Inverse of the following $2\times 2$ Matrix

Question

I was asked to find the inverse of the following Matrix $A$:

$$\begin{bmatrix}8&-3\\-4&-7\end{bmatrix}$$

I created a new Matrix, $[A|I]$, where $I$ is the $2\times 2$ identity matrix and row reduced it. I ended up with:

$$\begin{bmatrix}7/68&31/68\\-1/17&-2/17\end{bmatrix}$$

But apparently I'm wrong. Any ideas on how to find the inverse?

Answer 1

Formula for the inverse of a $2$ by $2$ matrix is

$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a \end{bmatrix}$$

Answer 2

Setting up the system $[A\mid I]$ and row reducing gives: $$\begin{bmatrix}8&-3&|&1&0\\-4&-7&|&0&1\end{bmatrix}\xrightarrow{(1)}\begin{bmatrix}0&-17&|&1&2\\-4&-7&|&0&1\end{bmatrix}\xrightarrow{(2)}\begin{bmatrix}0&1&|&-1/17&-2/17\\1&7/4&|&0&-1/4\end{bmatrix}$$ where the steps are $$\begin{align}&(1)\qquad R_2+R_1\mapsto R_1\\&(2)\qquad \begin{cases}-\frac{1}{17}R_1\mapsto R_1\\-\frac{1}{4}R_2\mapsto R_2\end{cases}\end{align}$$ I leave the final step to you, as it is fairly straight forward. Maybe you can spot your mistake from these first couple of steps. This method (i.e. row reducing the system $[A\mid I]$) is a very general method for finding the inverse, so it's good to familiarize yourself with it. However, in the trivial $2\times 2$ case, one often uses the formula that Siong Thye Goh gives.