Chapter 4 · Determinants

Exercise 4.4 · Q1: Adjoint of a $2\times2$ matrix

Exercise 4.4Adjoint of a matrix · Section 4.5.1

The question

Find adjoint of each of the matrices in Exercises 1 and 2.

A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

Exam answer

Minors:

\begin{aligned} M_{11} &= 4,\ \ M_{12} = 3, \\ M_{21} &= 2,\ \ M_{22} = 1 \end{aligned}

Cofactors, $A_{ij} = (-1)^{i+j}M_{ij}$ ⁠:

\begin{aligned} A_{11} &= 4,\ \ A_{12} = -3, \\ A_{21} &= -2,\ \ A_{22} = 1 \end{aligned}

\operatorname{adj}A = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix}

$\therefore\ \operatorname{adj}A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

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

The one idea this problem is built on

The adjoint is built in two moves: replace every entry by its cofactor, then transpose (turn rows into columns). Nothing is divided and no determinant of the whole matrix is needed. For a $2\times2$ matrix the two moves together come to a simple rule, which the textbook states after Example 12: swap the two diagonal entries and change the sign of the other two. We work from the definition and then confirm the answer with that rule.

Given, to find, and the plan

Given: The $2\times2$ matrix $A$ above, with $a_{11}=1$ ⁠, $a_{12}=2$ ⁠, $a_{21}=3$ ⁠, $a_{22}=4$ ⁠.
To find: $\operatorname{adj}A$ ⁠, which is again a $2\times2$ matrix.
Plan: Find the four minors $M_{ij}$ ⁠, attach the signs to get the cofactors $A_{ij}$ ⁠, place them in the matrix of cofactors, transpose it. Then check with the $2\times2$ shortcut and with Theorem 1, $A(\operatorname{adj}A)=|A|\,I$ ⁠.

Basics you need

1Positions in a matrix

$a_{ij}$ is the entry in row $i$ and column $j$ ⁠. So in $A$ ⁠, $a_{12}=2$ (row 1, column 2) and $a_{21}=3$ (row 2, column 1). Basics: rows, columns and positions

2Minor and cofactor

The minor $M_{ij}$ is the determinant left after deleting row $i$ and column $j$ ⁠. The cofactor is $A_{ij}=(-1)^{i+j}M_{ij}$ ⁠: the same number, with its sign changed when $i+j$ is odd. Lesson: minors and cofactors

3The sign pattern

$(-1)^{i+j}$ is $+1$ when $i+j$ is even and $-1$ when it is odd. For a $2\times2$ matrix the signs are $+\ -$ in row 1 and $-\ +$ in row 2. Basics: the + − + sign pattern

4Transpose

The transpose of a matrix has the rows of the matrix as its columns: the entry in position $(i,j)$ moves to position $(j,i)$ ⁠. Basics: transpose and identity

5Definition of the adjoint

$\operatorname{adj}A$ is the transpose of the matrix $[A_{ij}]$ of cofactors (Definition 3, Section 4.5.1). Lesson: the adjoint

Solution

Check the order and list the entries

$A$ has 2 rows and 2 columns, so it is a square matrix of order $2\times2$ ⁠. The adjoint is defined only for square matrices, and $\operatorname{adj}A$ has the same order, $2\times2$ ⁠. We need one cofactor for each of the four entries: $a_{11}=1$ ⁠, $a_{12}=2$ ⁠, $a_{21}=3$ and $a_{22}=4$ ⁠.

Every step below refers to entries by their position $(i,j)$ ⁠, and the sign of a cofactor depends only on that position. Writing the positions out first keeps the bookkeeping straight.

Positions are always given row first, then column. Basics: rows, columns and positions
Find the four minors

For each entry, delete its row and its column. In a $2\times2$ matrix this leaves a single entry, and the determinant of a $1\times1$ matrix $[k]$ is just $k$ ⁠. So each minor is simply the entry left over.
- $a_{11}=1$ ⁠: delete row 1 and column 1; left is $4$ ⁠. So $M_{11}=4$ ⁠.
- $a_{12}=2$ ⁠: delete row 1 and column 2; left is $3$ ⁠. So $M_{12}=3$ ⁠.
- $a_{21}=3$ ⁠: delete row 2 and column 1; left is $2$ ⁠. So $M_{21}=2$ ⁠.
- $a_{22}=4$ ⁠: delete row 2 and column 2; left is $1$ ⁠. So $M_{22}=1$ ⁠.
This is Definition 1 of Section 4.4 applied four times. The minor of each entry is the entry diagonally opposite to it: deleting a row and a column of a $2\times2$ grid always leaves the one cell that shares neither.

Determinant of a matrix of order one: $|[k]| = k$ (Section 4.2.1). Lesson: determinant of a square matrix

Attach the signs: the four cofactors

Element	Minor	Cofactor
$a_{11} = 1$ ⁠; delete row 1 and column 1	$M_{11} = 4$	$A_{11} = +(4) = 4$
$a_{12} = 2$ ⁠; delete row 1 and column 2	$M_{12} = 3$	$A_{12} = -(3) = -3$
$a_{21} = 3$ ⁠; delete row 2 and column 1	$M_{21} = 2$	$A_{21} = -(2) = -2$
$a_{22} = 4$ ⁠; delete row 2 and column 2	$M_{22} = 1$	$A_{22} = +(1) = 1$

So $A_{11}=4$ ⁠, $A_{12}=-3$ ⁠, $A_{21}=-2$ ⁠, $A_{22}=1$ ⁠.

The cofactor is the minor multiplied by

(-1)^{i+j}

(Definition 2, Section 4.4). For

(1,2)

and

(2,1)

the sum

i+j=3

is odd, so those two minors change sign. For

(1,1)

and

(2,2)

the sum is even, so they stay as they are.

(-1)^{\text{even}}=1

and

(-1)^{\text{odd}}=-1

⁠. Basics: the + − + sign pattern

Place each cofactor in its own position

The matrix of cofactors has $A_{ij}$ in row $i$ ⁠, column $j$ ⁠:
$[A_{ij}] = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$
This matrix is the halfway point of the definition. Each number sits where its element sat in $A$ ⁠: the cofactor of $a_{12}=2$ ⁠, which is $-3$ ⁠, goes in row 1, column 2.
Transpose to get $\operatorname{adj}A$

Row 1 of $[A_{ij}]$ ⁠, which is $(4,\ -3)$ ⁠, becomes column 1; row 2, which is $(-2,\ 1)$ ⁠, becomes column 2:
$\operatorname{adj}A = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$
Definition 3 says the adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself. The reason is Theorem 1: with the transpose, row $i$ of $A$ meets the cofactors of its own row $i$ in the product $A(\operatorname{adj}A)$ ⁠, which gives $|A|$ on the diagonal (see the check below).

Transposing moves the entry in position $(i,j)$ to position $(j,i)$ ⁠; diagonal entries stay put. Basics: transpose and identity
Confirm with the $2\times2$ shortcut

The Remark after Example 12 says: for a $2\times2$ matrix, $\operatorname{adj}A$ is obtained by interchanging $a_{11}$ and $a_{22}$ and changing the signs of $a_{12}$ and $a_{21}$ ⁠. Swapping 1 and 4 and negating 2 and 3 gives
$\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix},$
the same matrix as in Step 5.

The blue diagonal entries 1 and 4 change places; the gold entries 2 and 3 stay where they are but change sign. This is exactly what the transposed cofactor matrix of Steps 4 and 5 produced.

The shortcut is not a separate fact; it is what Steps 2 to 5 always produce. For a general $2\times2$ matrix with entries $a, b$ in row 1 and $c, d$ in row 2, the cofactors are $A_{11}=d$ ⁠, $A_{12}=-c$ ⁠, $A_{21}=-b$ ⁠, $A_{22}=a$ ⁠, and transposing the cofactor matrix gives $\begin{bmatrix} d & -c \\ -b & a\end{bmatrix} \;\longrightarrow\; \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}:$ diagonal swapped, the other two negated.

Use the shortcut to check your work, but be able to do the long way: for $3\times3$ matrices (Exercise 4.4 Q2) there is no such shortcut. Lesson: the adjoint

Result

\operatorname{adj}A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}

(Textbook answer: the same matrix, with entries $4, -2$ in row 1 and $-3, 1$ in row 2.)

Check it with numbers

Theorem 1 says $A(\operatorname{adj}A)=|A|\,I$ ⁠. Here $|A| = 1\cdot4-2\cdot3 = 4-6=-2$ ⁠, so the product should be $-2I$ ⁠. Row times column:

\begin{aligned} p_{11} &= 1\cdot 4 + 2\cdot (-3) \\ &= 4 - 6 = -2 \\ p_{12} &= 1\cdot (-2) + 2\cdot 1 \\ &= -2 + 2 = 0 \\ p_{21} &= 3\cdot 4 + 4\cdot (-3) \\ &= 12 - 12 = 0 \\ p_{22} &= 3\cdot (-2) + 4\cdot 1 \\ &= -6 + 4 = -2 \end{aligned}

A(\operatorname{adj}A) = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} = -2\,I.

Exactly $|A|\,I$ ⁠, as the theorem predicts, so the adjoint is right.

Common mistakes

Stopping at the cofactor matrix. The matrix of Step 4, with $-3$ in row 1 and $-2$ in row 2, is $[A_{ij}]$ ⁠, not $\operatorname{adj}A$ ⁠. The adjoint is its transpose.
Forgetting the signs. Using the minors $4, 3, 2, 1$ without $(-1)^{i+j}$ gives a matrix with no minus signs, which fails the check $A(\operatorname{adj}A)=|A|I$ ⁠.
Mixing up the shortcut. It is the diagonal pair that swaps and the off-diagonal pair that changes sign, not the other way round.
Confusing adjoint and inverse. No division by $|A|$ happens here. Dividing by $|A|=-2$ would give $A^{-1}$ ⁠, a different matrix.

What to take away

Adjoint = transpose of the matrix of cofactors. Minors first, then signs, then transpose.
For any $2\times2$ matrix, $\operatorname{adj}\begin{bmatrix} a & b \\ c & d\end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}.$
$A(\operatorname{adj}A)=|A|\,I$ is a quick and reliable check on any adjoint.

Every fact used, and where it comes from

Fact	Used in	Source
Minor $M_{ij}$ ⁠: delete row $i$ ⁠, column $j$	Step 2	Definition 1, Section 4.4 (Lesson: minors and cofactors)
Determinant of order one: $\|[k]\|=k$	Step 2	Section 4.2.1
Cofactor $A_{ij}=(-1)^{i+j}M_{ij}$	Step 3	Definition 2, Section 4.4
Adjoint = transpose of $[A_{ij}]$	Steps 4, 5	Definition 3, Section 4.5.1 (Lesson: the adjoint)
Transpose: $(i,j)$ entry moves to $(j,i)$	Step 5	Chapter 3, Definition 3, Section 3.5
$2\times2$ shortcut: swap diagonal, negate the rest	Step 6	Remark after Example 12, Section 4.5.1
$\|A\| = a_{11}a_{22}-a_{12}a_{21}$	Check	Section 4.2.2
$A(\operatorname{adj}A)=(\operatorname{adj}A)A=\|A\|\,I$	Check	Theorem 1, Section 4.5.1
Product of matrices, row times column	Check	Chapter 3, Section 3.4.5 (Basics: matrix facts)

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Exercise 4.4 · Q1: Adjoint of a 2×22\times22×2 matrix

Basics you need

Solution

Check the order and list the entries

Find the four minors

Attach the signs: the four cofactors

Place each cofactor in its own position

Transpose to get adj⁡A\operatorname{adj}AadjA

Confirm with the 2×22\times22×2 shortcut

Every fact used, and where it comes from

Exercise 4.4 · Q1: Adjoint of a $2\times2$ matrix

Transpose to get $\operatorname{adj}A$

Confirm with the $2\times2$ shortcut