Determinants • Topic 2 of 3

Minors, Cofactors, Adjoint & Area

To turn the expansion rule into a usable tool — and to reach the inverse in the next topic — we attach two numbers to every entry of a matrix: its minor and its cofactor.

Minors and cofactors

The minor $M_{ij}$ of the entry $a_{ij}$ is the determinant of the smaller matrix left after deleting row $i$ and column $j$. The cofactor attaches the checkerboard sign:

$$A_{ij}=(-1)^{i+j}M_{ij}.$$

The sign pattern for a $3\times3$ matrix is $\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}$. For example $A_{11}=+M_{11}$, $A_{12}=-M_{12}$, $A_{23}=-M_{23}$.

Determinant by cofactor expansion

A determinant equals the sum of the entries of any single line, each multiplied by its own cofactor:

$$|A|=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}\quad(\text{expansion along row }i).$$

A subtle but examinable fact: if you multiply the entries of one row by the cofactors of a different row, the sum is always $0$ (because that computes a determinant with two equal rows).

Adjoint of a matrix

The adjoint (adjugate) $\operatorname{adj}A$ is the transpose of the matrix of cofactors: replace each $a_{ij}$ by its cofactor $A_{ij}$, then transpose. Its defining property — the bridge to the inverse — is

$$A\,(\operatorname{adj}A)=(\operatorname{adj}A)\,A=|A|\,I.$$

For a $2\times2$ matrix there is a one-line shortcut: $\operatorname{adj}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ — swap the diagonal entries and negate the off-diagonal ones. Two standard results follow: for an $n\times n$ matrix, $|\operatorname{adj}A|=|A|^{\,n-1}$, and $\operatorname{adj}(\operatorname{adj}A)=|A|^{\,n-2}A$.

Area of a triangle

The area of a triangle with vertices $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})$ is

$$\text{Area}=\frac12\left|\,\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}\,\right|.$$

The outer bars are absolute value: area is never negative, even though the determinant itself can be. The expanded form is $\text{Area}=\tfrac12\,|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$.

Collinearity test. If the determinant is $0$, the "triangle" has zero area, which means the three points lie on one straight line:

$$\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}=0\iff (x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})\text{ are collinear.}$$

Deeper Insight — why $A(\operatorname{adj}A)=|A|I$ is the whole point: Look at the diagonal entries of the product $A(\operatorname{adj}A)$: each is a row of $A$ dotted with the matching cofactors — that is exactly the cofactor expansion, so each diagonal entry equals $|A|$. Every off-diagonal entry is a row of $A$ dotted with the cofactors of a different row, which is $0$ by the fact noted above. So the product is $|A|$ down the diagonal and $0$ elsewhere — precisely $|A|I$. Dividing through by $|A|$ (when it is non-zero) hands you the inverse formula in the next topic for free. Master this identity and the inverse, area formula and collinearity test all become one idea wearing three hats.

Minor M-one-one obtained by deleting row one and column one, with the checkerboard cofactor sign chart Minor M₁₁ — delete row 1 and column 1 a₁₁a₁₂a₁₃ a₂₁a₂₂a₂₃ a₃₁a₃₂a₃₃ M₁₁ = remaining 2×2 block M₁₁ = a₂₂a₃₃ − a₂₃a₃₂ Cofactor signs (−1)^(i+j) +−+ −+− +−+ A₁₁ = +M₁₁ Triangle on a coordinate plane with vertices one one, four one and two five, whose area is found by a determinant Area of a Triangle by Determinant x y O 1 2 4 1 5 (1, 1) (4, 1) (2, 5) Area = ½ | x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) | = ½ | 1(1−5) + 4(5−1) + 2(1−1) | = ½ | −4 + 16 + 0 | = ½ × 12 = 6 sq units
Solved Examples
1
Worked Example
Find the cofactor $A_{12}$ of $a_{12}$ in $A=\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$.
Solution

Delete row $1$, column $2$: $M_{12}=\begin{vmatrix}4&6\\7&9\end{vmatrix}=36-42=-6$. Cofactor $A_{12}=(-1)^{1+2}M_{12}=-(-6)=6$.

Answer: $A_{12}=6$

2
Worked Example
For $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$, find $\operatorname{adj}A$.
Solution

For a $2\times2$ matrix the adjoint swaps the diagonal and negates the off-diagonal: $\operatorname{adj}A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$. Check $A(\operatorname{adj}A)=|A|I=(5)I$.

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

3
Worked Example
Find the area of the triangle with vertices $(0,0),(4,0),(0,3)$.
Solution

$\text{Area}=\tfrac12\left|\,\begin{vmatrix}0&0&1\\4&0&1\\0&3&1\end{vmatrix}\,\right|=\tfrac12\,|\,0-0+1\cdot(12-0)\,|=\tfrac12\cdot12=6$ square units.

Answer: $6$ square units.

4
Worked Example
Show the points $(1,2),(2,4),(3,6)$ are collinear.
Solution

$\begin{vmatrix}1&2&1\\2&4&1\\3&6&1\end{vmatrix}$: columns $1$ and $2$ are proportional ($y=2x$), so the determinant is $0$, giving zero area — the points are collinear.

Answer: The area determinant is $0$, so the points are collinear.

5
Worked Example
Find all minors and cofactors of $A=\begin{bmatrix}1&-2\\4&3\end{bmatrix}$.
Solution

Minors: $M_{11}=3$, $M_{12}=4$, $M_{21}=-2$, $M_{22}=1$. Apply signs: $A_{11}=3$, $A_{12}=-4$, $A_{21}=-(-2)=2$, $A_{22}=1$. (Check $|A|=a_{11}A_{11}+a_{12}A_{12}=1\cdot3+(-2)(-4)=3+8=11$, and directly $1\cdot3-(-2)\cdot4=3+8=11$.)

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

6
Worked Example
Evaluate $\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}$ using cofactor expansion along the first row.
Solution

$=1\begin{vmatrix}5&6\\8&9\end{vmatrix}-2\begin{vmatrix}4&6\\7&9\end{vmatrix}+3\begin{vmatrix}4&5\\7&8\end{vmatrix}=1(45-48)-2(36-42)+3(32-35)=1(-3)-2(-6)+3(-3)=-3+12-9=0.$

Answer: $0$

7
Worked Example
Find $\operatorname{adj}A$ for $A=\begin{bmatrix}1&2&3\\0&1&4\\0&0&1\end{bmatrix}$ and verify the leading entry of $A(\operatorname{adj}A)$.
Solution

Cofactors: $A_{11}=\begin{vmatrix}1&4\\0&1\end{vmatrix}=1$, $A_{12}=-\begin{vmatrix}0&4\\0&1\end{vmatrix}=0$, $A_{13}=\begin{vmatrix}0&1\\0&0\end{vmatrix}=0$, $A_{21}=-\begin{vmatrix}2&3\\0&1\end{vmatrix}=-2$, $A_{22}=\begin{vmatrix}1&3\\0&1\end{vmatrix}=1$, $A_{23}=-\begin{vmatrix}1&2\\0&0\end{vmatrix}=0$, $A_{31}=\begin{vmatrix}2&3\\1&4\end{vmatrix}=5$, $A_{32}=-\begin{vmatrix}1&3\\0&4\end{vmatrix}=-4$, $A_{33}=\begin{vmatrix}1&2\\0&1\end{vmatrix}=1$. Transpose the cofactor matrix: $\operatorname{adj}A=\begin{bmatrix}1&-2&5\\0&1&-4\\0&0&1\end{bmatrix}$. Since $|A|=1$, $A(\operatorname{adj}A)=I$, consistent with $A(\operatorname{adj}A)=|A|I$.

Answer: $\operatorname{adj}A=\begin{bmatrix}1&-2&5\\0&1&-4\\0&0&1\end{bmatrix}$, and $A(\operatorname{adj}A)=I$.

8
Worked Example
If $A$ is a $3\times3$ matrix with $|A|=5$, find $|\operatorname{adj}A|$ and $|\operatorname{adj}(\operatorname{adj}A)|$.
Solution

$|\operatorname{adj}A|=|A|^{\,n-1}=5^{2}=25$. For the double adjoint, $\operatorname{adj}(\operatorname{adj}A)=|A|^{\,n-2}A=|A|A$, so $|\operatorname{adj}(\operatorname{adj}A)|=|A|^{n}\cdot|A|=|A|^{(n-1)^{2}}=5^{4}=625.$

Answer: $|\operatorname{adj}A|=25$ and $|\operatorname{adj}(\operatorname{adj}A)|=625$.

9
Worked Example
Find the value of $k$ for which the points $(k,2),(-3,-4),(7,-5)$ are collinear.
Solution

Set the area determinant to $0$: $\begin{vmatrix}k&2&1\\-3&-4&1\\7&-5&1\end{vmatrix}=0$. Expand along row $1$: $k(-4-(-5))-2(-3-7)+1(15+28)=k(1)-2(-10)+43=k+20+43=k+63$. So $k+63=0\Rightarrow k=-63.$

Answer: $k=-63$

10
Worked Example
Find the area of the triangle with vertices $(2,3),(4,7),(6,5)$.
Solution

$\text{Area}=\tfrac12\left|\,\begin{vmatrix}2&3&1\\4&7&1\\6&5&1\end{vmatrix}\,\right|$. Expand along row $1$: $2(7-5)-3(4-6)+1(20-42)=2(2)-3(-2)+(-22)=4+6-22=-12$. Area $=\tfrac12|-12|=6$ square units.

Answer: $6$ square units.

11
Worked Example
If $k$ square units is the area of the triangle with vertices $(0,0),(5,0),(0,k)$ and the area is $15$, find $k$.
Solution

$\text{Area}=\tfrac12\left|\,\begin{vmatrix}0&0&1\\5&0&1\\0&k&1\end{vmatrix}\,\right|=\tfrac12|1\cdot(5k-0)|=\tfrac{5k}{2}$ (taking $k>0$). Set $\tfrac{5k}{2}=15\Rightarrow k=6.$

Answer: $k=6$

12
Worked Example
For $A=\begin{bmatrix}2&-1\\3&4\end{bmatrix}$, verify that $A(\operatorname{adj}A)=|A|I$.
Solution

$|A|=2\cdot4-(-1)\cdot3=8+3=11$. $\operatorname{adj}A=\begin{bmatrix}4&1\\-3&2\end{bmatrix}$. Then $A(\operatorname{adj}A)=\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}4&1\\-3&2\end{bmatrix}=\begin{bmatrix}8+3&2-2\\12-12&3+8\end{bmatrix}=\begin{bmatrix}11&0\\0&11\end{bmatrix}=11I=|A|I.$

Answer: $A(\operatorname{adj}A)=11I=|A|I$ (verified).

Key Points

  • Minor $M_{ij}$ = determinant left after deleting row $i$ and column $j$; cofactor $A_{ij}=(-1)^{i+j}M_{ij}$ (checkerboard sign).
  • $|A|$ = sum of a line's entries times their own cofactors; entries times another line's cofactors sum to $0$.
  • $\operatorname{adj}A$ = transpose of the cofactor matrix; the $2\times2$ shortcut is swap-diagonal, negate-off-diagonal.
  • Defining identity: $A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I$.
  • $|\operatorname{adj}A|=|A|^{\,n-1}$ and $\operatorname{adj}(\operatorname{adj}A)=|A|^{\,n-2}A$ for $n\times n$.
  • Triangle area $=\tfrac12\big|\det\big|$ of the coordinate matrix (always take the absolute value).
  • The same determinant $=0$ is exactly the test that three points are collinear.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The cofactor $A_{ij}$ is defined as:
Explanation: The checkerboard sign $(-1)^{i+j}$ multiplies the minor.
Q2.$A(\operatorname{adj}A)$ equals:
Explanation: The defining property of the adjoint.
Q3.For a $3\times3$ matrix with $|A|=2$, $|\operatorname{adj}A|=$
Explanation: $|\operatorname{adj}A|=|A|^{n-1}=2^{2}=4$.
Q4.Three points are collinear when the area determinant equals:
Explanation: Zero area means the points lie on one line.
Practice more MCQs → 📝 Assignment · 35 marks