Chapter 4 · Determinants

Example 10: Minors and cofactors of $a_{11}$ ⁠, $a_{21}$

ExamplesMinors and cofactors · Section 4.4

The question

Find minors and cofactors of the elements $a_{11}$ ⁠, $a_{21}$ in the determinant

\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}.

Exam answer

For $a_{11}$ ⁠:

\begin{aligned} M_{11} &= \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} \\ &= a_{22}a_{33} - a_{23}a_{32} \\ A_{11} &= (-1)^{1+1}M_{11} \\ &= a_{22}a_{33} - a_{23}a_{32} \end{aligned}

For $a_{21}$ ⁠:

\begin{aligned} M_{21} &= \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} \\ &= a_{12}a_{33} - a_{13}a_{32} \\ A_{21} &= (-1)^{2+1}M_{21} \end{aligned}

$\therefore\ A_{21} = -a_{12}a_{33} + a_{13}a_{32}$ ⁠.

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

The one idea this problem is built on

The definitions work the same with letters as with numbers. The minor $M_{ij}$ is the $2\times2$ determinant left after deleting row $i$ and column $j$ ⁠; the cofactor is $A_{ij} = (-1)^{i+j}M_{ij}$ ⁠. For $a_{11}$ the index sum 2 is even, so $A_{11} = M_{11}$ ⁠; for $a_{21}$ the index sum 3 is odd, so $A_{21} = -M_{21}$ ⁠. Doing it in general gives formulas that hold for every $3\times3$ determinant.

Given, to find, and the plan

Given: A general $3\times3$ determinant with entries $a_{ij}$ ⁠.
To find: $M_{11}$ ⁠, $A_{11}$ ⁠, $M_{21}$ ⁠, $A_{21}$ ⁠, as expressions in the entries.
Plan: Delete the right row and column for each, expand the $2\times2$ determinant, attach the sign $(-1)^{i+j}$ ⁠. Then test the formulas on a numerical determinant and see where they are used.

Basics you need

1Minor (Definition 1, Section 4.4)

$M_{ij}$ ⁠: delete row $i$ and column $j$ ⁠, keep the rest in place, evaluate. Lesson: minors and cofactors

2Cofactor (Definition 2, Section 4.4)

$A_{ij} = (-1)^{i+j}M_{ij}$ ⁠. Basics: the + − + sign pattern

3The $2\times2$ rule

(main-diagonal product) $-$ (other-diagonal product). Basics: where the determinant comes from

4Reading double subscripts

$a_{32}$ is row 3, column 2. After deleting a row and a column, the entries keep their original subscripts. Basics: rows, columns and positions

Solution

Minor of $a_{11}$

Delete row 1 and column 1. What remains is rows 2, 3 and columns 2, 3:
$M_{11} = \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} = a_{22}a_{33} - a_{23}a_{32}.$
The main diagonal of the leftover block is $a_{22}, a_{33}$ and the other diagonal is $a_{32}, a_{23}$ ⁠. The $2\times2$ rule subtracts the second product from the first. The entries keep their original names, which is what makes the formula usable for any $3\times3$ determinant.

Definition 1, Section 4.4. Lesson: minors and cofactors
Cofactor of $a_{11}$
$\begin{aligned} A_{11} &= (-1)^{1+1}M_{11} = (+1)M_{11}\\ &= a_{22}a_{33} - a_{23}a_{32}. \end{aligned}$
$1 + 1 = 2$ is even, so $(-1)^2 = 1$ and the cofactor equals the minor.

Definition 2, Section 4.4. Basics: the + − + sign pattern
Minor of $a_{21}$

Delete row 2 and column 1. What remains is rows 1, 3 and columns 2, 3:
$M_{21} = \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} = a_{12}a_{33} - a_{13}a_{32}.$
Row 1 supplies $a_{12}, a_{13}$ (top), row 3 supplies $a_{32}, a_{33}$ (bottom). Row 1 stays above row 3, as in the original. Main diagonal $a_{12}a_{33}$ ⁠, other diagonal $a_{32}a_{13}$ ⁠.
Cofactor of $a_{21}$
$\begin{aligned} A_{21} &= (-1)^{2+1}M_{21} = (-1)M_{21}\\ &= -(a_{12}a_{33} - a_{13}a_{32})\\ &= -a_{12}a_{33} + a_{13}a_{32}. \end{aligned}$
$2 + 1 = 3$ is odd, so $(-1)^3 = -1$ ⁠: the cofactor is the negative of the minor. The minus sign multiplies the whole bracket, so both terms change sign.

$-(u - v) = -u + v$ ⁠. Basics: algebra the chapter leans on
Test the formulas on numbers

Take the determinant of Example 3, with rows $1, 2, 4$ ⁠; $-1, 3, 0$ ⁠; $4, 1, 0$ ⁠. Then $a_{12} = 2$ ⁠, $a_{13} = 4$ ⁠, $a_{22} = 3$ ⁠, $a_{23} = 0$ ⁠, $a_{32} = 1$ ⁠, $a_{33} = 0$ ⁠, and the formulas give
$\begin{aligned} M_{11} &= 3(0) - 0(1) = 0 = A_{11},\\ M_{21} &= 2(0) - 4(1) = -4,\\ A_{21} &= -(-4) = 4. \end{aligned}$
Deleting directly gives the same: the block left by row 1 and column 1 has rows $3, 0$ and $1, 0$ ⁠, value $0$ ⁠; the block left by row 2 and column 1 has rows $2, 4$ and $1, 0$ ⁠, value $0 - 4 = -4$ ⁠.

A general formula is only useful if it agrees with the direct definition in particular cases. Here it does, for both elements.
Where these cofactors are used

The book's Remark after this example shows that $\Delta$ is the sum of the elements of any row or column times their own cofactors. Down the first column, that is
$\Delta = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31},$
which is the expansion along $C_1$ from Section 4.2.3, with the signs now built into the cofactors. With the Example 3 numbers (deleting row 3 and column 1 leaves rows $2, 4$ and $3, 0$ ⁠, and the sign $(-1)^{3+1}$ is $+$ ⁠, so $A_{31} = 2(0) - 4(3) = -12$ ⁠):
$\begin{aligned} &1(0) + (-1)(4) + 4(-12)\\ &= -4 - 48 = -52, \end{aligned}$
the value found in Example 3.

Cofactors package "sign $\times$ minor" into one number per element, so every expansion becomes "elements times cofactors, added". The book's Remark says it expands "the determinant $\Delta$ ⁠, in Example 21", but this reprint has only Examples 1 to 19; the number is left over from an older edition. The $\Delta$ it expands is the general determinant of this example (Example 10), printed just above the Remark.

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

Result

\begin{aligned} M_{11} &= a_{22}a_{33} - a_{23}a_{32},\\ A_{11} &= a_{22}a_{33} - a_{23}a_{32},\\ M_{21} &= a_{12}a_{33} - a_{13}a_{32},\\ A_{21} &= -a_{12}a_{33} + a_{13}a_{32}. \end{aligned}

This agrees with the textbook's solution.

Check it with numbers

Take the determinant with rows $1, 2, 3$ ⁠; $4, 5, 6$ ⁠; $7, 8, 10$ ⁠. The formulas give $M_{11} = 5(10) - 6(8) = 2 = A_{11}$ ⁠, $M_{21} = 2(10) - 3(8) = -4$ ⁠, $A_{21} = 4$ ⁠, and $A_{31} = 2(6) - 3(5) = -3$ ⁠. Expanding along $C_1$ ⁠:

\begin{aligned} &1(2) + 4(4) + 7(-3)\\ &= 2 + 16 - 21 = -3. \end{aligned}

Expanding along $R_1$ directly:

\begin{aligned} &1(50 - 48) - 2(40 - 42)\\ &\quad + 3(32 - 35)\\ &= 2 + 4 - 9 = -3. \end{aligned}

The two agree.

Common mistakes

Forgetting the sign for $a_{21}$ ⁠. $2+1$ is odd, so $A_{21} = -M_{21}$ ⁠.
Changing only one sign. $-(a_{12}a_{33} - a_{13}a_{32})$ is $-a_{12}a_{33} + a_{13}a_{32}$ ⁠; both terms flip.
Swapping the rows of the leftover block. In $M_{21}$ ⁠, row 1 entries go on top, row 3 entries below. Swapping them changes the sign of the minor.

What to take away

$M_{ij}$ ⁠: delete row $i$ and column $j$ ⁠; $A_{ij} = (-1)^{i+j}M_{ij}$ ⁠.
In a $3\times3$ determinant the cofactor signs follow the chessboard $+, -, +$ ⁠; so $A_{11} = M_{11}$ and $A_{21} = -M_{21}$ ⁠.
Down the first column, $\Delta = a_{11}A_{11} + a_{21}A_{21}$ $+\ a_{31}A_{31}$ ⁠: expanding is "elements times their cofactors, added".

Every fact used, and where it comes from

Fact	Used in	Source
Minor $M_{ij}$	Steps 1, 3, 5	Section 4.4, Definition 1 (Lesson: minors and cofactors)
Cofactor $A_{ij} = (-1)^{i+j}M_{ij}$	Steps 2, 4, 5	Section 4.4, Definition 2
$2\times2$ rule	Steps 1, 3, 5	Section 4.2.2
$\Delta =$ elements of a row or column times their cofactors	Step 6, Check	Section 4.4, Remark after Example 10; Section 4.2.3 (Lesson: expanding a 3 × 3 determinant)
$\Delta = -52$ for the Example 3 determinant	Step 6	Example 3

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Example 10: Minors and cofactors of a11a_{11}a11​⁠, a21a_{21}a21​

Basics you need

Solution

Minor of a11a_{11}a11​

Cofactor of a11a_{11}a11​

Minor of a21a_{21}a21​

Cofactor of a21a_{21}a21​

Test the formulas on numbers

Where these cofactors are used

Every fact used, and where it comes from

Example 10: Minors and cofactors of $a_{11}$ ⁠, $a_{21}$

Minor of $a_{11}$

Cofactor of $a_{11}$

Minor of $a_{21}$

Cofactor of $a_{21}$