Further Maths/Lessons/Matrices/Inverting a 3×3 Matrix
Core Pure · Matrices · Year 1

Inverting a 3×3 Matrix

Cofactors, the adjugate, and one final division by the determinant — a longer process than the 2×2 case, but built from exactly the same ideas.

19 min read 4 worked examples

Cofactors and the adjugate matrix

Definition 22.1 — cofactor

The cofactor $C_{ij}$ of entry $a_{ij}$ is $C_{ij}=(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor — the determinant of the $2\times2$ matrix left after deleting row $i$ and column $j$. The matrix of all cofactors is the cofactor matrix $C$.

Definition 22.2 — adjugate and inverse

The adjugate of $A$ is the transpose of the cofactor matrix: $\text{adj}(A)=C^T$. The inverse is then $$A^{-1}=\frac{1}{\det(A)}\text{adj}(A)$$

Method 22.3 — the five-step process

1. Find $\det(A)$ by expanding along a row or column. 2. Form the matrix of minors $M$: replace every entry with the determinant left after deleting its row and column. 3. Turn $M$ into the matrix of cofactors $C$ by applying the checkerboard of signs $\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}$. 4. Transpose $C$ to get $\text{adj}(A)=C^T$. 5. Divide every entry of $\text{adj}(A)$ by $\det(A)$.

Example 1 · Full 3×3 inverse, worked in detail
Find $A^{-1}$ for $A=\begin{pmatrix}3&1&2\\0&2&1\\1&0&2\end{pmatrix}$.

Step 1 — the determinant. Expanding along the top row:

$\det(A)=3\begin{vmatrix}2&1\\0&2\end{vmatrix}-1\begin{vmatrix}0&1\\1&2\end{vmatrix}+2\begin{vmatrix}0&2\\1&0\end{vmatrix}=3(4-0)-1(0-1)+2(0-2)=12+1-4=9$.

Step 2 — the matrix of minors. Cover the row and column through each entry in turn and take the determinant of the $2\times2$ block left behind:

$$M=\begin{pmatrix}\begin{vmatrix}2&1\\0&2\end{vmatrix}&\begin{vmatrix}0&1\\1&2\end{vmatrix}&\begin{vmatrix}0&2\\1&0\end{vmatrix}\\[8pt]\begin{vmatrix}1&2\\0&2\end{vmatrix}&\begin{vmatrix}3&2\\1&2\end{vmatrix}&\begin{vmatrix}3&1\\1&0\end{vmatrix}\\[8pt]\begin{vmatrix}1&2\\2&1\end{vmatrix}&\begin{vmatrix}3&2\\0&1\end{vmatrix}&\begin{vmatrix}3&1\\0&2\end{vmatrix}\end{pmatrix}=\begin{pmatrix}4&-1&-2\\2&4&-1\\-3&3&6\end{pmatrix}$$

Step 3 — the matrix of cofactors. Multiply each entry of $M$ by the sign in the matching position of $\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}$:

$$C=\begin{pmatrix}4&1&-2\\-2&4&1\\-3&-3&6\end{pmatrix}$$

Step 4 — transpose. Rows become columns to give the adjugate:

$$C^T=\begin{pmatrix}4&-2&-3\\1&4&-3\\-2&1&6\end{pmatrix}$$

Step 5 — divide by the determinant.

$$A^{-1}=\frac19\begin{pmatrix}4&-2&-3\\1&4&-3\\-2&1&6\end{pmatrix}=\begin{pmatrix}\tfrac49&-\tfrac29&-\tfrac13\\[4pt]\tfrac19&\tfrac49&-\tfrac13\\[4pt]-\tfrac29&\tfrac19&\tfrac23\end{pmatrix}$$
Common mistake

Forgetting to transpose the matrix of cofactors before dividing by the determinant is the single most common error — $C$ on its own is not the adjugate, except in the rare case where it happens to be symmetric, which can mask the mistake.

Example 2 · The same five steps, a second matrix
Find $B^{-1}$ for $B=\begin{pmatrix}1&2&2\\2&1&0\\0&1&3\end{pmatrix}$.

Step 1 — the determinant.

$\det(B)=1\begin{vmatrix}1&0\\1&3\end{vmatrix}-2\begin{vmatrix}2&0\\0&3\end{vmatrix}+2\begin{vmatrix}2&1\\0&1\end{vmatrix}=1(3-0)-2(6-0)+2(2-0)=3-12+4=-5$.

Step 2 — the matrix of minors. Working through each position as in Example 1:

$$M=\begin{pmatrix}3&6&2\\4&3&1\\-2&-4&-3\end{pmatrix}$$

Step 3 — the matrix of cofactors. Apply the same checkerboard of signs to $M$:

$$C=\begin{pmatrix}3&-6&2\\-4&3&-1\\-2&4&-3\end{pmatrix}$$

Step 4 — transpose. Notice the off-diagonal entries swap places — $C^T$ is not the same matrix as $C$ here:

$$C^T=\begin{pmatrix}3&-4&-2\\-6&3&4\\2&-1&-3\end{pmatrix}$$

Step 5 — divide by the determinant. Since $\det(B)=-5$ is negative, every entry is divided by $-5$:

$$B^{-1}=\frac{1}{-5}\begin{pmatrix}3&-4&-2\\-6&3&4\\2&-1&-3\end{pmatrix}=\begin{pmatrix}-\tfrac35&\tfrac45&\tfrac25\\[4pt]\tfrac65&-\tfrac35&-\tfrac45\\[4pt]-\tfrac25&\tfrac15&\tfrac35\end{pmatrix}$$
Example 3 · Verifying a 3×3 inverse partially
Using $B$ and $B^{-1}$ from Example 2, verify that the top-left entry of $BB^{-1}$ is $1$.

Row 1 of $B$ is $(1,2,2)$; column 1 of $B^{-1}$ is $\dfrac{1}{-5}(3,-6,2)$.

Dot product: $\dfrac{1}{-5}\big[1(3)+2(-6)+2(2)\big]=\dfrac{1}{-5}(3-12+4)=\dfrac{-5}{-5}=1$ ✓

Example 4 · Recognising when a 3×3 matrix has no inverse
Show that $D=\begin{pmatrix}1&2&3\\2&4&6\\0&1&1\end{pmatrix}$ has no inverse.

$\det(D)=1(4-6)-2(2-0)+3(2-0)=-2-4+6=0$.

Since $\det(D)=0$, $D$ is singular and has no inverse (note also that row 2 is exactly twice row 1, which is another way to spot singularity by inspection).

Exam-style questions

Question 1 9 marks

Find $A^{-1}$ for $A=\begin{pmatrix}1&0&2\\0&1&1\\1&1&0\end{pmatrix}$.

$\det(A)=1(0-1)-0(0-1)+2(0-1)=-1-0-2=-3$ (M1, A1)

Cofactors: $C_{11}=-1$, $C_{12}=1$, $C_{13}=-1$, $C_{21}=2$, $C_{22}=-2$, $C_{23}=-1$, $C_{31}=-2$, $C_{32}=-1$, $C_{33}=1$ (M1, A1, A1)

$\text{adj}(A)=\begin{pmatrix}-1&2&-2\\1&-2&-1\\-1&-1&1\end{pmatrix}$ (M1, A1)

$A^{-1}=\dfrac{1}{-3}\begin{pmatrix}-1&2&-2\\1&-2&-1\\-1&-1&1\end{pmatrix}$ (A1, A1)

Question 2 4 marks

Show that $B=\begin{pmatrix}2&1&3\\4&2&6\\1&0&1\end{pmatrix}$ has no inverse.

$\det(B)=2(2-0)-1(4-6)+3(0-2)=4+2-6=0$ (M1, A1, A1)

Since $\det(B)=0$, $B$ has no inverse. (A1)

Question 3 4 marks

Given a $3\times3$ matrix $A$ with $\det(A)=5$, state the value of $\det(A^{-1})$ and the value of $\det(2A)$, giving reasons.

$\det(A^{-1})=\dfrac{1}{\det(A)}=\dfrac15$ (B1, B1)

For an $n\times n$ matrix, $\det(kA)=k^n\det(A)$; here $n=3$, so $\det(2A)=2^3(5)=40$ (B1, B1)

← Previous
Inverting a 2×2 Matrix