Use the determinant to test if the matrix \(\Mtrx{A}\)
\[ \Mtrx{A} = \left[ \begin{array}{cc} 3 & 1 \\ 4 & 7 \end{array} \right] \]is invertible.
We know that \(\Mtrx{A}\) is invertible exactly when its determinant does not vanish. So we compute
\[ \det(\Mtrx{A}) = \left[ \begin{array}{cc} 3 & 1 \\ 4 & 7 \end{array} \right]\ =\ 21 - 4 = 17 \neq 0 \]Therefore \(\Mtrx{A}\) is invertible.
Use the determinant to test if the matrix
\[ \Mtrx{B}\ \DefEq\ \left[ \begin{array}{rrrrr} -1 & 4 & 3 & 4 & 5 \\ 5 & -2 & 3 & -2 & 1 \\ 0 & 6 & 0 & 6 & 2 \\ -1 & -1 & 2 & -1 & 4 \\ 9 & 3 & 3 & 3 & -1 \end{array} \right] \]is invertible.
We need to test if \(\det(\Mtrx{B})\) is nonzero. To this end we observe that the 2nd and 4th columns of matrix \(\Mtrx{B}\) are equal:
\[ \left[ \begin{array}{rrrrr} -1 & {\color{red} 4} & 3 & {\color{red} 4} & 5 \\ 5 & {\color{red} -2} & 3 & {\color{red} -2} & 1 \\ 0 & {\color{red} 6} & 0 & {\color{red} 6} & 2 \\ -1 & {\color{red} -1} & 2 & {\color{red} -1} & 4 \\ 9 & {\color{red} 3} & 3 & {\color{red} 3} & -1 \end{array} \right] \]Therefore
| \(\det\, \Mtrx{B}\) | \(=\) | \(0\) |
Consequently, \(\Mtrx{B}\) is not invertible.
Use the the determinant to test if the \((3,3)\)-matrix \(\Mtrx{C}\)
\[ \Mtrx{C}\ \DefEq \left[ \begin{array}{rrr} 1 & 0 & 1 \\ 2 & 3 & 1 \\ 0 & 5 & 0 \end{array} \right] \]is invertible.
We use row/column operations to turn \(\Mtrx{C}\) into an upper triangular matrix. The effect on the determinant is recorded in the following computation
| \( \det\, \left[ \begin{array}{rrr} 1 & 0 & 1 \\ 2 & 3 & 1 \\ 0 & 5 & 0 \end{array} \right] \) | \(= \) | \( -\det\, \left[ \begin{array}{rrr} 2 & 3 & 1 \\ 1 & 0 & 1 \\ 0 & 5 & 0 \end{array} \right] \) |
| \(\) | \(= \) | \( \det\, \left[ \begin{array}{rrr} 2 & 3 & 1 \\ 0 & 5 & 0 \\ 1 & 0 & 1 \end{array} \right] \) |
| \(\) | \(= \) | \( \det\, \left[ \begin{array}{rrr} 1 & 3 & 1 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{array} \right] \) |
| \(\) | \(= \) | \(5\neq 0\) |
Therefore \(\Mtrx{C}\) is invertible.