2.2 Evaluating Determinants by Row Reduction: Class Notes

Theorem 2.2.1

Let $A$ be a square matrix. If $A$ has a row of zeros or a column of zeros, then $det (A) = 0$ .

Theorem 2.2.2

Let $A$ be a square matrix. Then $det (A) = det (A^{T})$ .

Theorem 2.2.3 Elementary Row Operations

Let $A$ be an $n \times n$ matrix.

a) If $B$ is the matrix that results when a single row or single column of $A$ is multiplied by a scalar $k$ , then $det (B) = k det (A)$ .

b) If $B$ is the matrix that results when two rows or two columns of $A$ are interchanged, then $det (B) = - det (A)$ .

c) If $B$ is the matrix that results when a multiple of one row of $A$ is added to another or when a multiple of one column is added to another, then $det (B) = det (A)$ .

Proof of (a)

Let $B$ be the matrix that results from multiplying the $i$ th row of $A$ by $k$ . By expanding along the $i$ th row,

$det (A) = \sum_{l = 1}^{n} a_{i l} C_{i l}$

The entries of row $i$ of $B$ are $b_{i l} = k a_{i l}$ , and $B$ has the same cofactors $C_{i l}$ along row $i$ as $A$ (they are computed with row $i$ deleted, and the other rows are unchanged). So

$det (B) = \sum_{l = 1}^{n} b_{i l} C_{i l} = \sum_{l = 1}^{n} k a_{i l} C_{i l} = k \sum_{l = 1}^{n} a_{i l} C_{i l} = k det (A)$

Consider the determinant below. Its second row is 2 times the second row of the determinant on the right, so a factor of 2 comes out:

$| \begin{matrix} 2 & 1 & 3 \\ 4 & 6 & 8 \\ 5 & 7 & 2 \end{matrix} | = 2 | \begin{matrix} 2 & 1 & 3 \\ 2 & 3 & 4 \\ 5 & 7 & 2 \end{matrix} |$

Note

Multiplying a matrix by 2 multiplies every entry: if $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , then $2 A = [\begin{matrix} 2 a & 2 b \\ 2 c & 2 d \end{matrix}]$ . But 2 times a determinant multiplies just one row:

$2 | \begin{matrix} a & b \\ c & d \end{matrix} | = 2 (a d - b c)$

So $det (2 A) = (2 a) (2 d) - (2 b) (2 c) = 4 (a d - b c)$ , not $2 det (A)$ .

Theorem 2.2.4

Let $E$ be an $n \times n$ elementary matrix.

a) If $E$ results from multiplying a single row of $I_{n}$ by a nonzero number $k$ , then $det (E) = k$ .

b) If $E$ results from interchanging two rows of $I_{n}$ , then $det (E) = - 1$ .

c) If $E$ results from adding a multiple of one row of $I_{n}$ to another, then $det (E) = 1$ .

This is a special case of Theorem 2.2.3, applied to $A = I_{n}$ , since $det (I_{n}) = 1$ .

#12

Evaluate the determinant of the matrix by first reducing the matrix to row echelon form and then using some combination of row operations and cofactor expansion.

$[\begin{matrix} 1 & - 3 & 0 \\ - 2 & 4 & 1 \\ 5 & - 2 & 2 \end{matrix}]$

Solution

Adding a multiple of one row to another doesn’t change the determinant (Theorem 2.2.3 (c)). $R_{2} \to R_{2} + 2 R_{1}$ and $R_{3} \to R_{3} - 5 R_{1}$ :

$| \begin{matrix} 1 & - 3 & 0 \\ - 2 & 4 & 1 \\ 5 & - 2 & 2 \end{matrix} | = | \begin{matrix} 1 & - 3 & 0 \\ 0 & - 2 & 1 \\ 0 & 13 & 2 \end{matrix} |$

$R_{3} \to R_{3} + \frac{13}{2} R_{2}$ :

$= | \begin{matrix} 1 & - 3 & 0 \\ 0 & - 2 & 1 \\ 0 & 0 & \frac{17}{2} \end{matrix} |$

Tip: Use $R_{3} \to R_{3} + (multiple) R_{2}$ , not $R_{3} \to 2 R_{3} + \dots$ : multiplying the row being replaced by 2 would also multiply the determinant by 2.

This is upper triangular, so by Theorem 2.1.2 the determinant is $1 (- 2) (\frac{17}{2}) = - 17$ .

Or, with row operations: factor $- 2$ out of row 2 and $\frac{17}{2}$ out of row 3 (Theorem 2.2.3 (a)):

$= (- 2) (\frac{17}{2}) | \begin{matrix} 1 & - 3 & 0 \\ 0 & 1 & - \frac{1}{2} \\ 0 & 0 & 1 \end{matrix} | = - 17$

For example, call the determinant with $- 2$ in row 2 $B$ and the one with 1 in row 2 $A$ (before factoring $\frac{17}{2}$ ). Row 2 of $B$ is row 2 of $A$ times $- 2$ , so $det (B) = - 2 det (A)$ .

Answer: $- 17$

#16

In #16 and 20, evaluate the determinant, given that $| \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} | = - 6$ .

$| \begin{matrix} g & h & i \\ d & e & f \\ a & b & c \end{matrix} |$

Solution

This is the given determinant with $R_{1} \leftrightarrow R_{3}$ . Interchanging two rows changes the sign (Theorem 2.2.3 (b)):

$| \begin{matrix} g & h & i \\ d & e & f \\ a & b & c \end{matrix} | = - (- 6)$

Answer: $6$

#20

$| \begin{matrix} a & b & c \\ 2 d & 2 e & 2 f \\ g + 3 a & h + 3 b & i + 3 c \end{matrix} |$

Solution

This comes from the given determinant by $R_{2} \to 2 R_{2}$ , which multiplies the determinant by 2, and $R_{3} \to R_{3} + 3 R_{1}$ , which doesn’t change the determinant. So

$| \begin{matrix} a & b & c \\ 2 d & 2 e & 2 f \\ g + 3 a & h + 3 b & i + 3 c \end{matrix} | = 2 (- 6)$

Answer: $- 12$

Theorem 2.2.5

If $A$ is a square matrix with two proportional rows or two proportional columns, then $det (A) = 0$ .

Because a row operation (adding a multiple of one of the proportional rows to the other) will yield a row of zeros (or a column of zeros, for proportional columns). That row operation doesn’t change the determinant, so $det (A) = 0$ by Theorem 2.2.1.

Proof of Theorem 2.2.3 (b)

Mathematical induction.

Base case: $2 \times 2$ . Let $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ . Then $det (A) = a d - b c$ . Let $A' = [\begin{matrix} c & d \\ a & b \end{matrix}]$ . Then $det (A') = c b - a d = - det (A)$ .

Inductive step: Assume $n > 2$ and that the row-interchange property holds for matrices smaller than $n \times n$ .

Let $A$ be an $n \times n$ matrix and $B$ be the matrix resulting from interchanging rows $i$ and $j$ of $A$ (or columns). Expanding along a row $k$ with $k \neq i$ and $k \neq j$ (row $k$ is the same in $A$ and $B$ ) uses cofactors

$(- 1)^{k + r} | A_{k r} | and (- 1)^{k + r} | B_{k r} |$

where $| A_{k r} |$ and $| B_{k r} |$ are determinants. $B_{k r}$ is $A_{k r}$ with two rows interchanged, and since these are of size $(n - 1) \times (n - 1)$ , the induction hypothesis gives $| B_{k r} | = - | A_{k r} |$ : the cofactors have opposite signs. Cofactor expansion yields $det (A) = - det (B)$ .

Proof of Theorem 2.2.3 (c)

Let ${\vec{a}}_{i} = [a_{i 1} a_{i 2} \dots a_{i n}]$ be the $i$ th row of $A$ . Let $B$ be the same as $A$ except that the $j$ th row ( $j \neq i$ ) results from adding $k {\vec{a}}_{i}$ to ${\vec{a}}_{j}$ , the $j$ th row of $A$ . Then

${\vec{b}}_{j} = [k a_{i 1} + a_{j 1} k a_{i 2} + a_{j 2} \dots k a_{i n} + a_{j n}]$

Since every row of $B$ other than row $j$ is the same as in $A$ , the minors $| A_{j r} |$ and $| B_{j r} |$ are equal for each $r$ , so $A$ and $B$ have the same cofactors $C_{j r}$ along row $j$ . Expanding along row $j$ :

$\begin{matrix} det (B) & = \sum_{r = 1}^{n} b_{j r} C_{j r} = \sum_{r = 1}^{n} (k a_{i r} + a_{j r}) C_{j r} \\ = \sum_{r = 1}^{n} k a_{i r} C_{j r} + \sum_{r = 1}^{n} a_{j r} C_{j r} \\ = k det (D) + det (A) \end{matrix}$

where $D$ is the matrix obtained by replacing the $j$ th row of $A$ with the $i$ th row of $A$ . Thus the $i$ th row of $D$ and the $j$ th row of $D$ are equal. By part (b), we can interchange rows $i$ and $j$ of $D$ to show

$det (D) = - det (D) \Rightarrow det (D) = 0 \Rightarrow det (B) = det (A)$