==>$\frac{{2 - 3a}}{{2(a - 2)}}$ $=$ $\frac{1}{2}$

==> $2 - 3a = a - 2$

==> $a = 1$

Again $c =$ $\frac{{cofactor\,\,of\,\,a\,\,in\,\,|A|}}{{|A|}}$

$=$ $\frac{{\left| {\,\begin{array}{*{20}{c}}0&2\\1&3\end{array}\,} \right|}}{{2(a - 2)}}$

$=$ $\frac{2}{{2(1 - 2)}}$

$=$ $- 1$

Alternative : $AA^{-1} = I$

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