Section 9.3 Examples
Subsection 9.3.1 Determinants by row reduction
As discussed in Warning 8.3.3, determinants by cofactor expansions are extremely inefficient for matrices larger than \(3\times 3\text{.}\) Here we provide an example of using the row reduction method to compute a determinant.
Example 9.3.1. Using row reduction to compute a determinant.
Let’s recompute the determinant of
\begin{equation*}
A = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
2 \amp 0 \amp 1 \amp 1 \\
0 \amp 1 \amp 0 \amp -3 \\
1 \amp -2 \amp -1 \amp 0
\end{array}\right],
\end{equation*}
the same matrix from Example 8.4.8.
First, let’s row reduce. For the purposes of describing our thinking in using the matrix reduction calculation to determine the determinant of \(A\text{,}\) we’ll label our matrices as we go.
\begin{align*}
A = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
2 \amp 0 \amp 1 \amp 1 \\
0 \amp 1 \amp 0 \amp -3 \\
1 \amp -2 \amp -1 \amp 0
\end{array}\right]
\begin{array}{l} \phantom{x} \\ R_2 - 2R_1 \\ \phantom{x} \\ R_4-R_1 \end{array}
\longrightarrow
A_1 = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
0 \amp 2 \amp -3 \amp -1 \\
0 \amp 1 \amp 0 \amp -3 \\
0 \amp -1 \amp -3 \amp -1
\end{array}\right]
\begin{array}{l} \phantom{x} \\ R_2 \leftrightarrow R_3 \\ \phantom{x} \end{array}\\
\longrightarrow
A_2 = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
0 \amp 1 \amp 0 \amp -3 \\
0 \amp 2 \amp -3 \amp -1 \\
0 \amp -1 \amp -3 \amp -1
\end{array}\right]
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_3 - 2R_2 \\ R_4+R_2 \end{array}
\longrightarrow
A_3 = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
0 \amp 1 \amp 0 \amp -3 \\
0 \amp 0 \amp -3 \amp 5 \\
0 \amp 0 \amp -3 \amp -4
\end{array}\right]
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ -\frac{1}{3}R_3 \\ \phantom{x} \end{array}\\
\longrightarrow
A_4 = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
0 \amp 1 \amp 0 \amp -3 \\
0 \amp 0 \amp 1 \amp -\frac{5}{3} \\
0 \amp 0 \amp -3 \amp -4
\end{array}\right]
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ R_4+3R_3 \end{array}
\longrightarrow
A_5 = \left[\begin{array}{rrrr}
1 \amp -1 \amp 2 \amp 1 \\
0 \amp 1 \amp 0 \amp -3 \\
0 \amp 0 \amp 1 \amp -\frac{5}{3} \\
0 \amp 0 \amp 0 \amp -9
\end{array}\right]
\end{align*}
We would need one more operation to get to REF, but we are already at upper triangular so we don’t need to bother. And notice that we didn’t bother clearing entries above leading ones, since our goal was to get to an upper triangular matrix, which only requires entries below leading ones to be cleared.
Now we’ll work backwards to determine \(\det A\text{.}\)
- \(A_5\)
- This last matrix is upper triangular, so its determinant is equal to the product of the diagonal entries: \(\det A_5 = 1\cdot 1\cdot 1\cdot (-9) = -9\text{.}\)
- \(A_4\)
- Matrix \(A_5\) was produced from \(A_4\) by an operation that does not change the determinant, so \(\det A_4\) must be \(-9\) as well.
- \(A_3\)
- Matrix \(A_4\) was produced from \(A_3\) by multiplying a row, so \(\det A_4 = -\frac{1}{3}\det A_3\text{.}\) Solving for \(\det A_3\text{,}\) we get \(\det A_3 = -3\cdot (-9) = 27\text{.}\)
- \(A_2\)
- Matrix \(A_3\) was produced from \(A_2\) by an operation that does not change the determinant, so \(\det A_2\) must be \(27\) as well.
- \(A_1\)
- Matrix \(A_2\) was produced from \(A_1\) by swapping rows, so these two determinants have opposite signs. Thus, \(\det A_1 = -27\text{.}\)
- \(A\)
- Matrix \(A_1\) was produced from \(A\) by a pair of operations, neither of which changes the determinant, so finally we have \(\det A = -27\text{.}\)
This analysis agrees with the calculation of \(\det A\) by cofactor expansion in Example 8.4.8.
Subsection 9.3.2 Matrices of determinant zero
Example 9.3.2. Recognizing \(\det A = 0\).
Here are a few examples of recognizing matrices that have determinant \(0\text{.}\)
- \(\displaystyle \left[\begin{array}{rrrr} 1 \amp -1 \amp 2 \amp 1 \\ 0 \amp 1 \amp 0 \amp -3 \\ 0 \amp 1 \amp 0 \amp -3 \\ 1 \amp 2 \amp 3 \amp 4 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{rrrr} 1 \amp -1 \amp 2 \amp 1 \\ 0 \amp 1 \amp 5 \amp 5 \\ 7 \amp 1 \amp 0 \amp -3 \\ -2 \amp 2 \amp -4 \amp -2 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{rrrr} 1 \amp -1 \amp 0 \amp 1 \\ 4 \amp 1 \amp 0 \amp -3 \\ -1 \amp 1 \amp 0 \amp -3 \\ 1 \amp 2 \amp 0 \amp 4 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{rrrr} 1 \amp -1 \amp 2 \amp -1 \\ 0 \amp 1 \amp 5 \amp 1 \\ 7 \amp 1 \amp 0 \amp 1 \\ -2 \amp 2 \amp -4 \amp 2 \end{array}\right]\)
The first matrix has two identical rows, the second matrix has two proportional rows (\(R_4 = -2R_1\)), the third matrix has a column of zeros, and the fourth matrix has two identical columns. So the determinant of each of these matrices is \(0\text{.}\)
