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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
Let $A$ is a symmetric and $ \,B$ is a skew symmetric matrix, such that  $A - B = \left[ {\begin{array}{*{20}{c}}
  1&2 \\ 
  3&4 
\end{array}} \right]$, then $|A|$ is
  • A
    $-\frac{3}{4}$
  • B
    $-\frac{1}{4}$
  • C
    $-\frac{11}{4}$
  • ✓
    $-\frac{9}{4}$
✓

Answer

Correct option: D.
$-\frac{9}{4}$
d
$A=A^{T}, B=-B^{T}$

$A-B=\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right]$         ........$(i)$

$\Rightarrow A^{T}+B^{T}=\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right] $

$\Rightarrow(A+B)^{T}=\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right]$

$ \Rightarrow A + B = \left[ {\begin{array}{*{20}{l}}
1&3\\
2&4
\end{array}} \right]$         ......$(ii)$

Adding $(i)$ and $(ii)$ $A=\left[\begin{array}{ll}{1} & {\frac{5}{2}} \\ {\frac{5}{2}} & {4}\end{array}\right]$

$\therefore|A|=4-\frac{25}{4}=\frac{-9}{4}$

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