Question

If $ A=\left[ \begin{matrix}
   \alpha & 0 \\
   1 & 1 \\
\end{matrix} \right] $ and $ B=\left[ \begin{matrix}
   1 & 0 \\
   5 & 1 \\
\end{matrix} \right] $ then value of $ \alpha $ for which $ {{A}^{2}}=B $ is
A. 1
B. -1
C. 4
D. no real values

Answer 1

Hint: To solve this question we will use the given condition $ {{A}^{2}}=B $ . First, we will calculate the value of $ {{A}^{2}} $ and then equate it to the value of B. Then by comparing both we will find the value of $ \alpha $ and then according to the value obtained we will choose the correct option.

Complete step by step answer:
We have been given that $ A=\left[ \begin{matrix}
   \alpha & 0 \\
   1 & 1 \\
\end{matrix} \right] $ and $ B=\left[ \begin{matrix}
   1 & 0 \\
   5 & 1 \\
\end{matrix} \right] $ and $ {{A}^{2}}=B $ .
We have to find the value of $ \alpha $ for which $ {{A}^{2}}=B $ .
Let us consider $ {{A}^{2}}=B $ .
Now, we have to find the value of $ {{A}^{2}} $
Now, we know that $ {{A}^{2}}=A\times A $ and we have $ A=\left[ \begin{matrix}
   \alpha & 0 \\
   1 & 1 \\
\end{matrix} \right] $
Now, substituting the value and solving further we get
 $ {{A}^{2}}=\left[ \begin{matrix}
   \alpha & 0 \\
   1 & 1 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   \alpha & 0 \\
   1 & 1 \\
\end{matrix} \right] $
As it is a $ 2\times 2 $ matrix the multiplication is done in the following manner:
 $ \Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   \alpha \times \alpha & 0 \\
   \alpha +1 & 1 \\
\end{matrix} \right] $
Now, simplifying further we get
 $ \Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   {{\alpha }^{2}} & 0 \\
   \alpha +1 & 1 \\
\end{matrix} \right] $
Now, we have given that $ {{A}^{2}}=B $
Substituting the values we get
 $ \Rightarrow \left[ \begin{matrix}
   {{\alpha }^{2}} & 0 \\
   \alpha +1 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   5 & 1 \\
\end{matrix} \right] $
Now, comparing both the sides we get
 $ \Rightarrow {{\alpha }^{2}}=1 $ and $ \alpha +1=5 $
Now, solving both the equations we get
 $ \Rightarrow \alpha =\pm 1 $ and $ \alpha =4 $
But different values of $ \alpha $ are not possible at the same time. It means $ \alpha $ has no real values.
So the correct answer is option D.

Note:
 Students must have knowledge of matrix multiplication to solve this question. Students may choose option C as the correct answer or may choose multiple correct answers. But it leads to the wrong answer. So, be careful while choosing the option.