Matrices And Determinants Ques 31
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
(2003, 1M)
(a) $1$
(b) $-1$
(c) $4$
(d) no real values
Show Answer
Answer:
Correct Answer: 31.(d)
Solution:
Formula:
Properties of matrix multiplication:
- $ \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.
BETA
