Question-2

Find the inverse of the following matrix:

\[ A=\begin{bmatrix}2 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & -1 \end{bmatrix} \]

---

Method-1

We reduce \(A\) to its RREF while simultaneously applying these operations to the identity matrix on the right:

\[ \begin{aligned} \begin{bmatrix}2 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & -1 \end{bmatrix} & & \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\\ \\\begin{bmatrix}0 & -1 & 2\\ 1 & 0 & -1\\ 0 & 1 & -1 \end{bmatrix} & & \begin{bmatrix}1 & -2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\\ \\\begin{bmatrix}1 & 0 & -1\\ 0 & -1 & 2\\ 0 & 1 & -1 \end{bmatrix} & & \begin{bmatrix}0 & 1 & 0\\ 1 & -2 & 0\\ 0 & 0 & 1 \end{bmatrix}\\ \\\begin{bmatrix}1 & 0 & -1\\ 0 & -1 & 2\\ 0 & 0 & 1 \end{bmatrix} & & \begin{bmatrix}0 & 1 & 0\\ 1 & -2 & 0\\ 1 & -2 & 1 \end{bmatrix}\\ \\\begin{bmatrix}1 & 0 & -1\\ 0 & 1 & -2\\ 0 & 0 & 1 \end{bmatrix} & & \begin{bmatrix}0 & 1 & 0\\ -1 & 2 & 0\\ 1 & -2 & 1 \end{bmatrix}\\ \\\begin{bmatrix}1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} & & \begin{bmatrix}0 & 1 & 0\\ 1 & -2 & 2\\ 1 & -2 & 1 \end{bmatrix}\\ \\\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} & & \begin{bmatrix}1 & -1 & 1\\ 1 & -2 & 2\\ 1 & -2 & 1 \end{bmatrix} \end{aligned} \]

We have:

\[ A^{-1}=\begin{bmatrix}1 & -1 & 1\\ 1 & -2 & 2\\ 1 & -2 & 1 \end{bmatrix} \]

Method-2

Alternatively, we have the cofactor matrix. For the sake of convenience, let us write down \(A\) again:

\[ A=\begin{bmatrix}2 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & -1 \end{bmatrix} \]

First for the nine minors:

\[ \begin{aligned} M_{11} & =\begin{vmatrix}0 & -1\\ 1 & -1 \end{vmatrix} & M_{12} & =\begin{vmatrix}1 & -1\\ 0 & -1 \end{vmatrix} & M_{13} & =\begin{vmatrix}1 & 0\\ 0 & 1 \end{vmatrix}\\ & =1 & & =-1 & & =1\\ \\M_{21} & =\begin{vmatrix}-1 & 0\\ 1 & -1 \end{vmatrix} & M_{22} & =\begin{vmatrix}2 & 0\\ 0 & -1 \end{vmatrix} & M_{23} & =\begin{vmatrix}2 & -1\\ 0 & 1 \end{vmatrix}\\ & =1 & & =-2 & & =2\\ \\M_{31} & =\begin{vmatrix}-1 & 0\\ 0 & -1 \end{vmatrix} & M_{32} & =\begin{vmatrix}2 & 0\\ 1 & -1 \end{vmatrix} & M_{33} & =\begin{vmatrix}2 & -1\\ 1 & 0 \end{vmatrix}\\ & =1 & & =-2 & & =1 \end{aligned} \]

And now the cofactor matrix:

\[ C=\begin{bmatrix}1 & 1 & 1\\ -1 & -2 & -2\\ 1 & 2 & 1 \end{bmatrix} \]

The determinant of \(A\) is \(1\). Finally, the inverse is:

\[ \begin{aligned} A^{-1} & =\cfrac{1}{\text{det}(A)}\text{adj}(A)\\ & =\cfrac{1}{\text{det}(A)}\cdot C^{T}\\ & =\begin{bmatrix}1 & -1 & 1\\ 1 & -2 & 2\\ 1 & -2 & 1 \end{bmatrix} \end{aligned} \]