Chapter 4 · Determinants

Exercise 4.1 · Q6: Find $|A|$

Exercise 4.1Determinant of a matrix of order 3 × 3 · Section 4.2.3

The question

A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix},

find $|A|$ ⁠.

Exam answer

Expanding along $R_1$ ⁠:

\begin{aligned} |A| &= 1\begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} - 1\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} \\ &\quad + (-2)\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} \\ &= 1(-9 + 12) - 1(-18 + 15) \\ &\quad - 2(8 - 5) \\ &= 3 + 3 - 6 \end{aligned}

$\therefore\ |A| = 0$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Expand the $3\times3$ determinant along a row or column: sign $\times$ element $\times$ the $2\times2$ determinant left after deleting that element's row and column, added up. No entry is 0, so every term must be computed. The answer turns out to be 0, and there is a reason: each entry of column 3 is minus the sum of the entries of columns 1 and 2 in the same row.

Given, to find, and the plan

Given: The $3\times3$ matrix $A$ with rows $1, 1, -2$ ⁠; $2, 1, -3$ ⁠; $5, 4, -9$ ⁠.
To find: $|A|$ ⁠.
Plan: (1) Choose a row to expand along and write its signs. (2) Compute the three $2\times2$ minors. (3) Form the signed sum. (4) Confirm down column 1. (5) Explain why the value is 0.

Basics you need

1Expanding a $3\times3$ determinant

Along row $i$ ⁠: $|A| = \sum_j (-1)^{i+j} a_{ij} M_{ij}$ ⁠, where the minor $M_{ij}$ is the $2\times2$ determinant left after deleting row $i$ and column $j$ ⁠. Any row or column gives the same value (Section 4.2.3). Lesson: expanding a 3 × 3 determinant

2The sign pattern

$(-1)^{i+j}$ is $+$ for $i + j$ even, $-$ for $i + j$ odd: row 1 and column 1 both read $+, -, +$ ⁠. Basics: the + − + sign pattern

3Negative entries in a $2\times2$ determinant

In $ad - bc$ ⁠, bracket negatives: $1\cdot(-9) - 4\cdot(-3) = -9 + 12$ ⁠. Basics: algebra the chapter leans on

Solution

Choose row 1

$A$ is $3\times3$ ⁠, so $|A|$ is a determinant of order 3. No entry is 0, so no line saves work; we expand along row 1, $1, 1, -2$ ⁠, with signs $+, -, +$ ⁠.

Remark (i) of Section 4.2.3 says to prefer the line with most zeros. With none, row 1 is the natural choice: its first two elements are 1, which keeps the arithmetic small.

Basics: the sign pattern
The three minors along row 1
$\begin{aligned} M_{11} &= \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} = 1(-9) - 4(-3) = 3\\ M_{12} &= \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} = 2(-9) - 5(-3) = -3\\ M_{13} &= \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} = 2\cdot4 - 5\cdot1 = 3 \end{aligned}$
$M_{11}$ ⁠: delete row 1 and column 1, leaving $1, -3$ and $4, -9$ ⁠; then $-9 - (-12) = 3$ ⁠. $M_{12}$ ⁠: delete row 1 and column 2, leaving $2, -3$ and $5, -9$ ⁠; then $-18 - (-15) = -3$ ⁠. $M_{13}$ ⁠: delete row 1 and column 3, leaving $2, 1$ and $5, 4$ ⁠; then $8 - 5 = 3$ ⁠.

$2\times2$ rule, Section 4.2.2. Lesson: determinant of a square matrix
Signed sum
$\begin{aligned} |A| &= (+1)(1)(3) + (-1)(1)(-3)\\ &\quad + (+1)(-2)(3)\\ &= 3 + 3 - 6 = 0. \end{aligned}$
So $|A| = 0$ .

The middle term is $(-1)\times1\times(-3) = +3$ ⁠: the sign from the pattern and the negative minor cancel. The last term is $1\times(-2)\times3 = -6$ ⁠. The three terms add to exactly 0.

Expansion along a row, Section 4.2.3. Lesson: expanding a 3 × 3 determinant
Confirm down column 1

Column 1 is $1, 2, 5$ with signs $+, -, +$ ⁠. $M_{11} = 3$ from Step 2, and
$\begin{aligned} M_{21} &= \begin{vmatrix} 1 & -2 \\ 4 & -9 \end{vmatrix} = -9 - (-8) = -1\\ M_{31} &= \begin{vmatrix} 1 & -2 \\ 1 & -3 \end{vmatrix} = -3 - (-2) = -1 \end{aligned}$ $\begin{aligned} |A| &= (+1)(1)(3) + (-1)(2)(-1)\\ &\quad + (+1)(5)(-1)\\ &= 3 + 2 - 5 = 0. \end{aligned}$
$M_{21}$ ⁠: delete row 2 and column 1, leaving $1, -2$ and $4, -9$ ⁠. $M_{31}$ ⁠: delete row 3 and column 1, leaving $1, -2$ and $1, -3$ ⁠. A second, independent expansion gives 0 again, so the value is not a slip.
Why the value had to be 0

Look at the columns. In every row, column 3 equals minus (column 1 + column 2):
$\begin{aligned} \text{row 1: } & -(1 + 1) = -2\\ \text{row 2: } & -(2 + 1) = -3\\ \text{row 3: } & -(5 + 4) = -9 \end{aligned}$
Now expand a determinant of that shape along row 1, with letters $p, q$ for the first row and the other entries as in $A$ ⁠:
$\begin{aligned} &\begin{vmatrix} p & q & -(p+q) \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{vmatrix}\\ &= p(3) - q(-3) - (p+q)(3)\\ &= 3p + 3q - 3p - 3q = 0. \end{aligned}$
The minors $3, -3, 3$ are the ones from Step 2 (they do not involve row 1). The calculation gives 0 for every $p, q$ with third entry $-(p+q)$ ⁠; our row 1 is the case $p = q = 1$ ⁠. This is an instance of a general fact that is not proved in this reprint (it belongs to the properties of determinants, beyond this chapter): whenever one column (or row) equals a sum of multiples of the others, the determinant is 0. Here it has been proved directly for this matrix. A square matrix with $|A| = 0$ is called singular (Definition 4, Section 4.5), and Theorem 4 of Section 4.5 shows it has no inverse.

Result

$|A| = 0.$ (Textbook answer: 0.)

Check it with numbers

The six-term formula (1) of Section 4.2.3:

\begin{aligned} |A| &= a_{11}a_{22}a_{33} - a_{11}a_{32}a_{23}\\ &\quad - a_{12}a_{21}a_{33} + a_{12}a_{31}a_{23}\\ &\quad + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22} \end{aligned}

\begin{aligned} &= 1\cdot1\cdot(-9) - 1\cdot4\cdot(-3)\\ &\quad - 1\cdot2\cdot(-9) + 1\cdot5\cdot(-3)\\ &\quad + (-2)\cdot2\cdot4 - (-2)\cdot5\cdot1\\ &= -9 + 12 + 18 - 15 - 16 + 10\\ &= 0 \quad\checkmark \end{aligned}

Also, the column relation of Step 5 says $A$ times the column $1, 1, 1$ gives zeros: row 1 gives $1 + 1 - 2 = 0$ ⁠, row 2 gives $2 + 1 - 3 = 0$ ⁠, row 3 gives $5 + 4 - 9 = 0$ ✓.

Common mistakes

Dropping the minus sign of the middle term: $3 - 3 - 6 = -6$ ⁠. The middle term is $(-1)(1)(-3) = +3$ ⁠.
$1(-9) - 4(-3) = -9 - 12$ ⁠. Subtracting $-12$ adds 12.
Doubting an answer of 0. A zero determinant is a real and important outcome (a singular matrix); confirm it with a second expansion rather than looking for an error.

What to take away

With no zeros, expand along any line, and check along another.
If one column (or row) equals a sum of multiples of the others, the determinant is 0. This is a fact from beyond this reprint (Step 5 proves it only for this matrix), so in this chapter use it to spot or confirm an answer, and still show the expansion.

Every fact used, and where it comes from

Fact	Used in	Source
Expansion along any row or column, signs $(-1)^{i+j}$	Steps 1–5	Section 4.2.3 (Lesson: expanding a 3 × 3 determinant)
Prefer the line with most zeros; sign by parity of $i+j$	Step 1	Remarks (i), (ii), Section 4.2.3
$\|A\| = a_{11}a_{22} - a_{21}a_{12}$ for order 2	Steps 2, 4	Section 4.2.2
Six-term formula	Check	Equation (1), Section 4.2.3
If one column is a sum of multiples of the others, the determinant is 0	Step 5, What to take away	Supplementary fact from beyond this reprint; proved on this page only for this matrix (Step 5)
A matrix with $\|A\| = 0$ is singular and has no inverse	Step 5	Definition 4 and Theorem 4, Section 4.5 (Lesson: inverse of a matrix)
All minors and values on this page	Steps 2–5, Check	Recomputed in the site's verification script

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Exercise 4.1 · Q6: Find ∣A∣|A|∣A∣

Basics you need

Solution

Choose row 1

The three minors along row 1

Signed sum

Confirm down column 1

Why the value had to be 0

Every fact used, and where it comes from

Exercise 4.1 · Q6: Find $|A|$