§ 20. Determinants of Order n

Definition and Properties of Determinants

Definition 20.1 : Determinant / Cofactor

Given \(A \in \mathrm{M}_{n \times n}(F)\), for \(n \geq 2\), denote the \((n-1) \times(n-1)\) matrix obtained from \(A\) by deleting row \(i\) and column \(j\) by \(\tilde{A}_{i j}\).

Let \(A \in \mathrm{M}_{n \times n}(F)\). If \(n=1\), so that \(A=\left(A_{11}\right)\), we define \(\operatorname{det}(A)=A_{11}\). For \(n \geq 2\), we define \(\operatorname{det}(A)\) recursively as

\[ \operatorname{det}(A)=\sum_{j=1}^{n}(-1)^{1+j} A_{1 j} \cdot \operatorname{det}\left(\tilde{A}_{1 j}\right) \]

The scalar \(\operatorname{det}(A)\) is called the determinant of \(A\) and is also denoted by \(|A|\). The scalar

\[ (-1)^{i+j} \operatorname{det}\left(\tilde{A}_{i j}\right) \]

is called the cofactor of the entry of \(A\) in row \(i\), column \(j\).

Letting

\[ c_{i j}=(-1)^{i+j} \operatorname{det}\left(\tilde{A}_{i j}\right) \]

denote the cofactor of the row \(i\), column \(j\) entry of \(A\), we can express the formula for the determinant of \(A\) as

\[ \operatorname{det}(A)=A_{11} c_{11}+A_{12} c_{12}+\cdots+A_{1 n} c_{1 n} . \]

Thus the determinant of \(A\) equals the sum of the products of each entry in row 1 of \(A\) multiplied by its cofactor. This formula is called cofactor expansion along the first row of \(A\).

Example 20.2 : Example of cofactor expansion

Let

\[ A=\left(\begin{array}{rrr} 1 & 3 & -3 \\ -3 & -5 & 2 \\ -4 & 4 & -6 \end{array}\right) \in \mathrm{M}_{3 \times 3}(\mathbb{R}) . \]

Using cofactor expansion along the first row of \(A\), we obtain

\[ \begin{aligned} \operatorname{det}(A)= & (-1)^{1+1} A_{11} \cdot \operatorname{det}\left(\tilde{A}_{11}\right)+(-1)^{1+2} A_{12} \cdot \operatorname{det}\left(\tilde{A}_{12}\right) \\ & \quad+(-1)^{1+3} A_{13} \cdot \operatorname{det}\left(\tilde{A}_{13}\right) \\ = & (-1)^{2}(1) \cdot \operatorname{det}\left(\begin{array}{rr} -5 & 2 \\ 4 & -6 \end{array}\right)+(-1)^{3}(3) \cdot \operatorname{det}\left(\begin{array}{rr} -3 & 2 \\ -4 & -6 \end{array}\right) \\ & \quad+(-1)^{4}(-3) \cdot \operatorname{det}\left(\begin{array}{rr} -3 & -5 \\ -4 & 4 \end{array}\right) \\ = & 1[-5(-6)-2(4)]-3[-3(-6)-2(-4)]-3[-3(4)-(-5)(-4)] \\ = & 1(22)-3(26)-3(-32) \\ = & 40 \end{aligned} \]

Example 20.3 : Determinant of the identity matrix

The determinant of the \(n \times n\) identity matrix is \(1\). We prove this assertion by mathematical induction on \(n\). The result is clearly true for the \(1 \times 1\) identity matrix. Assume that the determinant of the \((n-1) \times(n-1)\) identity matrix is \(1\) for some \(n \geq 2\), and let \(I\) denote the \(n \times n\) identity matrix. Using cofactor expansion along the first row of \(I\), we obtain

\[ \begin{aligned} \operatorname{det}(I)= & (-1)^{2}(1) \cdot \operatorname{det}\left(\tilde{I}_{11}\right)+(-1)^{3}(0) \cdot \operatorname{det}\left(\tilde{I}_{12}\right)+\cdots \\ & \quad+(-1)^{1+n}(0) \cdot \operatorname{det}\left(\tilde{I}_{1 n}\right) \\ = & 1(1)+0+\cdots+0 \\ = & 1 \end{aligned} \]

because \(\tilde{I}_{11}\) is the \((n-1) \times(n-1)\) identity matrix. This shows that the determinant of the \(n \times n\) identity matrix is 1, and so the determinant of any identity matrix is 1 by the principle of mathematical induction.

Theorem 20.4 : Linearity of determinant in each row

The determinant of an \(n \times n\) matrix is a linear function of each row when the remaining rows are held fixed. That is, for \(1 \leq r \leq n\), we have

\[ \operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r-1} \\ u+k v \\ a_{r+1} \\ \vdots \\ a_{n} \end{array}\right)=\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r-1} \\ u \\ a_{r+1} \\ \vdots \\ a_{n} \end{array}\right)+k \operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r-1} \\ v \\ a_{r+1} \\ \vdots \\ a_{n} \end{array}\right) \]

whenever \(k\) is a scalar and \(u, v\), and each \(a_{i}\) are row vectors in \(F^{n}\).

Proof

The proof is by mathematical induction on \(n\). The result is immediate if \(n=1\). Assume that for some integer \(n \geq 2\) the determinant of any \((n-1) \times(n-1)\) matrix is a linear function of each row when the remaining rows are held fixed. Let \(A\) be an \(n \times n\) matrix with rows \(a_{1}, a_{2}, \ldots, a_{n}\), respectively, and suppose that for some \(r\) (\(1 \leq r \leq n\)), we have \(a_{r}=u+k v\) for some \(u, v \in F^{n}\) and some scalar \(k\). Let \(u=\left(b_{1}, b_{2}, \ldots, b_{n}\right)\) and \(v=\left(c_{1}, c_{2}, \ldots, c_{n}\right)\), and let \(B\) and \(C\) be the matrices obtained from \(A\) by replacing row \(r\) of \(A\) by \(u\) and \(v\), respectively. We must prove that \(\operatorname{det}(A)=\operatorname{det}(B)+k \operatorname{det}(C)\).

First, we consider \(r=1\). For \(1 \leq j \leq n\), the matrices \(\tilde{A}_{1 j}, \tilde{B}_{1 j}\), and \(\tilde{C}_{1 j}\) are obtained by deleting row \(1\) and column \(j\) from \(A, B\), and \(C\), respectively. Since \(B\) and \(C\) differ from \(A\) only in row \(1\), it follows that

\[ \tilde{A}_{1 j}=\tilde{B}_{1 j}=\tilde{C}_{1 j} \]

for every \(j\). Also \(A_{1 j}=b_{j}+k c_{j}\), \(B_{1 j}=b_{j}\), and \(C_{1 j}=c_{j}\). Therefore, using the cofactor expansion along the first row,

\[ \begin{aligned} \operatorname{det}(A) & =\sum_{j=1}^{n}(-1)^{1+j} A_{1 j}\cdot \operatorname{det}\left(\tilde{A}_{1 j}\right) \\ & =\sum_{j=1}^{n}(-1)^{1+j} (b_{j}+k c_{j})\cdot \operatorname{det}\left(\tilde{B}_{1 j}\right) \\ & =\sum_{j=1}^{n}(-1)^{1+j} b_{j}\cdot \operatorname{det}\left(\tilde{B}_{1 j}\right) +k \sum_{j=1}^{n}(-1)^{1+j} c_{j}\cdot \operatorname{det}\left(\tilde{C}_{1 j}\right) \\ & =\operatorname{det}(B)+k \operatorname{det}(C) . \end{aligned} \]

For \(r>1\) and \(1 \leq j \leq n\), the rows of \(\tilde{A}_{1 j}, \tilde{B}_{1 j}\), and \(\tilde{C}_{1 j}\) are the same except for row \(r-1\). Moreover, row \(r-1\) of \(\tilde{A}_{1 j}\) is

\[ \left(b_{1}+k c_{1}, \ldots, b_{j-1}+k c_{j-1}, b_{j+1}+k c_{j+1}, \ldots, b_{n}+k c_{n}\right) \]

which is the sum of row \(r-1\) of \(\tilde{B}_{1 j}\) and \(k\) times row \(r-1\) of \(\tilde{C}_{1 j}\). Since \(\tilde{B}_{1 j}\) and \(\tilde{C}_{1 j}\) are \((n-1) \times(n-1)\) matrices, we have

\[ \operatorname{det}\left(\tilde{A}_{1 j}\right)=\operatorname{det}\left(\tilde{B}_{1 j}\right)+k \operatorname{det}\left(\tilde{C}_{1 j}\right) \]

by the induction hypothesis. Thus since \(A_{1 j}=B_{1 j}=C_{1 j}\), we have

\[ \begin{aligned} \operatorname{det}(A) & =\sum_{j=1}^{n}(-1)^{1+j} A_{1 j} \cdot \operatorname{det}\left(\tilde{A}_{1 j}\right) \\ & =\sum_{j=1}^{n}(-1)^{1+j} A_{1 j} \cdot\left[\operatorname{det}\left(\tilde{B}_{1 j}\right)+k \operatorname{det}\left(\tilde{C}_{1 j}\right)\right] \\ & =\sum_{j=1}^{n}(-1)^{1+j} A_{1 j} \cdot \operatorname{det}\left(\tilde{B}_{1 j}\right)+k \sum_{j=1}^{n}(-1)^{1+j} A_{1 j} \cdot \operatorname{det}\left(\tilde{C}_{1 j}\right) \\ & =\operatorname{det}(B)+k \operatorname{det}(C) \end{aligned} \]

This shows that the theorem is true for \(n \times n\) matrices, and so the theorem is true for all square matrices by mathematical induction.

Corollary 20.5 : Zero row gives zero determinant.

If \(A \in \mathrm{M}_{n \times n}(F)\) has a row consisting entirely of zeros, then \(\operatorname{det}(A)=0\).

Proof

Let the zero row be row \(r\) of \(A\). Let \(u\) be any row vector in \(F^{n}\). Since the zero row can be written as \(0\cdot u\), we have row \(r\) equal to \(0\cdot u\). Apply Theorem 20.4 with \(v=u\) and \(k=0\) (so that \(u+k v=u\) and \(0\cdot u\) gives the zero row). More directly, using the linearity in row \(r\) from Theorem 20.4, we have

\[ \operatorname{det}(A)=\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r-1} \\ 0\cdot u \\ a_{r+1} \\ \vdots \\ a_{n} \end{array}\right)=0\cdot \operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r-1} \\ u \\ a_{r+1} \\ \vdots \\ a_{n} \end{array}\right)=0 . \]

Cofactor Expansion along Any Row

Lemma 20.6 : Row equal to a standard basis vector

Let \(B \in \mathrm{M}_{n \times n}(F)\), where \(n \geq 2\). If row \(i\) of \(B\) equals \(e_{k}\) for some \(k\) (\(1 \leq k \leq n\)), then \(\operatorname{det}(B)=(-1)^{i+k} \operatorname{det}\left(\tilde{B}_{i k}\right)\).

Proof

The proof is by mathematical induction on \(n\). The lemma is easily proved for \(n=2\). Assume that for some integer \(n \geq 3\), the lemma is true for \((n-1) \times(n-1)\) matrices, and let \(B\) be an \(n \times n\) matrix in which row \(i\) of \(B\) equals \(e_{k}\) for some \(k\) (\(1 \leq k \leq n\)). The result follows immediately from the definition of the determinant if \(i=1\). Suppose therefore that \(1<i \leq n\). For each \(j \neq k\) (\(1 \leq j \leq n\)), let \(C_{i j}\) denote the \((n-2) \times(n-2)\) matrix obtained from \(B\) by deleting rows 1 and \(i\) and columns \(j\) and \(k\). For each \(j\), row \(i-1\) of \(\tilde{B}_{1 j}\) is the following vector in \(F^{n-1}\):

\[ \begin{cases} e_{k-1} & \text { if } j<k \\ 0 & \text { if } j=k \\ e_{k} & \text { if } j>k \end{cases} \]

Hence by the induction hypothesis and the corollary to Theorem 20.4, we have

\[ \operatorname{det}\left(\tilde{B}_{1 j}\right)= \begin{cases}(-1)^{(i-1)+(k-1)} \operatorname{det}\left(C_{i j}\right) & \text { if } j<k \\ 0 & \text { if } j=k \\ (-1)^{(i-1)+k} \operatorname{det}\left(C_{i j}\right) & \text { if } j>k\end{cases} \]

Therefore

\[ \begin{aligned} & \operatorname{det}(B)= \sum_{j=1}^{n}(-1)^{1+j} B_{1 j} \cdot \operatorname{det}\left(\tilde{B}_{1 j}\right) \\ &= \sum_{j<k}(-1)^{1+j} B_{1 j} \cdot \operatorname{det}\left(\tilde{B}_{1 j}\right)+\sum_{j>k}(-1)^{1+j} B_{1 j} \cdot \operatorname{det}\left(\tilde{B}_{1 j}\right) \\ &= \sum_{j<k}(-1)^{1+j} B_{1 j} \cdot\left[(-1)^{(i-1)+(k-1)} \operatorname{det}\left(C_{i j}\right)\right] +\sum_{j>k}(-1)^{1+j} B_{1 j} \cdot\left[(-1)^{(i-1)+k} \operatorname{det}\left(C_{i j}\right)\right] \\ &=(-1)^{i+k}\left[\sum_{j<k}(-1)^{1+j} B_{1 j} \cdot \operatorname{det}\left(C_{i j}\right)\right. \left.+\sum_{j>k}(-1)^{1+(j-1)} B_{1 j} \cdot \operatorname{det}\left(C_{i j}\right)\right] \end{aligned} \]

Because the expression inside the preceding bracket is the cofactor expansion of \(\tilde{B}_{i k}\) along the first row, it follows that

\[ \operatorname{det}(B)=(-1)^{i+k} \operatorname{det}\left(\tilde{B}_{i k}\right) \]

This shows that the lemma is true for \(n \times n\) matrices, and so the lemma is true for all square matrices by mathematical induction.

Theorem 20.7 : Cofactor expansion along any row

The determinant of a square matrix can be evaluated by cofactor expansion along any row. That is, if \(A \in \mathrm{M}_{n \times n}(F)\), then for any integer \(i\) (\(1 \leq i \leq n\)),

\[ \operatorname{det}(A)=\sum_{j=1}^{n}(-1)^{i+j} A_{i j} \cdot \operatorname{det}\left(\tilde{A}_{i j}\right) \]

Proof

Cofactor expansion along the first row of \(A\) gives the determinant of \(A\) by definition. So the result is true if \(i=1\). Fix \(i>1\). Row \(i\) of \(A\) can be written as \(\sum_{j=1}^{n} A_{i j} e_{j}\). For \(1 \leq j \leq n\), let \(B_{j}\) denote the matrix obtained from \(A\) by replacing row \(i\) of \(A\) by \(e_{j}\). Then by Theorem 20.4 and the lemma, we have

\[ \operatorname{det}(A)=\sum_{j=1}^{n} A_{i j} \operatorname{det}\left(B_{j}\right)=\sum_{j=1}^{n}(-1)^{i+j} A_{i j} \cdot \operatorname{det}\left(\tilde{A}_{i j}\right) \]

Corollary 20.8 : Identical rows give zero determinant.

If \(A \in \mathrm{M}_{n \times n}(F)\) has two identical rows, then \(\operatorname{det}(A)=0\).

Proof

The proof is by mathematical induction on \(n\). When \(n=2\), it is easily shown that \(\operatorname{det}(A)=0\). Assume that for some integer \(n \geq 3\), it is true for \((n-1) \times(n-1)\) matrices, and let rows \(r\) and \(s\) of \(A \in \mathrm{M}_{n \times n}(F)\) be identical for \(r \neq s\). Because \(n \geq 3\), we can choose an integer \(i\) (\(1 \leq i \leq n\)) other than \(r\) and \(s\). Now

\[ \operatorname{det}(A)=\sum_{j=1}^{n}(-1)^{i+j} A_{i j} \cdot \operatorname{det}\left(\tilde{A}_{i j}\right) \]

by Theorem 20.7. Since each \(\tilde{A}_{i j}\) is an \((n-1) \times(n-1)\) matrix with two identical rows, the induction hypothesis implies that each \(\operatorname{det}\left(\tilde{A}_{i j}\right)=0\), and hence \(\operatorname{det}(A)=0\). This completes the proof for \(n \times n\) matrices, and so the lemma is true for all square matrices by mathematical induction.

Elementary Row Operations and Determinants

Theorem 20.9 : Row interchange changes the sign.

If \(A \in \mathrm{M}_{n \times n}(F)\) and \(B\) is a matrix obtained from \(A\) by interchanging any two rows of \(A\), then \(\operatorname{det}(B)=-\operatorname{det}(A)\).

Proof

Let the rows of \(A \in \mathrm{M}_{n \times n}(F)\) be \(a_{1}, a_{2}, \ldots, a_{n}\), and let \(B\) be the matrix obtained from \(A\) by interchanging rows \(r\) and \(s\), where \(r<s\). Thus

\[ A=\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r} \\ \vdots \\ a_{s} \\ \vdots \\ a_{n} \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{s} \\ \vdots \\ a_{r} \\ \vdots \\ a_{n} \end{array}\right) \]

Consider the matrix obtained from \(A\) by replacing rows \(r\) and \(s\) by \(a_{r}+a_{s}\). By the corollary to Theorem 20.7 and Theorem 20.4, we have

\[ \begin{aligned} 0 & =\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r}+a_{s} \\ \vdots \\ a_{r}+a_{s} \\ \vdots \\ a_{n} \end{array}\right)=\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r} \\ \vdots \\ a_{r}+a_{s} \\ \vdots \\ a_{n} \end{array}\right)+\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{s} \\ \vdots \\ a_{r}+a_{s} \\ \vdots \\ a_{n} \end{array}\right) \\ & =\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r} \\ \vdots \\ a_{r} \\ \vdots \\ a_{n} \end{array}\right)+\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{r} \\ \vdots \\ a_{s} \\ \vdots \\ a_{n} \end{array}\right)+\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{s} \\ \vdots \\ a_{r} \\ \vdots \\ a_{n} \end{array}\right)+\operatorname{det}\left(\begin{array}{c} a_{1} \\ \vdots \\ a_{s} \\ \vdots \\ a_{s} \\ \vdots \\ a_{n} \end{array}\right) \\ & =0+\operatorname{det}(A)+\operatorname{det}(B)+0 . \end{aligned} \]

Therefore \(\operatorname{det}(B)=-\operatorname{det}(A)\).

Theorem 20.10 : Row replacement leaves the determinant unchanged.

Let \(A \in \mathrm{M}_{n \times n}(F)\), and let \(B\) be a matrix obtained by adding a multiple of one row of \(A\) to another row of \(A\). Then \(\operatorname{det}(B)=\operatorname{det}(A)\).

Proof

Suppose that \(B\) is the \(n \times n\) matrix obtained from \(A\) by adding \(k\) times row \(r\) to row \(s\), where \(r \neq s\). Let the rows of \(A\) be \(a_{1}, a_{2}, \ldots, a_{n}\), and the rows of \(B\) be \(b_{1}, b_{2}, \ldots, b_{n}\). Then \(b_{i}=a_{i}\) for \(i \neq s\) and \(b_{s}=a_{s}+k a_{r}\). Let \(C\) be the matrix obtained from \(A\) by replacing row \(s\) with \(a_{r}\). Applying Theorem 20.4 to row \(s\) of \(B\), we obtain

\[ \operatorname{det}(B)=\operatorname{det}(A)+k \operatorname{det}(C)=\operatorname{det}(A) \]

because \(\operatorname{det}(C)=0\) by the corollary to Theorem 20.7.

Corollary 20.11 : Rank less than \(n\) implies zero determinant.

If \(A \in \mathrm{M}_{n \times n}(F)\) has rank less than \(n\), then \(\operatorname{det}(A)=0\).

Proof

If the rank of \(A\) is less than \(n\), then the rows \(a_{1}, a_{2}, \ldots, a_{n}\) of \(A\) are linearly dependent. By Exercise 5.16, some row of \(A\), say, row \(r\), is a linear combination of the other rows. So there exist scalars \(c_{i}\) such that

\[ a_{r}=c_{1} a_{1}+\cdots+c_{r-1} a_{r-1}+c_{r+1} a_{r+1}+\cdots+c_{n} a_{n} \]

Let \(B\) be the matrix obtained from \(A\) by adding \(-c_{i}\) times row \(i\) to row \(r\) for each \(i \neq r\). Then row \(r\) of \(B\) consists entirely of zeros, and so \(\operatorname{det}(B)=0\). But by Theorem 20.10, \(\operatorname{det}(B)=\operatorname{det}(A)\). Hence \(\operatorname{det}(A)=0\).

Concept 20.12 : Elementary row operations and determinants

The following rules summarize the effect of an elementary row operation on the determinant of a matrix \(A \in \mathrm{M}_{n \times n}(F)\).

• (a) If \(B\) is a matrix obtained by interchanging any two rows of \(A\), then \(\operatorname{det}(B)=-\operatorname{det}(A)\).
• (b) If \(B\) is a matrix obtained by multiplying a row of \(A\) by a nonzero scalar \(k\), then \(\operatorname{det}(B)=k \operatorname{det}(A)\).
• (c) If \(B\) is a matrix obtained by adding a multiple of one row of \(A\) to another row of \(A\), then \(\operatorname{det}(B)=\operatorname{det}(A)\).

Theorem 20.13 : Determinant of an upper triangular matrix

The determinant of an upper triangular matrix is the product of its diagonal entries.

Proof

The proof is by mathematical induction on \(n\).

If \(n=1\), the result is immediate.

Assume that the result holds for all \((n-1)\times(n-1)\) upper triangular matrices, where \(n\ge 2\). Let \(A=(a_{ij})\) be an \(n\times n\) upper triangular matrix. By cofactor expansion along the first column, we have

\[ \operatorname{det}(A)=\sum_{i=1}^{n}(-1)^{i+1}a_{i1}\operatorname{det}(\tilde{A}_{i1}). \]

Since \(A\) is upper triangular, we have \(a_{i1}=0\) for all \(i\ge 2\). Hence the sum reduces to

\[ \operatorname{det}(A)=a_{11}\operatorname{det}(\tilde{A}_{11}). \]

The matrix \(\tilde{A}_{11}\) is obtained from \(A\) by deleting row \(1\) and column \(1\), and it is an \((n-1)\times(n-1)\) upper triangular matrix with diagonal entries \(a_{22},a_{33},\ldots,a_{nn}\). By the induction hypothesis,

\[ \operatorname{det}(\tilde{A}_{11})=\prod_{i=2}^{n} a_{ii}. \]

Therefore,

\[ \operatorname{det}(A)=a_{11}\prod_{i=2}^{n} a_{ii}=\prod_{i=1}^{n} a_{ii}. \]

Example 20.14 : Calculating determinant with row reduction

Concept 20.12 is used to transform the matrix into upper triangular matrices, for which the determinant can be easily calculated using Theorem 20.13.

\[ C=\left(\begin{array}{rrrr} 2 & 0 & 0 & 1 \\ 0 & 1 & 3 & -3 \\ -2 & -3 & -5 & 2 \\ 4 & -4 & 4 & -6 \end{array}\right) \]

This matrix can be transformed into an upper triangular matrix by means of the following sequence of elementary row operations of type 3:

\[ \begin{aligned} & \left(\begin{array}{rrrr} 2 & 0 & 0 & 1 \\ 0 & 1 & 3 & -3 \\ -2 & -3 & -5 & 2 \\ 4 & -4 & 4 & -6 \end{array}\right) \longrightarrow\left(\begin{array}{rrrr} 2 & 0 & 0 & 1 \\ 0 & 1 & 3 & -3 \\ 0 & -3 & -5 & 3 \\ 0 & -4 & 4 & -8 \end{array}\right) \rightarrow\left(\begin{array}{rrrr} 2 & 0 & 0 & 1 \\ 0 & 1 & 3 & -3 \\ 0 & 0 & 4 & -6 \\ 0 & 0 & 16 & -20 \end{array}\right) \\ & \longrightarrow\left(\begin{array}{rrrr} 2 & 0 & 0 & 1 \\ 0 & 1 & 3 & -3 \\ 0 & 0 & 4 & -6 \\ 0 & 0 & 0 & 4 \end{array}\right) \text { . } \end{aligned} \]

Thus \(\operatorname{det}(C)=2 \cdot 1 \cdot 4 \cdot 4=32\).

Exercise

Exercise 20.25

Prove that \(\operatorname{det}(k A)=k^{n} \operatorname{det}(A)\) for any \(A \in \mathrm{M}_{n \times n}(F)\).

Exercise 20.27

Prove that if \(A \in \mathrm{M}_{n \times n}(F)\) has two identical columns, then \(\operatorname{det}(A)=0\).

Exercise 20.28

Compute \(\operatorname{det}\left(E_{i}\right)\) if \(E_{i}\) is an elementary matrix of type \(i\).

Exercise 20.29

Prove that if \(E\) is an elementary matrix, then \(\operatorname{det}\left(E^{t}\right)=\operatorname{det}(E)\).