Solution - Matrix core operations

$$\in v\left(\left[\begin{array}{cc}2& 3\\ 2& -3\end{array}\right]\right)$$

Build the augmented matrix [A|I], then apply row operations until the left side becomes I.

$$\in v\left(\left[\begin{array}{cc}2& 3\\ 2& -3\end{array}\right]\right)$$

When left side is I, the right side is the inverse A^-1.

$$\in v\left(\left[\begin{array}{cc}2& 3\\ 2& -3\end{array}\right]\right)$$

R1 <- 1/2R1

$$\left[\begin{array}{cccc}1& 1.5& 0.5& 0\\ 2& -3& 0& 1\end{array}\right]$$

R2 <- R2 - 2R1

$$\left[\begin{array}{cccc}1& 1.5& 0.5& 0\\ 0& -6& -1& 1\end{array}\right]$$

R2 <- -1/6R2

$$\left[\begin{array}{cccc}1& 1.5& 0.5& 0\\ 0& 1& 0.166667& -0.166667\end{array}\right]$$

R1 <- R1 - 3/2R2

$$\left[\begin{array}{cccc}1& 0& 0.25& 0.25\\ 0& 1& 0.166667& -0.166667\end{array}\right]$$

Track both sides of [A|I]: the same row operations transform I into A^-1.

$$\left[\begin{array}{cc}0.25& 0.25\\ 0.166667& -0.166667\end{array}\right]$$

The right block of [A|I] is the inverse matrix after reducing the left block to identity.

$$\left[\begin{array}{cc}0.25& 0.25\\ 0.166667& -0.166667\end{array}\right]$$

Study tip: multiply A by A^-1 to check that the product is the identity matrix.

$$\left[\begin{array}{cc}0.25& 0.25\\ 0.166667& -0.166667\end{array}\right]$$

Present the final matrix or scalar outcome in canonical form for stable routing and review.