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 (Procedure 9.2.2).
Example 9.3.1. Using row reduction to compute a determinant.
Let’s recompute the determinant of
\begin{equation*}
A = \begin{abmatrix}{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{abmatrix},
\end{equation*}
the same matrix from Example 8.4.8.
We will carry out the first two steps of Procedure 9.2.2 simultaneously, displaying the determinant of each matrix below it. For the purposes of describing how the sequence of determinant values progresses afterward, we’ll label our matrices as we go. As in the procedure, let \(d\) represent the unknown determinant value of \(A\text{.}\)
\begin{align*}
\amp \phantom{\text{(i) }} \det A = d
\amp \amp\qquad
\amp \phantom{\longrightarrow} \quad \amp
A =
\begin{abmatrix}{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{abmatrix}
\begin{array}{l} \\ R_2 - 2 R_1 \\ \\ R_4 - R_1 \end{array}\\
\amp \text{(i) } \det A_1 = d
\amp \amp
\amp \longrightarrow \quad \amp
A_1 =
\begin{abmatrix}{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{abmatrix}
\begin{array}{l} \phantom{X} \\ R_2 \leftrightarrow R_3 \\ \phantom{X} \end{array}\\
\amp \text{(ii) } \det A_2 = -d
\amp \amp
\amp \longrightarrow \quad \amp
A_2 =
\begin{abmatrix}{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{abmatrix}
\begin{array}{l} \\ \\ R_3 - 2 R_2 \\ R_4 + R_2 \end{array}\\
\amp \text{(iii) } \det A_3 = -d
\amp \amp
\amp \longrightarrow \quad \amp
A_3 =
\begin{abmatrix}{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{abmatrix}
\begin{array}{l} \phantom{X} \\ \phantom{X} \\ -\frac{1}{3} R_3 \\ \phantom{X} \end{array}\\
\amp \text{(iv) } \det A_4 = \tfrac{d}{3}
\amp \amp
\amp \longrightarrow \quad \amp
A_4 =
\begin{abmatrix}{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{abmatrix}
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ R_4 + 3 R_3 \end{array}\\
\amp \text{(v) } \det A_5 = \tfrac{d}{3}
\amp \amp
\amp \longrightarrow \quad \amp
A_5 =
\begin{abmatrix}{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{abmatrix}
\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.
Here is how we arrived at our sequence of determinant formulas:
- Matrix \(A_1\) was produced from \(A\) by a pair of row combination operations, but that type of operation does not change the determinant. So \(\det A_1 = \det A = d\text{.}\)
- Matrix \(A_2\) was produced from \(A_1\) by swapping rows, so the determinants of those two matrices have opposite signs: \(\det A_2 = - \det A_1 = -d\text{.}\)
- Matrix \(A_3\) was produced from \(A_2\) by another pair of row combination operations, so the determinants of those two matrices are equal: \(\det A_3 = \det A_2 = -d\text{.}\)
- Matrix \(A_4\) was produced from \(A_3\) by multiplying a row, and so the determinants are related by the same multiplication factor: \(\det A_4 = -\frac{1}{3} \det A_3 = \frac{d}{3} \text{.}\)
- Matrix \(A_5\) was produced from \(A_4\) by another row combination operations, so \(\det A_5 = \det A_4 = \frac{d}{3} \text{.}\)
But since \(A_5\) is an upper triangular matrix, we easily calculate its determinant as the product of its diagonal entries:
\begin{equation*}
\det A_5 = 1 \cdot 1 \cdot 1 \cdot (-9) = -9 \text{.}
\end{equation*}
Equating this to our formula for \(\det A_5\) from our row operation analysis, we obtain
\begin{gather*}
\frac{d}{3} = -9 \\
d = -27 \text{.}
\end{gather*}
Therefore, \(\det A = -27\text{,}\) which agrees with the result of the calculation of \(\det A\) by cofactor expansion in Example 8.4.8.
Example 9.3.2. A determinant of a matrix whose entries involve a variable.
Suppose we would like to know the values of the variable \(x\) that make the determinant of the matrix below equal to \(0\text{.}\)
\begin{equation*}
C =
\begin{abmatrix}{rrrr}
x + 6 \amp 3 \amp 4 \amp -7 \\
2 \amp x + 5 \amp -4 \amp -13 \\
-3 \amp -3 \amp x - 1 \amp 7 \\
6 \amp 6 \amp 0 \amp x - 14
\end{abmatrix}
\end{equation*}
Let \(d\) represent the determinant of this matrix. We will reduce to upper triangular, but we will not strictly follow Procedure 2.3.2 to do so. In particular, we will not worry too much about obtaining leading ones, but will focus more on obtaining zeros. The reasons for this are:
- A matrix with entries that contain a variable are complicated enough, so we will avoid introducing fractions into our matrix. To do this, we will multiply other rows to match a leading value instead of dividing to obtain a leading one.
- We should not attempt to obtain a leading one by dividing a row by an expression involving \(x\text{,}\) since there may be values of \(x\) for which that operation would actually be dividing by zero.
We will not explain each choice of row operation or the progression of determinant values — see if you can fill in those details for yourself. We will also simplify/factor entries as we go to make it easier to see potential simplifying row operations.
\begin{gather*}
\det C = d \\
\phantom{\longrightarrow} \quad C =
\begin{bmatrix}
x + 6 \amp 3 \amp 4 \amp -7 \\
2 \amp x + 5 \amp -4 \amp -13 \\
-3 \amp -3 \amp x - 1 \amp 7 \\
6 \amp 6 \amp 0 \amp x - 14
\end{bmatrix}
\begin{array}{l} \phantom{X} \\ R_1 \leftrightarrow R_2 \end{array}\\
\\
\det C_1 = -d \\
\longrightarrow \quad
C_1 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
x + 6 \amp 3 \amp 4 \amp -7 \\
-3 \amp -3 \amp x - 1 \amp 7 \\
6 \amp 6 \amp 0 \amp x - 14
\end{bmatrix}
\begin{array}{l} \\ 2 R_2 \\ 2 R_3 \\ \phantom{X} \end{array}\\
\\
\det C_2 = -4 d \\
\longrightarrow \quad
C_2 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
2 (x + 6) \amp 6 \amp 8 \amp -14 \\
-6 \amp -6 \amp 2 (x - 1) \amp 7 \\
6 \amp 6 \amp 0 \amp x - 14
\end{bmatrix}
\begin{array}{l} \\ R_2 - (x + 6) R_1 \\ R_3 + 3 R_1 \\ R_4 - 3 R_1 \end{array}\\
\\
\det C_3 = -4 d \\
\longrightarrow \quad
C_3 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
0 \amp - (x + 8) (x + 3) \amp 4 (x + 8) \amp 13 x + 64 \\
0 \amp 3 (x + 3) \amp 2 (x - 7) \amp -25 \\
0 \amp -3 (x + 3) \amp 12 \amp x + 25
\end{bmatrix}
\begin{array}{l} \\ 3 R_2 \\ \phantom{X} \\ \phantom{X} \end{array}\\
\\
\det C_4 = -12 d \\
\longrightarrow \quad
C_4 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
0 \amp -3 (x + 8) (x + 3) \amp 12 (x + 8) \amp 39 x + 192 \\
0 \amp 3 (x + 3) \amp 2 (x - 7) \amp -25 \\
0 \amp -3 (x + 3) \amp 12 \amp x + 25
\end{bmatrix}
\begin{array}{l} \\ R_2 + (x + 8) R_3 \\ \\ R_4 + R_3 \end{array}\\
\\
\det C_5 = -12 d \\
\longrightarrow \quad
C_5 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
0 \amp 0 \amp 2 (x - 1) (x + 8) \amp 14 x - 8 \\
0 \amp 3 (x + 3) \amp 2 (x - 7) \amp -25 \\
0 \amp 0 \amp 2 (x - 1) \amp x
\end{bmatrix}
\begin{array}{l} \phantom{X} \\ R_2 \leftrightarrow R_3 \phantom{X} \end{array}\\
\\
\det C_6 = 12 d \\
\longrightarrow \quad
C_6 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
0 \amp 3 (x + 3) \amp 2 (x - 7) \amp -25 \\
0 \amp 0 \amp 2 (x - 1) (x + 8) \amp 14 x - 8 \\
0 \amp 0 \amp 2 (x - 1) \amp x
\end{bmatrix}
\begin{array}{l} \\ \\ R_3 \leftrightarrow R_4 \end{array}\\
\\
\det C_7 = -12 d \\
\longrightarrow \quad
C_7 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
0 \amp 3 (x + 3) \amp 2 (x - 7) \amp -25 \\
0 \amp 0 \amp 2 (x - 1) \amp x \\
0 \amp 0 \amp 2 (x - 1) (x + 8) \amp 14 x - 8
\end{bmatrix}
\begin{array}{l} \\ \\ \\ R_4 - (x + 8) R_3 \end{array}\\
\\
\det C_8 = -12 d \\
\longrightarrow \quad
C_8 =
\begin{bmatrix}
2 \amp x + 5 \amp -4 \amp -13 \\
0 \amp 3 (x + 3) \amp 2 (x - 7) \amp -25 \\
0 \amp 0 \amp 2 (x - 1) \amp x \\
0 \amp 0 \amp 0 \amp - (x - 2) (x - 4)
\end{bmatrix}
\end{gather*}
We can now calculate the determinant of upper triangular matrix \(C_6\) as the product of its diagonal entries:
\begin{align*}
\det C_8 \amp = 2 \cdot 3 (x + 3) \cdot 2 (x - 1) \cdot \bbrac{- (x - 2) (x - 4)} \\
\amp = -12 (x + 3) (x - 1) (x - 2) (x - 4) \text{.}
\end{align*}
Since we also know \(\det C_8 = -12 d\text{,}\) where \(d\) represents \(\det C\text{,}\) we must have
\begin{equation*}
\det C = (x + 3) (x - 1) (x - 2) (x - 4) \text{,}
\end{equation*}
and so we have \(\det C = 0\) precisely when \(x\) is one of the values \(-3, 1, 2, 4\text{.}\)
Subsection 9.3.2 Matrices of determinant zero
Example 9.3.3. Recognizing \(\det A = 0\).
Here are a few examples of recognizing matrices that have determinant \(0\text{.}\)
- \(\displaystyle \begin{abmatrix}{rrrr} 1 \amp -1 \amp 2 \amp 1 \\ 0 \amp 1 \amp 0 \amp -3 \\ 0 \amp 1 \amp 0 \amp -3 \\ 1 \amp 2 \amp 3 \amp 4 \end{abmatrix}\)
- \(\displaystyle \begin{abmatrix}{rrrr} 1 \amp -1 \amp 2 \amp 1 \\ 0 \amp 1 \amp 5 \amp 5 \\ 7 \amp 1 \amp 0 \amp -3 \\ -2 \amp 2 \amp -4 \amp -2 \end{abmatrix}\)
- \(\displaystyle \begin{abmatrix}{rrrr} 1 \amp -1 \amp 0 \amp 1 \\ 4 \amp 1 \amp 0 \amp -3 \\ -1 \amp 1 \amp 0 \amp -3 \\ 1 \amp 2 \amp 0 \amp 4 \end{abmatrix}\)
- \(\displaystyle \begin{abmatrix}{rrrr} 1 \amp -1 \amp 2 \amp -1 \\ 0 \amp 1 \amp 5 \amp 1 \\ 7 \amp 1 \amp 0 \amp 1 \\ -2 \amp 2 \amp -4 \amp 2 \end{abmatrix}\)
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 \(0\text{.}\)
