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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
Let $d \in R$, and  $A = \left[ {\begin{array}{*{20}{c}} { - 2}&{4 + d}&{\left( {\sin \,\theta } \right) - 2}\\ 1&{\left( {\sin \,\theta } \right) + 2}&d\\ 5&{\left( {2\sin \,\theta } \right) - d}&{\left( { - \sin \,\theta } \right) + 2 + 2d} \end{array}} \right]$, $\theta  \in \left[ {0,2\pi } \right]$. If the minimum value of det $(A)$ is $8$, then a value of $d$ is
  • ✓
    $-5$
  • B
    $-7$
  • C
    $2\left( {\sqrt 2  + 1} \right)$
  • D
    $2\left( {\sqrt 2  + 2} \right)$
✓

Answer

Correct option: A.
$-5$
a
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
{ - 2}&{4 + d}&{\left( {\sin \theta  - 2} \right)}\\
1&{\left( {\sin \theta } \right) + 2}&d\\
5&{\left( {2\sin \theta } \right) - d}&{\left( { - \sin \theta } \right) + 2 + 2d}
\end{array}} \right|$

$ = \left| {\begin{array}{*{20}{c}}
{ - 2}&{4 + d}&{\left( {\sin \theta  - 2} \right)}\\
1&{\left( {\sin \theta } \right)}&d\\
1&0&0
\end{array}} \right|$            (New ${R_3} = {R_3} - 2{R_2} + {R_1}$)

$ = \left( {4 + d} \right)d - {\sin ^2}\theta  + 4 = {\left( {d + 2} \right)^2} - {\sin ^2}\theta $

Because minimum value of $\left| A \right| = 8 \Rightarrow {\left( {d + 2} \right)^2} = 9 \Rightarrow d = 1$ or $-5$

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