Question

Find the inverse of $A=\left[ \begin{matrix} \cos \theta & -\sin \theta & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$
(i) By elementary row transformation
(ii) By elementary column transformation

Answer 1

Firstly, we have to check whether the inverse of the given matrix exists or not. For this, we have to find the determinant of the matrix A and check whether it is equal to 0 or not. If the determinant is not equal to 0, then the inverse exists. Else, the inverse does not exist. Then, we have to perform elementary row operations on the matrices that we will get by substituting the given matrix and identity matrix in the property $A{{A}^{-1}}=I$ . The elementary row operation has to be done such that we have to transform the matrix A into an identity matrix. Then, the inverse of A will be the RHS of this equation. Similarly, we have to perform column operations.

Complete step by step answer:
We have to find the inverse of $A=\left[ \begin{matrix} \cos \theta & -\sin \theta & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ by elementary row and column transformation. Firstly, we have to check whether the inverse of the matrix A exists or not. For this, we have to find the determinant of the matrix A. We know that for a matrix $A=\left[ \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right]$ , the determinant is given by
$\Delta =\left| A \right|=\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=a\left( ei-hf \right)-b\left( di-gf \right)+c\left( dh-ef \right)$
Therefore, we can find the determinant of the given matrix as follows.

& \Rightarrow \left| A \right|=\left| \begin{matrix} \cos \theta & -\sin \theta & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right|=\cos \theta \left( \cos \theta \times 1-0\times 0 \right)-\left( -\sin \theta \right)\left( \sin \theta \times 1-0\times 0 \right)+0\left( \sin \theta \times 0-0\times \cos \theta \right) \\\ & \Rightarrow \left| A \right|={{\cos }^{2}}\theta +{{\sin }^{2}}\theta +0 \\\ & \Rightarrow \left| A \right|={{\cos }^{2}}\theta +{{\sin }^{2}}\theta \\\ \end{aligned}$$ We know that $${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$$ . Therefore, the above determinant becomes $$\Rightarrow \left| A \right|=1\ne 0$$ We can see that the determinant is not equal to 0. Hence, ${{A}^{-1}}$ or the inverse of A exists. Now, let us find the inverse by elementary row transformation. (i) Let us consider $A{{A}^{-1}}=I$ . Let us substitute the values in this property. We know that $I=\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ . Therefore, we can write $\Rightarrow \left[ \begin{matrix} \cos \theta & -\sin \theta & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ To find ${{A}^{-1}}$ we have to convert the matrix A in the above equation to identity matrix, I. Let us multiply the first row by $\cos \theta $ . $\Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & -\cos \theta \sin \theta & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{R}_{1}}\to {{R}_{1}}\cos \theta $ Now, we have to transform the first row by multiplying the second row by $\sin \theta $ and add this with the first row. $\begin{aligned} & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & -\cos \theta \sin \theta +\cos \theta \sin \theta & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\sin \theta \\\ & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & 0 & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right] \\\ \end{aligned}$ We know that $${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$$ . Therefore, the above matrix becomes $\Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ Now, we have to transform the second row by subtracting the product of first row and $\sin \theta $ from the second row. $\begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ \sin \theta -\sin \theta & \cos \theta -0 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ -\cos \theta \sin \theta & 1-{{\sin }^{2}}\theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\sin \theta \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ -\cos \theta \sin \theta & 1-{{\sin }^{2}}\theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]...\left( i \right) \\\ \end{aligned}$ We know that $$\begin{aligned} & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\\ & \Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\\ \end{aligned}$$ Let us substitute this result in the equation (i). $\Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ -\cos \theta \sin \theta & {{\cos }^{2}}\theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ Now, we have to multiply the second row by $\dfrac{1}{\cos \theta }$ . $\begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & \dfrac{\cos \theta }{\cos \theta } & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ \dfrac{-\cos \theta \sin \theta }{\cos \theta } & \dfrac{{{\cos }^{2}}\theta }{\cos \theta } & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{R}_{2}}\to \left( \dfrac{1}{\cos \theta } \right){{R}_{2}} \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ -\sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right] \\\ \end{aligned}$ Therefore, the inverse of matrix A, ${{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ -\sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ . Now, let us find the inverse of matrix A using elementary column transformation. (ii) Let us consider $A{{A}^{-1}}=I$ . Let us substitute the values in this property. We know that $I=\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ . Therefore, we can write $\Rightarrow \left[ \begin{matrix} \cos \theta & -\sin \theta & 0 \\\ \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ We have to multiply the first column by $\cos \theta $ . $\Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & -\sin \theta & 0 \\\ \cos \theta \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{C}_{1}}\to {{C}_{1}}\cos \theta $ Now, we have to transform column one by subtracting the product of the second column and $\sin \theta $ from the first column. $\begin{aligned} & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & -\sin \theta & 0 \\\ \cos \theta \sin \theta -\cos \theta \sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & 0 & 0 \\\ -\sin \theta & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{C}_{1}}\to {{C}_{1}}-{{C}_{2}}\sin \theta \\\ & \Rightarrow \left[ \begin{matrix} 1 & -\sin \theta & 0 \\\ 0 & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & 0 & 0 \\\ -\sin \theta & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right] \\\ \end{aligned}$ We have to transform the second column by adding the second column and the product of first column and $\sin \theta $ . $\begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & -\sin \theta +\sin \theta & 0 \\\ 0 & \cos \theta +0 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta \cos \theta & 0 \\\ -\sin \theta & 1-{{\sin }^{2}}\theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{C}_{2}}\to {{C}_{2}}+{{C}_{1}}\sin \theta \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta \cos \theta & 0 \\\ -\sin \theta & {{\cos }^{2}}\theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right] \\\ \end{aligned}$ Now, we have to multiply the second column by $\dfrac{1}{\cos \theta }$ . $$\begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & \dfrac{\cos \theta }{\cos \theta } & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \dfrac{\sin \theta \cos \theta }{\cos \theta } & 0 \\\ -\sin \theta & \dfrac{{{\cos }^{2}}\theta }{\cos \theta } & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]\text{ }{{C}_{2}}\to {{C}_{2}}\left( \dfrac{1}{\cos \theta } \right) \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ -\sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right] \\\ \end{aligned}$$ Therefore, the inverse of matrix A, ${{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta & 0 \\\ -\sin \theta & \cos \theta & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$ . **Note:** Students must be through with the properties of matrices and determinants. They must deeply learn how the row and column operations are performed. They can perform either row or column operation but not both. Students must note that the inverse of a matrix exists only if its determinant is a non-zero value.