Section 3.1 The Cofactor Expansion

Question: We know the determinant of a [latex]2 \times 2[/latex] matrix. How about the determinant of an [latex]n \times n[/latex] matrix?

We know that if a [latex]2 \times 2[/latex] matrix is invertible if and only if det[latex]A \neq 0[/latex]. If we want to define the determinant of an [latex]n \times n[/latex] matrix then we would want the similar property as the [latex]2 \times 2[/latex] matrix case.

Recall that a [latex]2 \times 2[/latex] matrix is invertible if and only if it is row equivalent to the [latex]2 \times 2[/latex] identity matrix. Here is how we operate the row reduction on a [latex]2 \times 2[/latex] matrix:

[latex]A = \begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix} \sim \begin{bmatrix}a_{11} & a_{12}\\a_{11}a_{21} & a_{11}a_{22}\end{bmatrix} \sim \begin{bmatrix}a_{11} & a_{12}\\ & a_{11}a_{22} - a_{12}a_{21}\end{bmatrix}[/latex]

When [latex]A[/latex] is equivalent to identity matrix, we must have det[latex]A[/latex] is nonzero. Similarly, we can have the row operation on an [latex]3 \times 3[/latex] matrix:

[latex]A = \begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{bmatrix} \sim \begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{11}a_{21} & a_{11}a_{22} & a_{11}a_{23}\\a_{11}a_{31} & a_{11}a_{32} & a_{11}a_{33}\end{bmatrix} \sim \begin{bmatrix}a_{11} & a_{12} & a_{13}\\0 & a_{11}a_{22} - a_{21}a_{12} & a_{11}a_{23} - a_{21}a_{13}\\0 & a_{11}a_{32} - a_{31}a_{12} & a_{11}a_{33} - a_{31}a_{13}\end{bmatrix} \sim \begin{bmatrix}a_{11} & a_{12} & a_{13}\\0 & a_{11}a_{22} - a_{21}a_{12} & a_{11}a_{23} - a_{21}a_{13}\\0 & 0 & a_{11}\triangle \end{bmatrix}[/latex]

where [latex]\triangle = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} - a_{13}a_{22}a_{31}[/latex]. If [latex]A[/latex] is equivalent to the [latex]3 \times 3[/latex] identity matrix then [latex]\triangle[/latex] must be nonzero. It is naturally to define the determinant of the [latex]3 \times 3[/latex] matrix [latex]A[/latex] as [latex]\triangle[/latex]. But how about [latex]4 \times 4[/latex] matrix?

Notice that

[latex]\triangle = a_{11}[/latex]det[latex]\begin{bmatrix}a_{22} & a_{23}\\a_{32} & a_{33}\end{bmatrix} - a_{12}[/latex]det[latex]\begin{bmatrix}a_{21} & a_{23}\\a_{31} & a_{33}\end{bmatrix} + a_{13}[/latex]det[latex]\begin{bmatrix}a_{21} & a_{22}\\a_{31} & a_{32}\end{bmatrix} = a_{11}[/latex]det[latex]A_{11} - a_{12}[/latex]det[latex]A_{12} + a_{13}[/latex]det[latex]A_{13}[/latex]

where [latex]A_{11}[/latex] is coming from deleting first row and first column of [latex]A[/latex] or the row and the column of the [latex]a_{11}, A_{12}[/latex] is coming from deleting first row and the second column of [latex]A[/latex] or the row and the column of a_{12}, and [latex]A_{13}[/latex] is coming from deleting first row and third column of [latex]A[/latex] or the row and the column of [latex]a_{13}[/latex].

We use [latex]\triangle = a_{11}[/latex]det[latex]A_{11} - a_{12}[/latex]det[latex]A_{12} + a_{13}[/latex]det[latex]A_{13}[/latex] to extend to the determinant of [latex]n \times n[/latex] matrix [latex]A[/latex].

Definition: Let [latex]n \geq 2[/latex], the determinant of an [latex]n \times n[/latex] matrix [latex]A = \begin{bmatrix}a_{ij}\end{bmatrix}[/latex] is the following

det[latex]A = a_{11}[/latex]det[latex]A_{11} - a_{12}[/latex]det[latex]A_{12} + a_{13}[/latex]det[latex]A_{13} - a_{14}[/latex]det[latex]A_{14} + \cdots + (-1)^{1+n}a_{1n}[/latex]det[latex]A_{1n} = \sum_{n}^{j = 1}(-1)^{1j}[/latex]det[latex]A_{1j}[/latex]

where [latex]A_{1j}[/latex] is the submatrix of [latex]A[/latex] by deleting the first row and [latex]j[/latex]-th column of [latex]A[/latex] or the row and the column of [latex]a_{1j}[/latex].

Example 1: Compute the determinant of [latex]A = \begin{bmatrix}1 & -2 & 3\\0 & -1 & 4\\2 & 4 & -2\end{bmatrix}[/latex].

Exercise 1: Compute the determinant of [latex]A = \begin{bmatrix}0 & 2 & -2\\1 & 0 & 5\\-1 & 4 & -2\end{bmatrix}[/latex].

Definition: Given [latex]A=\begin{bmatrix}a_{ij}\end{bmatrix}[/latex], the ([latex]i, j[/latex])-cofactor of [latex]A[/latex] is the number [latex]C_{ij} = (-1)^{i+j}[/latex]det[latex]A_{ij}[/latex] where [latex]A_{ij}[/latex] is the submtrix of [latex]A[/latex] by deleting [latex]i[/latex]-th row and [latex]j[/latex]-th column of [latex]A[/latex]. Then

det[latex]A = a_{11}C_{11} + a_{12}C_{12} + \cdots + a_{1n}C_{1n}[/latex]

is called a cofactor expansion across the first row of [latex]A[/latex].

Theorem: The determinant of an [latex]n \times n[/latex] matrix [latex]A[/latex] can be computed by a cofactor expansion across any row or down any column. The expansion across the [latex]i[/latex]-th row is the following:

det[latex]A = a_{i1}C_{i1} + a_{i2}C_{i2} + \cdots + a_{in}C_{in}[/latex]

The cofactor expansion down the [latex]j[/latex]-th column is

det[latex]A = a_{1j}C_{1j} + a_{2j}C_{2j} + \cdots + a_{nj}C_{nj}[/latex]

Example 2: Use cofactor expansion across third row to compute det[latex]A[/latex] and then use the cofactor expansion across second column to compute det[latex]A[/latex], where [latex]A = \begin{bmatrix}1 & 0 & -1\\0 & 2 & 3\\3 & 4 & -2\end{bmatrix}[/latex].

Exercise 2: Use cofactor expansion across third row to compute det[latex]A[/latex] and then use the cofactor expansion across second column to compute det[latex]A[/latex], where [latex]A = \begin{bmatrix}2 & -1 & 1\\1 & 0 & 3\\0 & 4 & -2\end{bmatrix}[/latex].

Example 3: Compute determinant [latex]A = \begin{bmatrix}2 & 1 & -2 & 1 & 4\\-1 & 0 & 1 & -1 & 2\\0 & 0 & 3 & 0 & 0\\1 & 0 & 2 & -3 & 1\\2 & 0 & 1 & 3 & 0\end{bmatrix}[/latex].

Exercise 3: Compute determinant [latex]A = \begin{bmatrix}2 & 0 & -2 & 1 & 0\\-1 & 0 & 1 & -1 & 2\\3 & 0 & 0 & 0 & 0\\1 & 1 & 2 & -3 & -2\\2 & 0 & 1 & 3 & 1\end{bmatrix}[/latex].

Definition: A matrix is called a triangular matrix if [latex]A[/latex] has either upper diagonal or lower diagonal entries that are nonzero.

Theorem: If [latex]A[/latex] is a triangular matrix then det[latex]A[/latex] is the product of the entries on the main diagonal.

Fact: For an [latex]3 \times 3[/latex] matrix [latex]A = \begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{bmatrix}[/latex], one can compute the determinant as the following: 1: write the first two columns of [latex]A[/latex] after the third column of [latex]A[/latex] then the determinant of [latex]A[/latex] is the sum of products of diagonal minus the sum of products of anti-diagonal.

Example 4: Compute determinant [latex]A[/latex] using the fact above. [latex]A = \begin{bmatrix}1 & 2 & 3\\-2 & -3 & 4\\0 & 3 & 5\end{bmatrix}[/latex].

Exercise 4: Compute determinant [latex]A[/latex] using the fact above. [latex]A = \begin{bmatrix}0 & 1 & 3\\-2 & -3 & 3\\1 & 2 & 4\end{bmatrix}[/latex].

Group Work 1: Mark each statement True or False. Justify each answer. All matrices are [latex]n \times n[/latex] matrices.

a. The determinant of an [latex]n \times n[/latex] matrix is defined by the [latex](n-1) \times (n-1)[/latex] submatrices.

b. The [latex](i, j)[/latex]-cofactor of a matrix [latex]A = \begin{bmatrix}a_{ij}\end{bmatrix}[/latex] is the matrix [latex]A_{ij}[/latex] obtained by deleting from [latex]A[/latex] its [latex]i[/latex]-th column and [latex]j[/latex]-th row.

c. det[latex](2A) = 2[/latex] det[latex]A[/latex] when [latex]A[/latex] is a [latex]2 \times 2[/latex] matrix.

d. If [latex]A[/latex] has a row or column consisting of zeros then det[latex]A = 0[/latex].

e. The cofactor expansion of det[latex]A[/latex] down a column is the negative of the cofactor down a row.

f. The determinant of a triangular matrix is the sum of the diagonal matrix.

g. det[latex](-A)[/latex] = det[latex]A[/latex].

Group Work 2: Compute the determinant.

[latex]E_{1} = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\k & 0 & 1\end{bmatrix}[/latex]

[latex]E_{2} = \begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{bmatrix}[/latex]

[latex]E_{3} = \begin{bmatrix}1 & 0 & 0\\0 & c & 0\\0 & 0 & 1\end{bmatrix}[/latex]

Group Work 3: Compute det[latex]AB[/latex], det[latex]A[/latex] and det[latex]B[/latex]. Find a relationship between them.

[latex]A = \begin{bmatrix}1 & 3 & 0\\1 & 2 & -2\\0 & 0 & 1\end{bmatrix}[/latex]

[latex]B = \begin{bmatrix}-2 & 5 & -2\\1 & 0 & 4\\3 & 0 & 1\end{bmatrix}[/latex]

Group Work 4: n each case either prove the statement or give an example showing that it is false. All matrices are [latex]n \times n[/latex] matrices.

a. det[latex](A+B) =[/latex]det[latex]A +[/latex]det[latex]B[/latex].

b. If det[latex]A = 0[/latex], then [latex]A[/latex] has two equal rows.

c. If [latex]R[/latex] is the reduced row-echelon form of [latex]A[/latex], then det[latex]A=[/latex]det[latex]R[/latex].

d. If det[latex]A=[/latex]det[latex]B[/latex] then [latex]A = B[/latex].

Group Work 5: Find the relationship between det[latex]A[/latex] and det[latex]cA[/latex] where [latex]c[/latex] is a scalar.

Group Work 6: If [latex]B[/latex] is obtained by multiply a number [latex]c[/latex] to one row of the matrix [latex]A[/latex]. Find the relationship between det[latex]A[/latex] and det[latex]B[/latex].

Group Work 7: If [latex]B[/latex] is obtained by switching two rows of the matrix [latex]A[/latex], find the relationship between det[latex]A[/latex] and det[latex]B[/latex].