Chapter 2 - Matrix Algebra - 2.2 Exercises - Page 112: 25

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$A=\begin{bmatrix} a&b\\ c&d\\ \end{bmatrix} $ $\begin{bmatrix} a&b\\ c&d\\ \end{bmatrix} $$\begin{bmatrix} d\\ -c\\ \end{bmatrix} $=$\begin{bmatrix} ad-bc\\ cd-dc\\ \end{bmatrix} $=$\begin{bmatrix} ad-bc\\ 0\\ \end{bmatrix} $ If $ad-bc=0$, the equation will be $\begin{bmatrix} a&b\\ c&d\\ \end{bmatrix} $$\begin{bmatrix} d\\ -c\\ \end{bmatrix} $=$\begin{bmatrix} 0\\ 0\\ \end{bmatrix} $ This means $\begin{bmatrix} d\\ -c\\ \end{bmatrix} $ is a solution to augmented matrix $\begin{bmatrix} a&b&0\\ c&d&0\\ \end{bmatrix} $ If $d\ne0$ and $c\ne0$, there is a nontrivial solution, which means the columns of A are linearly dependent, which means they can't have a pivot in each column, meaning A can't be reduced to the identity matrix, meaning it is not invertible.