Home
Articles
Vectors & Linear Algebra
Determinant and inverse

// Article · Linear algebra

How to Find the Determinant and Inverse Matrix (2×2 and 3×3) with Examples

Q: What does a zero determinant mean?

The matrix is singular and has no inverse, and the system Amathbf{x}=mathbf{b} does not have a unique solution: it has either no solution or infinitely many. The rows or columns of the matrix are linearly dependent.

Q: Why do we divide by the determinant when finding the inverse?

Because A operatorname{adj}(A)=det(A) I, so we must divide by det(A) to get the identity matrix. That is impossible when det(A)=0.

The determinant tells you whether a matrix is invertible, and the inverse is used to solve systems of equations such as the currents in a multi-loop circuit. This article walks through the hand calculations with examples checked by multiplying back.

By Ajarn ModUpdated 16 Oct 2026

30-second summary

2×2: $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$
3×3: expand along the first row with cofactors (alternating signs $+\,-\,+$ )
Inverse: $A^{-1}=\dfrac{1}{\det A}\operatorname{adj}(A)$ , which exists only when $\det A\ne0$

The 2×2 determinant

\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc

Example: $\det\begin{pmatrix}4&7\\2&6\end{pmatrix}=4\cdot6-7\cdot2=24-14=10$

The 3×3 determinant by cofactors

Expand along the first row: multiply each entry by the 2×2 determinant that remains when its row and column are removed, with alternating signs $+,-,+$ .

\det A=a_{11}M_{11}-a_{12}M_{12}+a_{13}M_{13}

Example

$A=\begin{pmatrix}2&1&3\\0&-1&4\\1&2&5\end{pmatrix}$

$\det A=2\begin{vmatrix}-1&4\\2&5\end{vmatrix}-1\begin{vmatrix}0&4\\1&5\end{vmatrix}+3\begin{vmatrix}0&-1\\1&2\end{vmatrix}$
$=2(-5-8)-1(0-4)+3(0+1)=-26+4+3$

$\det A=-19$ (not zero, so the inverse exists)

For 3×3 you can also use the Sarrus rule: copy the first two columns to the right, add the products along the three downward diagonals and subtract the products along the three upward diagonals. The result is the same.

The 2×2 inverse (shortcut formula)

\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

Swap $a$ and $d$ , change the signs of $b$ and $c$ , then divide by the determinant.

Example

$N=\begin{pmatrix}4&7\\2&6\end{pmatrix}$

$N^{-1}=\frac{1}{10}\begin{pmatrix}6&-7\\-2&4\end{pmatrix}=\begin{pmatrix}0.6&-0.7\\-0.2&0.4\end{pmatrix}$
Check: the first row of $N$ times the first column of $N^{-1}$
$4(0.6)+7(-0.2)=2.4-1.4=1$

And $4(-0.7)+7(0.4)=0$ , $2(0.6)+6(-0.2)=0$ , $2(-0.7)+6(0.4)=1$ , so $NN^{-1}=I$ ✓

The 3×3 inverse with the adjugate

Steps: (1) find the cofactor matrix $C$ ; (2) swap rows and columns (transpose) to get $\operatorname{adj}(A)=C^{T}$ ; (3) divide by $\det A$ .

Continuing the example

A

above

Find $A^{-1}$

Cofactors $C_{ij}=(-1)^{i+j}M_{ij}$
$C=\begin{pmatrix}-13&4&1\\1&7&-3\\7&-8&-2\end{pmatrix}$
Transpose
$\operatorname{adj}(A)=C^{T}=\begin{pmatrix}-13&1&7\\4&7&-8\\1&-3&-2\end{pmatrix}$
Divide by $\det A=-19$
$A^{-1}=\frac{1}{19}\begin{pmatrix}13&-1&-7\\-4&-7&8\\-1&3&2\end{pmatrix}$

Check: the first row of $A$ times the first column of $A^{-1}$ : $\dfrac{1}{19}\big(2(13)+1(-4)+3(-1)\big)=\dfrac{19}{19}=1$ ✓ and the first row times the second column: $\dfrac{1}{19}\big(2(-1)+1(-7)+3(3)\big)=0$ ✓

What it is used for

Solving systems of equations: $A\mathbf{x}=\mathbf{b}\Rightarrow\mathbf{x}=A^{-1}\mathbf{b}$ , for example the loop currents in a multi-loop electric circuit.
Checking for a unique solution: if $\det A\ne0$ the system has exactly one solution.
Eigenvalues: found from $\det(A-\lambda I)=0$ , used in the stability analysis of state-space systems.

Keep learning in the library

The full lessons are in Thai, with English technical terms and the same formulas.

Frequently asked questions

What does a zero determinant mean?

The matrix is singular and has no inverse, and the system $A\mathbf{x}=\mathbf{b}$ does not have a unique solution: it has either no solution or infinitely many. The rows or columns of the matrix are linearly dependent.

Can the Sarrus rule be used for a 4×4 matrix?

No. The Sarrus rule (drawing diagonals) works only for 3×3. Larger matrices use cofactor expansion or row operations down to a triangular matrix.

Why do we divide by the determinant when finding the inverse?

Because $A\,\operatorname{adj}(A)=\det(A)\,I$ , so we must divide by $\det(A)$ to get the identity matrix. That is impossible when $\det(A)=0$ .

Which method is preferred for finding an inverse in practice?

By hand, use the shortcut formula for 2×2, and the adjugate or row operations (Gauss–Jordan) for 3×3. In computer programs one usually solves $A\mathbf{x}=\mathbf{b}$ by factorisation without forming the inverse explicitly, for better accuracy.