Section 8.4 Examples
Subsection 8.4.1 Determinants of \(2 \times 2\) matrices
Example 8.4.1. Determinant of a \(2 \times 2\) matrix.
Here is a \(2 \times 2\) matrix along with its determinant calculation pattern.
\(\displaystyle A = \begin{bmatrix} 1 \amp 2 \\ 3 \amp 4 \end{bmatrix} \)
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
A = \begin{bmatrix} 1 \amp 2 \\ 3 \amp 4 \end{bmatrix} \text{.}
\end{equation*}
The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
Using this pattern, we calculate
\begin{equation*}
\det A = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2 \text{.}
\end{equation*}
Watch out for double negatives! The next example illustrates the occurrence of a double negative in a determinant calculation.
Example 8.4.2. Another \(2 \times 2\) determinant.
\(\displaystyle A = \begin{abmatrix}{rr} 1 \amp 2 \\ -3 \amp 4 \end{abmatrix} \)
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
A = \begin{abmatrix}{rr} 1 \amp 2 \\ -3 \amp 4 \end{abmatrix} \text{.}
\end{equation*}
The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
\begin{equation*}
\det A = 1 \cdot 4 - 2 \cdot (-3) = 4 + 6 = 10 \text{.}
\end{equation*}
Subsection 8.4.2 Minors and cofactors of \(3 \times 3\) matrices
Subsubsection 8.4.2.1 Minors
A minor determinant is just a one-size-smaller determinant. To obtain that smaller matrix, we remove one row and one column. Usually we specify which to remove by focusing on a single entry and removing the row and column that contain the entry.
Example 8.4.3. Minor determinants in a \(3 \times 3\) matrix.
Let’s compute a couple of minor determinants in the matrix from Discovery 8.7:
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{.}
\end{equation*}
The notation \(M_{11}\) means the minor associated to the \((1,1)\) entry, so we should remove both the first row and the first column, leaving behind a \(2 \times 2\) matrix.
A diagram illustrating removing a row and a column in a \(3 \times 3\) matrix to create a \(2 \times 2\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{11}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{,}
\end{equation*}
so the first row has been crossed out by a horizontal line running through its entries and the first column has been crossed out by a vertical line running through its entries.
\(\displaystyle
M_{11} =
\begin{avmatrix}{rr}
-2 \amp 1 \\
1 \amp -1
\end{avmatrix}\)
We can now compute this minor determinant using the \(a d - b c\) pattern for \(2 \times 2\) determinants.
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
\begin{abmatrix}{rr}
-2 \amp 1 \\
1 \amp -1
\end{abmatrix}\text{.}
\end{equation*}
The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
\begin{align*}
M_{11} \amp = (-2) \cdot (-1) - 1 \cdot 1 \\
\amp = 2 - 1 \\
\amp = 1
\end{align*}
Now let’s try the \(M_{23}\) minor determinant. This time we should remove the second row and the third column.
A diagram illustrating removing a row and a column in a \(3 \times 3\) matrix to create a \(2 \times 2\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{23}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{,}
\end{equation*}
so the second row has been crossed out by a horizontal line running through its entries and the third column has been crossed out by a vertical line running through its entries.
\(\displaystyle
M_{23} =
\begin{vmatrix}
3 \amp 1 \\
0 \amp 1
\end{vmatrix}\)
Again, from here we compute this minor determinant using the \(a d - b c\) pattern.
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
\begin{bmatrix}
3 \amp 1 \\
0 \amp 1
\end{bmatrix}\text{.}
\end{equation*}
The matrix has entries \(3, 1\) in the first row and entries \(0, 1\) in the second row. The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
\begin{align*}
M_{23} \amp = 3 \cdot 1 - 1 \cdot 0 \\
\amp = 3 - 0 \\
\amp = 3
\end{align*}
Subsubsection 8.4.2.2 Cofactors
A cofactor just takes a minor determinant and (sometimes) flips its sign: when the corresponding entry is at an “even” position then the cofactor is equal to the minor determinant value, and when the corresponding entry is at an “odd” position then the sign is flipped.
Example 8.4.4. Cofactors in a \(3 \times 3\) matrix.
Let’s continue Example 8.4.3 above. The minor determinant \(M_{11}\) corresponds to the \((1,1)\) entry in the matrix, which is at an “even” position since \(1 + 1 = 2\) is even. So the corresponding cofactor value is equal to the minor determinant value:
\begin{equation*}
C_{11} = M_{11} = 1 \text{.}
\end{equation*}
But the minor determinant \(M_{23}\) corresponds to the \((2,3)\) entry in the matrix, which is at an “odd” position since \(2 + 3 = 5\) is odd. So the corresponding cofactor value is equal to the negative of the minor determinant value:
\begin{equation*}
C_{23} = - M_{23} = -3 \text{.}
\end{equation*}
Subsection 8.4.3 Determinants of \(3 \times 3\) matrices
For a \(3 \times 3\) matrix, we choose a single row or column and perform a cofactor expansion. It’s usually best to choose the row or column with the most zeros, since for a zero entry the “entry times cofactor” part of the expansion for that entry will be zero no matter the value of the cofactor, and we don’t actually have to calculate that cofactor. Also, we will use our cofactor sign patterns from Subsection 8.3.4 (see Pattern (8.3.1)), instead of calculating \((-1)^{i+j}\) explicitly.
Example 8.4.5. Determinant of a \(3 \times 3\) matrix along a row.
Let’s compute the determinant of the matrix from Discovery 8.7:
\begin{equation*}
A =
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{.}
\end{equation*}
Any of the first row or column or the third row or column would be good choices as they all contain a zero entry. Let’s choose the third row, since it also contains some \(1\)s, which will simplify things a bit:
\begin{equation*}
\begin{abmatrix}{rrr}
3\phantom{{}^+} \amp 1\phantom{{}^-} \amp 0\phantom{{}^+} \\
-2\phantom{{}^+} \amp -2\phantom{{}^-} \amp 1\phantom{{}^+} \\
{\color{red}\mathbf{0^+}} \amp {\color{red}\mathbf{1^-}} \amp {\color{red}\mathbf{{-1}^+}}
\end{abmatrix}\text{.}
\end{equation*}
Notice how we have also annotated that row with the cofactor sign pattern.
Proceed with computing \(M_{32}\) and \(M_{33}\text{.}\) (Since the \((3,1)\) entry is \(0\text{,}\) there is no need to compute \(M_{31}\text{.}\))
\(M_{32}\text{:}\)
A diagram illustrating removing a row and a column in a \(3 \times 3\) matrix to create a \(2 \times 2\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{32}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{,}
\end{equation*}
so the third row has been crossed out by a horizontal line running through its entries and the second column has been crossed out by a vertical line running through its entries.
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
\begin{abmatrix}{rr}
3 \amp 0 \\
-2 \amp 1
\end{abmatrix}\text{.}
\end{equation*}
The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
\(M_{33}\text{:}\)
A diagram illustrating removing a row and a column in a \(3 \times 3\) matrix to create a \(2 \times 2\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{33}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{.}
\end{equation*}
so the third row has been crossed out by a horizontal line running through its entries and the third column has been crossed out by a vertical line running through its entries.
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
\begin{abmatrix}{rr}
3 \amp 1 \\
-2 \amp -2
\end{abmatrix}\text{.}
\end{equation*}
The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
Now compute the cofactor expansion along that third row:
\begin{align*}
\det A \amp = a_{31} C_{31} + a_{32} C_{32} + a_{33} C_{33} \\
\amp = a_{31} M_{31} - a_{32} M_{32} + a_{33} M_{33} \\
\amp = 0 - 1 \cdot \bigl[ 3 \cdot 1 - 0 \cdot (-2) \bigr] + (-1) \cdot \bigl[ 3 \cdot(-2) - 1 \cdot(-2) \bigr] \\
\amp = -3 + (-1)(-4) \\
\amp = 1 \text{.}
\end{align*}
Note that the plus sign between the first two terms in the initial expansion became a minus sign when cofactors were replaced by minors due to the cofactor sign pattern in the chosen third row of the matrix.
Example 8.4.6. Determinant of a \(3 \times 3\) matrix along a column.
Let’s again compute the determinant of the matrix from Discovery 8.7, but this time along the middle column.
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp {\color{red}\mathbf{1^-}} \amp 0 \\
-2 \amp {\color{red}\mathbf{-2^+}} \amp 1 \\
0 \amp {\color{red}\mathbf{1^-}} \amp -1
\end{abmatrix}
\end{equation*}
Here we need to compute all three associated minor determinants.
\(M_{12}\text{:}\)
A diagram illustrating removing a row and a column in a \(3 \times 3\) matrix to create a \(2 \times 2\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{12}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{,}
\end{equation*}
so the first row has been crossed out by a horizontal line running through its entries and the second column has been crossed out by a vertical line running through its entries.
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
\begin{abmatrix}{rr}
-2 \amp 1 \\
0 \amp -1
\end{abmatrix}\text{.}
\end{equation*}
The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
\(M_{22}\text{:}\)
A diagram illustrating removing a row and a column in a \(3 \times 3\) matrix to create a \(2 \times 2\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{22}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{,}
\end{equation*}
so the second row has been crossed out by a horizontal line running through its entries and the second column has been crossed out by a vertical line running through its entries.
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
\begin{abmatrix}{rr}
3 \amp 0 \\
0 \amp -1
\end{abmatrix}\text{.}
\end{equation*}
The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
\(M_{32}\text{:}\)
A diagram illustrating removing a row and a column in a \(3 \times 3\) matrix to create a \(2 \times 2\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{32}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{abmatrix}\text{,}
\end{equation*}
so the third row has been crossed out by a horizontal line running through its entries and the second column has been crossed out by a vertical line running through its entries.
A diagram illustrating the determinant calculation pattern for the \(2 \times 2\) matrix
\begin{equation*}
\begin{abmatrix}{rr}
3 \amp 0 \\
-2 \amp 1
\end{abmatrix}\text{.}
\end{equation*}
The matrix has entries \(3, 0\) in the first row and entries \(-2, 1\) in the second row. The crisscross pattern is illustrated by an arrow marked with a positive sign running through the diagonal \((1,1)\) and \((2,2)\) entries and another arrow marked with a negative sign running through the antidiagonal \((1,2)\) and \((2,1)\) entries.
Now compute the cofactor expansion along that second column:
\begin{align*}
\det A \amp = a_{12} C_{12} + a_{22} C_{22} + a_{32} C_{32} \\
\amp = - a_{12} M_{12} + a_{22} M_{22} - a_{32} M_{32} \\
\amp = - 1 \cdot \bigl[ (-2) \cdot (-1) - 1 \cdot 0 \bigr] \\
\amp \phantom{=} + (-2) \cdot \bigl[ 3 \cdot (-1) - 0 \cdot 0 \bigr] \\
\amp \phantom{=} - 1 \cdot \bigl[ 3 \cdot 1 - 0 \cdot (-2) \bigr] \\
\amp = -2 + 6 - 3 \\
\amp = 1 \text{.}
\end{align*}
Again, note that minus signs in the second line of the calculation appear due to the cofactor sign pattern in the chosen second row of the matrix.
In the end, we got the same result as our first calculation, which is not a coincidence — see Theorem 8.5.1.
Subsection 8.4.4 Minors and cofactors of \(4 \times 4\) matrices
Applying the one-size-smaller pattern again, a minor determinant in a \(4 \times 4\) matrix is the determinant of a \(3 \times 3\) matrix obtained by removing one row and one column. And again cofactor values are equal to minor determinant values, except that we flip the signs for values associated to “odd” positions with the \(4 \times 4\) matrix.
Example 8.4.7.
Consider the matrix
\begin{equation*}
\begin{abmatrix}{rrrr}
-1 \amp 3 \amp 1 \amp 0 \\
-5 \amp 6 \amp 7 \amp 8 \\
2 \amp -2 \amp -2 \amp 1 \\
2 \amp 0 \amp 1 \amp -1
\end{abmatrix}\text{.}
\end{equation*}
To compute the \(M_{21}\) minor determinant, we remove the second row and the first column.
A diagram illustrating removing a row and a column in a \(4 \times 4\) matrix to create a \(3 \times 3\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{21}\) minor determinant of the matrix
\begin{equation*}
\begin{abmatrix}{rrrr}
-1 \amp 3 \amp 1 \amp 0 \\
-5 \amp 6 \amp 7 \amp 8 \\
2 \amp -2 \amp -2 \amp 1 \\
2 \amp 0 \amp 1 \amp -1
\end{abmatrix}\text{,}
\end{equation*}
so the second row has been crossed out by a horizontal line running through its entries and the first column has been crossed out by a vertical line running through its entries.
\(\displaystyle
M_{21} = \begin{avmatrix}{rrr}
3 \amp 1 \amp 0 \\
-2 \amp -2 \amp 1 \\
0 \amp 1 \amp -1
\end{avmatrix}\)
You might recognize this \(3 \times 3\) matrix as the same as the one from the examples in Subsection 8.4.3, so we already know its determinant. Also, the \((2,1)\) entry in the original \(4 \times 4\) matrix is in an “odd” position since \(2 + 1 = 3\) is odd, so must flip the sign to obtain the \(C_{21}\) cofactor value from the \(M_{21}\) minor determinant value:
\begin{align*}
M_{21} \amp = 1 \text{,} \amp C_{21} \amp = - M_{21} = -1 \text{.}
\end{align*}
Subsection 8.4.5 Determinants of \(4 \times 4\) matrices
Finally, here is a \(4 \times 4\) example. We’ll do one with a few zeros, so that it doesn’t get too out of hand.
Example 8.4.8. Determinant of a \(4 \times 4\) matrix.
In the matrix below we’ll choose to compute the determinant by cofactor expansion along the third row, as that row has two zero entries.
\begin{equation*}
A = \begin{abmatrix}{rrrr}
1\phantom{{}^+} \amp -1\phantom{{}^-} \amp 2\phantom{{}^+} \amp 1\phantom{{}^-} \\
2\phantom{{}^+} \amp 0\phantom{{}^-} \amp 1\phantom{{}^+} \amp 1\phantom{{}^-} \\
{\color{red}\mathbf{0^+}} \amp {\color{red}\mathbf{1^-}} \amp {\color{red}\mathbf{0^+}} \amp {\color{red}\mathbf{{-3}^-}} \\
1\phantom{{}^+} \amp -2\phantom{{}^-} \amp -1\phantom{{}^+} \amp 0\phantom{{}^-}
\end{abmatrix}\text{.}
\end{equation*}
Because of the two zeros in the third row, we will only need to compute minor determinants \(M_{32}\) and \(M_{34}\text{.}\)
\(M_{32}\text{:}\)
A diagram illustrating removing a row and a column in a \(4 \times 4\) matrix to create a \(3 \times 3\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{32}\) minor determinant of the matrix
\begin{equation*}
\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}\text{,}
\end{equation*}
so the third row has been crossed out by a horizontal line running through its entries and the second column has been crossed out by a vertical line running through its entries.
\(\displaystyle
M_{32} = \begin{avmatrix}{rrr}
1 \amp 2 \amp 1 \\
2 \amp 1 \amp 1 \\
1 \amp -1 \amp 0
\end{avmatrix}\)
\(M_{34}\text{:}\)
A diagram illustrating removing a row and a column in a \(4 \times 4\) matrix to create a \(3 \times 3\) submatrix for a minor determinant calculation. The goal is to calculate the \(M_{34}\) minor determinant of the matrix
\begin{equation*}
\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}\text{,}
\end{equation*}
so the third row has been crossed out by a horizontal line running through its entries and the fourth column has been crossed out by a vertical line running through its entries.
\begin{equation*}
M_{34} = \begin{avmatrix}{rrr}
1 \amp -1 \amp 2 \\
2 \amp 0 \amp 1 \\
1 \amp -2 \amp -1
\end{avmatrix}
\end{equation*}
Now we’ll compute the two minor determinants. Remember that these are entirely new determinant calculations, so that when we choose a row or column to expand along, the pattern of cofactor signs is specific to that \(3 \times 3\) matrix and does not come from the original \(4 \times 4\) matrix. After choosing a row or column, we expand as a combination of \(2 \times 2\) determinants, and then apply the crisscross \(a d - b c\) pattern in each.
\begin{align*}
M_{32}
\amp = \begin{avmatrix}{rrr}
1\phantom{{}^+} \amp 2\phantom{{}^-} \amp 1\phantom{{}^+} \\
2\phantom{{}^+} \amp 1\phantom{{}^-} \amp 1\phantom{{}^+} \\
{\color{red}\mathbf{1^+}} \amp {\color{red}\mathbf{{-1}^-}} \amp {\color{red}\mathbf{0^+}}
\end{avmatrix}\\
\amp =
1 \cdot \begin{avmatrix}{rr}
2 \amp 1 \\
1 \amp 1
\end{avmatrix}
- (-1) \cdot \begin{avmatrix}{rr}
1 \amp 1 \\
2 \amp 1
\end{avmatrix}
+ 0\\
\amp = ( 2 \cdot 1 - 1 \cdot 1 ) + ( 1 \cdot 1 - 2 \cdot 1 ) \\
\amp = 1 + (-1) \\
\amp = 0 \text{,} \\
\\
M_{34}
\amp = \begin{avmatrix}{rrr}
1\phantom{{}^-} \amp -1\phantom{{}^+} \amp 2\phantom{{}^-} \\
{\color{red}\mathbf{2^-}} \amp {\color{red}\mathbf{0^+}} \amp {\color{red}\mathbf{1^-}} \\
1\phantom{{}^-} \amp -2\phantom{{}^+} \amp -1\phantom{{}^-}
\end{avmatrix}\\
\amp =
-2 \cdot \begin{avmatrix}{rr}
-1 \amp 2 \\
-2 \amp -1
\end{avmatrix}
+ 0
- 1 \cdot \begin{avmatrix}{rr}
1 \amp -1 \\
1 \amp -2
\end{avmatrix}\\
\amp = -2 \bigl[ (-1) \cdot (-1) - 2 \cdot (-2) \bigr] - \bigl[ 1 \cdot (-2) - (-1) \cdot 1 \bigr] \\
\amp = -2 \cdot 5 - (-1) \\
\amp - -9 \text{.}
\end{align*}
Finally, we can compute \(\det A\) via cofactor expansion along our chosen row:
\begin{align*}
\det A \amp = a_{31} C_{31} + a_{32} C_{32} + a_{33} C_{33} + a_{34} C_{34} \\
\amp = 0 \cdot M_{31} - 1 \cdot M_{32} + 0 \cdot M_{33} - (-3) \cdot M_{34} \\
\amp = - 1 \cdot 0 + 3 \cdot (-9) \\
\amp = - 27 \text{.}
\end{align*}
