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# If $A=\begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}$ and $B=\begin{vmatrix}1 &0\\5 &1\end{vmatrix}$then the value of $\alpha$ for which $A^2=B$ is

$(a)\;1\qquad(b)\;-1\qquad(c)\;4\qquad(d)\;No\;real\;value$

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## 1 Answer

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Given that :
$A=\begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}$ and $B=\begin{vmatrix}1 &0\\5 &1\end{vmatrix}$
$A^2=B$
$\Rightarrow \begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}\begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}=\begin{vmatrix}1 &0\\5 &1\end{vmatrix}$
$\Rightarrow \begin{vmatrix}\alpha^2&0\\\alpha+1&1\end{vmatrix}=\begin{vmatrix}1 &0\\5&1\end{vmatrix}$
$\Rightarrow \alpha^2=1$ and $\alpha+1=5$
$\Rightarrow \alpha=\pm 1$ and $\alpha=4$
There is no common value.
There is no real value of $\alpha$ for which $A^2=B$
Hence (d) is the correct answer.
answered Nov 20, 2013

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