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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
Let $A =\left[\begin{array}{cc}1 & -1 \\ 2 & \alpha\end{array}\right]$ and $B =\left[\begin{array}{ll}\beta & 1 \\ 1 & 0\end{array}\right], \alpha, \beta \in R$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $( A + B )^{2}= A ^{2}+\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $( A + B )^{2}= B ^{2}$. Then $\left|\alpha_{1}-\alpha_{2}\right|$ is equal to.
  • ✓
    $2$
  • B
    $22$
  • C
    $3$
  • D
    $8$
✓

Answer

Correct option: A.
$2$
a
$A+B=\left[\begin{array}{cc}\beta+1 & 0 \\ 3 & \alpha\end{array}\right]$

$(A+B)^{2}=\left[\begin{array}{cc}\beta+1 & 0 \\ 3 & \alpha\end{array}\right]\left[\begin{array}{cc}\beta+1 & 0 \\ 3 & \alpha\end{array}\right]$

$=\left[\begin{array}{cc}(\beta+1)^{2} & 0 \\ 3(\beta+1)+3 \alpha & \alpha^{2}\end{array}\right]$

$A^{2}=\left[\begin{array}{cc}1 & -1 \\ 2 & \alpha\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & \alpha\end{array}\right]$

$=\left[\begin{array}{cc}-1 & -1-\alpha \\ 2+2 \alpha & \alpha^{2}-2\end{array}\right]$

$\therefore\left[\begin{array}{cc}1 & -\alpha+1 \\ 2 \alpha+4 & \alpha^{2}\end{array}\right]=\left[\begin{array}{cc}(\beta+1)^{2} & 0 \\ 3(\alpha+\beta+1) & \alpha^{2}\end{array}\right]$

$\alpha=1]=\alpha_{1}$

$B ^{2}=\left[\begin{array}{ll}\beta & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}\beta & 1 \\ 1 & 0\end{array}\right]$

$=\left[\begin{array}{cc}\beta^{2}+1 & \beta \\ \beta & 1\end{array}\right]=\left[\begin{array}{cc}(\beta+1)^{2} & 0 \\ 3(\beta+1)+3 \alpha & \alpha^{2}\end{array}\right]$

$\therefore \beta=0, \alpha=-1=\alpha_{2}$

$\left|\alpha_{1}-\alpha_{2}\right|=|1-(-1)|=2$

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