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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 $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{\frac{{ - 1}}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right]$ , then $(BB^TA)^5$ is equal to
  • A
    $\left[ {\begin{array}{*{20}{c}}
    {2 + \sqrt 3 }&1\\
    { - 1}&{2 - \sqrt 3 }
    \end{array}} \right]$
  • B
    $\frac{1}{2}\left[ {\begin{array}{*{20}{c}}
    1&5\\
    0&1
    \end{array}} \right]$
  • ✓
    $\left[ {\begin{array}{*{20}{c}}
    1&5\\
    0&1
    \end{array}} \right]$
  • D
    $\left[ {\begin{array}{*{20}{c}}
    5&1\\
    0&1
    \end{array}} \right]$
✓

Answer

Correct option: C.
$\left[ {\begin{array}{*{20}{c}}
1&5\\
0&1
\end{array}} \right]$
c
$\mathrm{BB}^{\mathrm{T}}=\left[\begin{array}{cc}{\frac{\sqrt{3}}{2}} & {\frac{1}{2}} \\ {-\frac{1}{2}} & {\frac{\sqrt{3}}{2}}\end{array}\right]\left[\begin{array}{cc}{\frac{\sqrt{3}}{2}} & {-\frac{1}{2}} \\ {\frac{1}{2}} & {\frac{\sqrt{3}}{2}}\end{array}\right]$

$=\left[\begin{array}{cc}{\frac{3}{4}+\frac{1}{4}} & {-\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}} \\ {-\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}} & {-\frac{1}{4}+\frac{3}{4}}\end{array}\right]=\left[\begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right]$

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