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# If a matrix is given by $A=\left[ \begin{matrix} i & 0 \\ 0 & i \\\end{matrix} \right]$, then find the value of ${{A}^{2}}$.

Last updated date: 13th Jul 2024
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Hint: For any general 2 x 2 matrix, $M=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]$, ${{M}^{2}}=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]\times \left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+bc & ab+bd \\ ca+dc & cb+{{d}^{2}} \\ \end{matrix} \right]$. So, to find ${{A}^{2}}$, use this formula of ${{M}^{2}}$ by considering a = i, b = 0, c = 0 and d = i.

We are given a 2 x 2 matrix, $A=\left[ \begin{matrix} i & 0 \\ 0 & i \\ \end{matrix} \right]$
Here, we have to find ${{A}^{2}}$.
Let us take the matrix given in the question,
$A=\left[ \begin{matrix} i & 0 \\ 0 & i \\ \end{matrix} \right]$ which is a 2 x 2 matrix as it has 2 rows and 2 columns.
Let us taken the general 2 x 2 matrix, that is
$M=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]$
Now, we know that ${{M}^{2}}=M\times M....\left( i \right)$
By, putting the general 2 x 2 matrix in place of M in the right hand side (RHS) of the equation (i), we get,
${{M}^{2}}={{\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]}_{2\times 2}}\times {{\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]}_{2\times 2}}$
We know that matrix multiplication is carried out by multiplying rows of the first matrix to columns of the second matrix. Therefore, we get above expression as,
${{M}^{2}}=\left[ \begin{matrix} a\times a+b\times c & a\times b+b\times d \\ c\times a+d\times c & c\times b+d\times d \\ \end{matrix} \right].....\left( ii \right)$
By simplifying the above expression, we get,
${{M}^{2}}=\left[ \begin{matrix} {{a}^{2}}+bc & ab+bd \\ ca+dc & cb+{{d}^{2}} \\ \end{matrix} \right]$
Now, we will compare the matrix given in the question, that is $A=\left[ \begin{matrix} i & 0 \\ 0 & i \\ \end{matrix} \right]$ with general 2 x 2 matrix $M=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]$ , so we will get,
\begin{align} & a=i \\ & b=0 \\ & c=0 \\ & d=i \\ \end{align}
Now, to get the value of ${{A}^{2}}$, we will put these values of a, b, c and d in equation (ii). So we will get,
${{A}^{2}}=\left[ \begin{matrix} {{\left( i \right)}^{2}}+0\times 0 & i\times \left( 0 \right)+\left( 0 \right)\times i \\ 0\times i+i\times 0 & 0\times 0+{{\left( i \right)}^{2}} \\ \end{matrix} \right]$
By simplifying the above expression, we get,
${{A}^{2}}=\left[ \begin{matrix} {{i}^{2}} & 0 \\ 0 & {{i}^{2}} \\ \end{matrix} \right]$
As we know that it is an imaginary number and its value is $\sqrt{-1}$. Therefore, we get ${{i}^{2}}=-1$.
Hence, we finally get ${{A}^{2}}$ as $\left[ \begin{matrix} {{i}^{2}} & 0 \\ 0 & {{i}^{2}} \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right]$.

Note: Students must note that to perform the matrix multiplication, the number of columns in the first matrix should be equal to the number of rows in the second matrix. Students should also remember that matrix multiplication is only carried out by multiplying rows of the first matrix by columns of the second matrix unlike in determinant. In determinant, multiplication can be carried out by multiplying row to row and column to column as well. Therefore, students must not confuse between multiplication of two matrices or two determinants. Also, take special care in taking the values of the variables a, b, c and d.