Question

If ${I_3}$ is the identity matrix of order $3$, then ${\left( {{I_3}} \right)^{ - 1}}$ is
A. 0
B. $3{I_3}$
C. ${I_3}$
D. Not necessarily exists.

Answer 1

Hint:
We begin by writing the identity matrix of order 3 and then find its determinant to check if the inverse exists or not. Next, find the cofactors of ${I_3}$. Use the cofactors to write the adjoint of the given matrix. Next, determine the inverse by using the formula, ${A^{ - 1}} = \dfrac{{Adj\left( A \right)}}{{\left| A \right|}}$.

Complete step by step solution:
We will first write the identity matrix of order 3.
${I_3} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
We will find the inverse of the above matrix.
We shall now find the determinant of the above matrix to check if the inverse exists or not.
We will expand the determinant along the first row of the given matrix.
$
  \left| {{I_3}} \right| = 1\left( {1 - 0} \right) - 0\left( {0 - 0} \right) + 0\left( {0 - 0} \right) \\
   \Rightarrow \left| {{I_3}} \right| = 1
$
Since, the determinant is non-zero, the inverse of the matrix exists.
Now, we will find the cofactors of ${I_3}$
${a_{ij}} = {\left( { - 1} \right)^{i + j}}\left( \text{Minor of }{a_{ij}} \right)$
${a_{11}} = {\left( { - 1} \right)^{1 + 1}}\left( {1 - 0} \right) = 1$
${a_{12}} = {\left( { - 1} \right)^{1 + 2}}\left( {0 - 0} \right) = 0$
${a_{13}} = {\left( { - 1} \right)^{1 + 3}}\left( {0 - 0} \right) = 0$
${a_{21}} = {\left( { - 1} \right)^{1 + 2}}\left( {0 - 0} \right) = 0$
${a_{22}} = {\left( { - 1} \right)^{2 + 2}}\left( {1 - 0} \right) = 1$
${a_{23}} = {\left( { - 1} \right)^{3 + 2}}\left( {0 - 0} \right) = 0$
${a_{31}} = {\left( { - 1} \right)^{1 + 3}}\left( {0 - 0} \right) = 0$
${a_{32}} = {\left( { - 1} \right)^{3 + 2}}\left( {0 - 0} \right) = 0$
${a_{33}} = {\left( { - 1} \right)^{3 + 3}}\left( {1 - 0} \right) = 1$
Now, we will write the co-factors in a matrix which will be the adjoint of the matrix.
$Adj\left( {{I_3}} \right) = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
Now, we know that the inverse of the matrix is $\dfrac{{Adj\left( A \right)}}{{\left| A \right|}}$
On substituting the values of determinant and adjoint of ${I_3}$, we will get,
${\left( {{I_3}} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = {I_3}$

Hence, option C is correct.

Note:
If the determinant is zero, then we say the matrix is a singular matrix and its inverse does not exist and if the determinant is zero, then the matrix is a non-singular matrix. Also, the inverse of any identity matrix of any order is the same as the identity matrix of that order.