Question

The inverse of the matrix $\left[ {\begin{array}{*{20}{l}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ is
A. $\left[ {\begin{array}{*{20}{l}}0&0&1\\0&1&0\\1&0&0\end{array}} \right]$
В. $\left[ {\begin{array}{*{20}{l}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{l}}0&1&0\\0&0&1\\1&0&0\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{l}}1&0&0\\0&0&1\end{array}} \right]$

Answer 1

Hint: Using the knowledge that multiplying a matrix by its inverse yields the identity matrix, we begin the problem-solving process. We write in the matrices in the specified order using the identity matrix definition. In our case, we are given an identity matrix. First we have to find the adjacent matrix and then we have to find the transpose of the adjacent matrix to get the desired answer.

Formula Used: ${A^{ - 1}} = \dfrac{{adj(A)}}{{|A|}};|A| \ne 0$


Complete step by step solution: We are provided the matrix in the question that
$\left[ {\begin{array}{*{20}{l}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
And we are asked to determine the inverse of the given matrix
Now, we have to determine the cofactors of the given matrix, we have
${C_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right| = 1$
${C_{12}} = {( - 1)^{1/2}}\left| {\begin{array}{*{20}{l}}0&0\\0&1\end{array}} \right| = 0$
${C_{13}} = {( - 1)^{1/3}}\left| {\begin{array}{*{20}{l}}0&1\\0&0\end{array}} \right| = 0$
$4{C_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{*{20}{l}}0&0\\0&1\end{array}} \right| = 04$
${C_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right| = 1$
${C_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{*{20}{l}}1&0\\0&0\end{array}} \right| = 0$
${C_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{*{20}{l}}0&0\\1&0\end{array}} \right| = 0$
${C_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{*{20}{l}}1&0\\0&0\end{array}} \right| = 0$
${C_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right| = 1$
Now, we have to determine the transpose of the cofactor matrix, we get
Here, columns change into rows, and rows change into columns.
Therefore, the transpose matrix is
$\left[ {\begin{array}{*{20}{l}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
Now, we have to add the identity matrix to it and run row operations to try to move the identity matrix to the left in order to discover the inverse matrix. The inverse matrix will then appear to the right.
But an identity matrix is already present in the matrix to the left. Thus, the inverse matrix also qualifies as an identity matrix.
So, on turning the left matrix to right matrix we get
$\left[ {\begin{array}{*{20}{l}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
Therefore, the inverse of the matrix $\left[ {\begin{array}{*{20}{l}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ is $\left[ {\begin{array}{*{20}{l}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$


Option ‘B’ is correct

Note: Although it is a challenging task, it is possible to estimate the inverse of a 3 by 3 matrix by doing certain procedures. Even though it could appear challenging, the best way to get beyond it is to repeatedly use a sample problem to solve the question. So, student should follow the correct concept to determine the desired solution.