Question

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

Answer 1

Hint: In this Question, we have to find the value of $\alpha$ in the given matrix. To find the value of $\alpha$, first of all, we have to multiply matrix A by itself as given in the question and then that required matrix has to be equated with matrix B which will give the value of $\alpha$.

Formula Used: A matrix is denoted by $A=\left[ {{a}_{ij}} \right]$ where $i$ represents rows and $j$ represents columns
Then, the multiplication of two $2\times 2$ matrices is
If A=$\left[ {{a}_{ij}} \right]$ is an $m\times n$ matrix and B=$\left[ {{b}_{ij}} \right]$ is $n\times p$ matrix, the product $AB$ becomes an $m\times p$ matrix
Consider matrix
A=$\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
and matrix
B=$\left( \begin{matrix}
   {{b}_{11}} & {{b}_{12}} \\
   {{b}_{21}} & {{b}_{22}} \\
\end{matrix} \right)$
Then the product of these two matrices is written as
$\left( \begin{matrix}
   {{a}_{11}}{{b}_{11}}+{{a}_{12}}{{b}_{21}} & {{a}_{11}}{{b}_{12}}+{{a}_{12}}{{b}_{22}} \\
   {{a}_{21}}{{b}_{11}}+{{a}_{22}}{{b}_{21}} & {{a}_{21}}{{b}_{12}}+{{a}_{22}}{{b}_{22}} \\
\end{matrix} \right)$

Complete step by step solution: The given two matrices are
A= $\left( \begin{matrix}
   \alpha & 0 \\
   1 & 1 \\
\end{matrix} \right)$
and
B= $\left( \begin{matrix}
   1 & 0 \\
   5 & 1 \\
\end{matrix} \right)$
For finding the square of a matrix, we have to multiply each element of row of Matrix A to each element of column of Matrix A.
\begin{align}
  & {{A}^{2}}=\left( \begin{matrix}
   \alpha \times \alpha +0\times 1 & \alpha \times 0+0\times 1 \\
   1\times \alpha +1\times 1 & 1\times 0+1\times 1 \\
\end{matrix} \right) \\
 & \text{ }=\left( \begin{matrix}
   {{\alpha }^{2}}+0 & 0+0 \\
   \alpha +1 & 0+1 \\
\end{matrix} \right) \\
 & \text{ }=\left( \begin{matrix}
   {{\alpha }^{2}} & 0 \\
   \alpha +1 & 1 \\
\end{matrix} \right) \\
\end{align}
It is given that, ${{A}^{2}}=B$
Thus,
\begin{align}
  & {{A}^{2}}=B \\
 & \Rightarrow \left( \begin{matrix}
   {{\alpha }^{2}} & 0 \\
   \alpha +1 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 \\
   5 & 1 \\
\end{matrix} \right) \\
\end{align}
Since both the matrices have same number of rows and columns, we can equate each element of the two matrices. I.e.,
${{\alpha }^{2}}=1;\alpha +1=5$
On simplifying these two expressions, we get $\alpha =4,1,-1$.
But \alpha should be one value as in the place of an element we can put only one number. Therefore, we can see clearly that there is no real value of \alpha.

Option ‘D’ is correct

Note: Here, we may go wrong with the multiplication of a matrix with itself. We have to keep in mind that we have to multiply the row of Matrix A by the Column of matrix A. If two matrixes are of the same order, then we can able to equate the respective elements of both the matrices on either side of the equal symbol.