Chapter 2 - Matrix Algebra - 2.2 Exercises - Page 112: 26

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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 $ad-bc\ne0$, the system of equations represented by the augmented matrix $\begin{bmatrix} a&b&ad-bc\\ c&d&0\\ \end{bmatrix} $ has only the trivial solution. This means the columns of A are linearly dependent, which means it has a pivot in each column, meaning it can be reduced to the identity matrix, meaning it is invertible.