Chapter 4 · Determinants

Exercise 4.3 · Q3: Evaluate using cofactors of the second row

Exercise 4.3Minors and cofactors · Section 4.4

The question

Using Cofactors of elements of second row, evaluate

\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}.

Exam answer

Cofactors of the elements of $R_2$ ⁠:

\begin{aligned} A_{21} &= -\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} \\ &= -(9 - 16) = 7 \\ A_{22} &= \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} \\ &= 15 - 8 = 7 \\ A_{23} &= -\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} \\ &= -(10 - 3) = -7 \end{aligned}

Expanding along $R_2$ ⁠:

\begin{aligned} \Delta &= a_{21}A_{21} + a_{22}A_{22} \\ &\quad + a_{23}A_{23} \\ &= 2(7) + 0(7) + 1(-7) \\ &= 14 + 0 - 7 \end{aligned}

$\therefore\ \Delta = 7$ ⁠.

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

The one idea this problem is built on

A determinant equals the sum of the elements of any row multiplied by their own cofactors. The question chooses the second row, $2, 0, 1$ ⁠, so

\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}.

We need the three cofactors of that row; the 0 in the middle means one of them does not even matter.

Given, to find, and the plan

Given: The $3\times3$ determinant $\Delta$ above.
To find: The value of $\Delta$ ⁠, using the cofactors of the elements of $R_2$ ⁠.
Plan: (1) Read off $a_{21}, a_{22}, a_{23}$ ⁠. (2) Find $M_{21}, M_{22}, M_{23}$ and then $A_{21}, A_{22}, A_{23}$ ⁠. (3) Form $a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$ ⁠. (4) Check by expanding along another row.

Basics you need

1Minor and cofactor

$M_{ij}$ is the determinant left after deleting row $i$ and column $j$ ⁠; $A_{ij} = (-1)^{i+j}M_{ij}$ (Section 4.4, Definitions 1 and 2). Lesson: minors and cofactors

2Signs along the second row

$(-1)^{2+1} = -1$ ⁠, $(-1)^{2+2} = +1$ ⁠, $(-1)^{2+3} = -1$ ⁠: the second row of the chessboard is $-, +, -$ ⁠. Basics: the + − + sign pattern

3Expansion by cofactors

$\Delta$ = sum of the products of the elements of any row (or column) with their corresponding cofactors (Section 4.4, Remark after Example 10). Lesson: expanding a 3 × 3 determinant

Solution

The elements of the second row

$a_{21} = 2$ ⁠, $a_{22} = 0$ ⁠, $a_{23} = 1$ ⁠.

These are the multipliers in the expansion. Since $a_{22} = 0$ ⁠, the term $a_{22}A_{22}$ will be 0 whatever $A_{22}$ is. The question asks for the cofactors of all elements of the row, so we still find $A_{22}$ ⁠, but it cannot affect $\Delta$ ⁠.
The three minors

Delete $R_2$ ⁠; each minor uses rows $(5, 3, 8)$ and $(1, 2, 3)$ with one column removed:
$\begin{aligned} M_{21} &= \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = 3\cdot3 - 8\cdot2 = -7,\\ M_{22} &= \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 5\cdot3 - 8\cdot1 = 7,\\ M_{23} &= \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = 5\cdot2 - 3\cdot1 = 7. \end{aligned}$
For $M_{21}$ delete $C_1$ ⁠, leaving columns 2 and 3: $3, 8$ over $2, 3$ ⁠. For $M_{22}$ delete $C_2$ ⁠, leaving columns 1 and 3. For $M_{23}$ delete $C_3$ ⁠, leaving columns 1 and 2.

$2\times2$ rule, Section 4.2.2. Basics: where the determinant comes from
The three cofactors
$\begin{aligned} A_{21} &= (-1)^{2+1}M_{21} = (-1)(-7) = 7,\\ A_{22} &= (-1)^{2+2}M_{22} = (+1)(7) = 7,\\ A_{23} &= (-1)^{2+3}M_{23} = (-1)(7) = -7. \end{aligned}$
The second row has signs $-, +, -$ ⁠. The outer two cofactors change the sign of their minors; the middle one keeps it.

Definition 2, Section 4.4. Basics: the + − + sign pattern
Multiply and add
$\begin{aligned} \Delta &= a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}\\ &= 2(7) + 0(7) + 1(-7)\\ &= 14 + 0 - 7 = 7. \end{aligned}$
This is the expansion of $\Delta$ along $R_2$ ⁠, written with cofactors: the signs are already inside the $A_{ij}$ ⁠, so we just multiply and add.

Section 4.4, Remark after Example 10. Lesson: minors and cofactors

Result

$A_{21} = 7$ ⁠, $A_{22} = 7$ ⁠, $A_{23} = -7$ ⁠, and $\Delta = 7$ ⁠.

(Textbook answer: 7.)

Check it with numbers

Expand along $R_1$ instead, with signs $+, -, +$ ⁠:

\begin{aligned} \Delta &= 5(0\cdot3 - 1\cdot2)\\ &\quad - 3(2\cdot3 - 1\cdot1)\\ &\quad + 8(2\cdot2 - 0\cdot1)\\ &= 5(-2) - 3(5) + 8(4)\\ &= -10 - 15 + 32 = 7. \end{aligned}

The same value. ✓ Also, the elements of $R_1$ times the cofactors of $R_2$ must give 0: $5(7) + 3(7) + 8(-7)$ is $35 + 21 - 56 = 0$ ⁠. ✓

Common mistakes

Using the first-row signs $+, -, +$ on the second row. Row 2 starts with $-$ ⁠.
Adding minors instead of cofactors. $2(-7) + 0(7) + 1(7) = -7$ is wrong because the signs were left out.
Applying the sign twice. Once the cofactors include $(-1)^{i+j}$ ⁠, the sum $a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$ has only plus signs.

What to take away

$\Delta = a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3}$ for any row $i$ ⁠; the same holds for any column.
Signs along row 2 are $-, +, -$ ⁠.
A zero element saves work: its cofactor does not affect the value.

Every fact used, and where it comes from

Fact	Used in	Source
Minor and cofactor	Steps 2, 3	Section 4.4, Definitions 1 and 2 (Lesson: minors and cofactors)
$2\times2$ rule	Step 2; Check	Section 4.2.2
Determinant $=$ elements of a row times their cofactors	Step 4	Section 4.4, Remark after Example 10
Expansion along $R_1$ gives the same value	Check	Section 4.2.3
Row times another row's cofactors $= 0$	Check	Section 4.4, Note

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Exercise 4.3 · Q3: Evaluate using cofactors of the second row

Basics you need

Solution

The elements of the second row

The three minors

The three cofactors

Multiply and add

Every fact used, and where it comes from