Matrices and Determinants 1 Question 9
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9. If $A=\begin{array}{cc}\alpha & 0 \ 1 & 1\end{array}$ and $B=\begin{array}{cc}1 & 0 \ 5 & 1\end{array}$, then value of $\alpha$ for which $A^{2}=B$, is
======= ####9. If $A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]$ then value of $\alpha$ for which $A^2=B$, is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed
(2003, 1M)
(a) 1
(b) -1
(c) 4
(d) no real values
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Answer:
Correct Answer: 9. (d)
Solution:
- $ \begin{array}{ll} \text { Given, } & A=\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right], B=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right] \\ \Rightarrow & A^2=\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} \alpha^2 & 0 \\ \alpha+1 & 1 \end{array}\right] \end{array} $
Also, given, $A^2=B$
$ \begin{aligned} & \Rightarrow\left[\begin{array}{ll} \alpha^2 & 0 \\ \alpha+1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right] \\ & \Rightarrow \alpha^2=1 \text { and } \alpha+1=5 \end{aligned} $
Which is not possible at the same time.
$\therefore$ No real values of $\alpha$ exists.
