A 3 by 3 matrix is one of the most fundamental objects in linear algebra. This guide covers everything you need to master the 3×3 array in 2026: how to compute its determinant, find its inverse, understand eigenvalues, and apply it to real problems. We’ll include step-by-step examples, common mistakes to avoid, and expert tips to save you time.
Table of Contents
- What Is a 3 by 3 Matrix?
- Properties of 3×3 Matrices
- How to Find the Determinant of a 3×3 Matrix
- Step-by-Step Solved Example: Determinant Using Cofactor Expansion
- How to Check if a 3×3 Matrix is Invertible (Non-Singular)
- How to Find the Inverse of a 3×3 Matrix
- Inverse of a 3×3 Matrix Using the Adjugate Method
- Using 3×3 Matrices with Cramer’s Rule
- Eigenvalues and Eigenvectors of a 3×3 Matrix
- Real-World Applications of the 3×3 Array
- Practice Problems with Solutions
- Computational Tools for 3×3 Matrices
- Common Mistakes with 3 by 3 Matrices
- Frequently Asked Questions
🔑 Key Takeaways
- A 3 by 3 matrix has 3 rows and 3 columns — it’s a square matrix.
- The determinant of a 3×3 block requires a specific formula; it’s zero if the matrix is singular.
- Inverse exists only when the determinant is nonzero — use the cofactor method or row reduction.
- Eigenvalues of a 3×3 array are found by solving the characteristic polynomial, which is cubic.
- This square matrix is essential in computer graphics, physics, and solving linear systems of three equations.
- Properties like det(AB)=det(A)det(B) apply, and row operations affect the determinant predictably.
- The adjugate method provides a formula for the inverse using minors, cofactors, and transpose.
- Cramer’s rule uses determinants to solve 3×3 linear systems efficiently when the matrix is invertible.
What Is a 3 by 3 Matrix?
A 3 by 3 matrix is a rectangular array with three rows and three columns. We usually write it as:
$$ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$
The entries are typically real numbers. In linear algebra, such a matrix represents a linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. For example, rotating a 3D object by 90 degrees around the $z$-axis uses a specific rotation matrix.
You’ll encounter 3 by 3 matrices in systems of three equations, computer graphics, physics (inertia tensors), and data science (covariance matrices). Understanding how to manipulate this 3×3 array is a core skill.
Properties of 3×3 Matrices
Knowing the properties of a 3 by 3 matrix makes computations faster and builds a deeper understanding of linear algebra. Here are the most important ones, with examples.
1. Determinant of a Product
For any two 3×3 matrices A and B, $$\det(AB) = \det(A) \cdot \det(B).$$ This property is critical when studying transformations: the volume scaling factor of a composition is the product of individual scaling factors.
Example: Let $A = \begin{bmatrix}1&0&2\\0&1&3\\0&0&4\end{bmatrix}$ and $B = \begin{bmatrix}2&0&0\\0&3&0\\0&0&5\end{bmatrix}$. Then $\det(A)=4$, $\det(B)=30$, so $\det(AB)=4\cdot30=120$. Direct computation of AB gives diagonal entries 2,3,20 → determinant $2\cdot3\cdot20=120$, confirming the property.
2. Effect of Row Swapping
Swapping two rows of a 3 by 3 matrix changes the sign of the determinant. This is used in row reduction and explains why the determinant is alternating. Similarly, swapping columns also flips the sign.
Example: $A = \begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$ has $\det(A)=0$. Swap row1 and row2: $A’ = \begin{bmatrix}4&5&6\\1&2&3\\7&8&9\end{bmatrix}$, $\det(A’)=0$ (still zero, but sign would have flipped if nonzero). Try with $B = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}$, $\det(B)=6$. Swap row1 and row3: $B’ = \begin{bmatrix}0&0&3\\0&2&0\\1&0&0\end{bmatrix}$, $\det(B’)=-6$.
3. Scalar Multiplication
If you multiply a single row (or column) of a 3×3 matrix by a scalar $c$, the determinant is multiplied by $c$. Multiplying the entire matrix by $c$ multiplies the determinant by $c^3$ (since all three rows are scaled).
Example: $A = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}$, $\det(A)=6$. Multiply row1 by 5: $A’ = \begin{bmatrix}5&0&0\\0&2&0\\0&0&3\end{bmatrix}$, $\det(A’)=5\cdot2\cdot3=30 = 5\cdot6$. Multiply entire matrix by 2: $2A = \begin{bmatrix}2&0&0\\0&4&0\\0&0&6\end{bmatrix}$, $\det(2A)=2^3\cdot6=48$.
4. Adding a Multiple of One Row to Another
Adding a scalar multiple of one row to another row does not change the determinant. This is the fundamental operation in Gaussian elimination that preserves the determinant (up to sign from row swaps).
Example: $A = \begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$, $\det(A)=0$. Add $2\cdot$row1 to row2: $\begin{bmatrix}1&2&3\\6&9&12\\7&8&9\end{bmatrix}$, determinant still 0.
These properties matter because they allow us to simplify determinants before computing them, check invertibility, and understand the geometry of transformations.
How to Find the Determinant of a 3×3 Matrix
The determinant of a 3 by 3 matrix is a single number that tells you if the matrix is invertible (nonzero) and describes the volume scaling factor of the linear transformation. Let’s use the matrix:
$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} $$
The formula for the determinant of this 3×3 array is:
$$ \det(A) = a_{11}(a_{22}a_{33} – a_{23}a_{32}) – a_{12}(a_{21}a_{33} – a_{23}a_{31}) + a_{13}(a_{21}a_{32} – a_{22}a_{31}) $$
Plugging in the numbers:
\begin{align*} \det(A) &= 1(5\cdot9 – 6\cdot8) – 2(4\cdot9 – 6\cdot7) + 3(4\cdot8 – 5\cdot7) \\ &= 1(45-48) – 2(36-42) + 3(32-35) \\ &= -3 -2(-6) + 3(-3) = -3 +12 -9 = 0 \end{align*}
Here the determinant is zero, so this particular matrix is singular — it has no inverse. In practice, a zero determinant means the rows are linearly dependent (in this case, the matrix is known to be rank 2).
Step-by-Step Solved Example: Determinant Using Cofactor Expansion
Cofactor expansion (also called Laplace expansion) is a general method that works for any $n \times n$ matrix. For a 3 by 3 matrix, it produces the same formula as the direct one but gives you a systematic way to choose a row or column for simplification. Let’s work through a full example.
Example 1: Basic Cofactor Expansion
Compute the determinant of $$A = \begin{bmatrix} 4 & 0 & 1 \\ -2 & 3 & 5 \\ 1 & -1 & 2 \end{bmatrix}$$ using cofactor expansion along the first row.
Step 1: For each element in row 1, compute its minor (determinant of the 2×2 matrix after removing row i and column j).
$$ M_{11} = \det\begin{bmatrix}3 & 5\\-1 & 2\end{bmatrix}= (3)(2) – (5)(-1)=6+5=11 $$
$$ M_{12} = \det\begin{bmatrix}-2 & 5\\1 & 2\end{bmatrix}= (-2)(2) – (5)(1)= -4 -5 = -9 $$
$$ M_{13} = \det\begin{bmatrix}-2 & 3\\1 & -1\end{bmatrix}= (-2)(-1) – (3)(1)= 2 – 3 = -1 $$
Step 2: Apply signs $(-1)^{i+j}$ to get cofactors: $C_{11}=+11$, $C_{12}=-(-9)=9$, $C_{13}=+(-1)=-1$.
Step 3: Multiply by the original elements and sum:
$$ \det(A) = 4\cdot 11 + 0\cdot 9 + 1\cdot(-1) = 44 + 0 – 1 = 43 $$
The determinant is 43. Let’s verify with the Rule of Sarrus: Copy first two columns to the right:
$$ \begin{bmatrix}4 & 0 & 1 & 4 & 0\\ -2 & 3 & 5 & -2 & 3\\ 1 & -1 & 2 & 1 & -1\end{bmatrix} $$
Downward diagonals: $(4\cdot3\cdot2)=24$, $(0\cdot5\cdot1)=0$, $(1\cdot(-2)\cdot(-1))=2$. Sum = 26.
Upward diagonals: $(1\cdot3\cdot1)=3$, $(4\cdot5\cdot(-1))=-20$, $(0\cdot(-2)\cdot2)=0$. Sum = -17.
Total: $26 – (-17) = 43$ ✓. Both methods agree. Cofactor expansion is more systematic when you want to choose a row with zeros (here row 1 had a zero, making one term vanish). Sarrus is a quick visual check but only works for 3×3.
Example 2: Determinant with Fractions and Decimals
Find the determinant of $$B = \begin{bmatrix} 1.5 & 0 & -2.5 \\ 0.25 & 0.5 & 0 \\ 3 & -1 & 2 \end{bmatrix}$$ using cofactor expansion along the second row (because it has a zero).
Step 1: Row 2 elements: $b_{21}=0.25$, $b_{22}=0.5$, $b_{23}=0$.
Minors for row 2:
$$ M_{21} = \det\begin{bmatrix}0 & -2.5\\-1 & 2\end{bmatrix}= (0)(2) – (-2.5)(-1)= -2.5 $$
$$ M_{22} = \det\begin{bmatrix}1.5 & -2.5\\3 & 2\end{bmatrix}= (1.5)(2) – (-2.5)(3)= 3 + 7.5 = 10.5 $$
$$ M_{23} = \det\begin{bmatrix}1.5 & 0\\3 & -1\end{bmatrix}= (1.5)(-1) – (0)(3)= -1.5 $$
Step 2: Cofactors: $C_{21}=(-1)^{2+1}(-2.5)= -(-2.5)=2.5$, $C_{22}=(-1)^{2+2}(10.5)=10.5$, $C_{23}=(-1)^{2+3}(-1.5)= -(-1.5)=1.5$.
Step 3: $\det(B)=0.25\cdot2.5 + 0.5\cdot10.5 + 0\cdot1.5 = 0.625 + 5.25 + 0 = 5.875$
You can verify using Sarrus or a calculator. Cofactor expansion with decimals works the same, just handle fractions carefully.
How to Check if a 3×3 Matrix is Invertible (Non-Singular)
A 3 by 3 matrix is invertible (non-singular) if and only if its determinant is nonzero. Here are quick tests:
- Determinant ≠ 0 → invertible.
- Rank = 3 → all rows and columns are linearly independent.
- Rows/columns are linearly independent → no row is a linear combination of the other two.
- The matrix has full rank (3).
If determinant = 0, the matrix is singular (not invertible). For example, the matrix
$$ S = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} $$
has determinant 0. Why? Row3 = Row2 + (Row1 – Row2) actually Row3 = Row2 + (Row1 – Row2) = Row1? No, check: row3 = row2 + row1 – something. More clearly, row3 = 2*row2 – row1 gives (2*4-1, 2*5-2, 2*6-3) = (7,8,9) yes. So rows are linearly dependent, rank < 3, determinant = 0, no inverse. Geometrically, the three column vectors lie in a plane (zero volume).
Another singular example: $T = \begin{bmatrix} 2 & -1 & 0 \\ 4 & -2 & 0 \\ 1 & 3 & 5 \end{bmatrix}$ has $\det(T)=0$ because the first two columns are multiples (column2 = -0.5 * column1). Quick check: compute determinant to confirm zero.
Always compute the determinant first – it’s the fastest way to check invertibility for a 3 by 3 matrix.
How to Find the Inverse of a 3×3 Matrix
Only a 3 by 3 matrix with nonzero determinant has an inverse. The inverse of such a matrix $A$ is denoted $A^{-1}$ and satisfies $AA^{-1} = I$, where $I$ is the $3 \times 3$ identity matrix.
There are two main methods: the cofactor method (adjugate) and row reduction (Gauss-Jordan). I’ll show the cofactor method because it’s explicit and builds intuition.
Consider this invertible 3×3 block:
$$ B = \begin{bmatrix} 2 & -1 & 0 \\ 4 & 3 & -2 \\ 1 & 0 & 5 \end{bmatrix} $$
First compute its determinant:
$$ \det(B) = 2(3\cdot5 – (-2)\cdot0) – (-1)(4\cdot5 – (-2)\cdot1) + 0(4\cdot0 – 3\cdot1) = 2\cdot15 + 1\cdot(20+2) = 30 + 22 = 52 $$
Since $\det(B) = 52 \neq 0$, the inverse exists. Next, find the matrix of minors for this 3×3 array. For each element, calculate the determinant of the $2 \times 2$ submatrix after removing its row and column. For example, the minor for $b_{11}=2$ is:
$$ M_{11} = \det\begin{bmatrix} 3 & -2 \\ 0 & 5 \end{bmatrix} = 15 $$
Then apply the cofactor sign pattern (+, –, +; –, +, –; +, –, +) to get the cofactor matrix. Then transpose to form the adjugate. Finally, divide by the determinant:
$$ B^{-1} = \frac{1}{\det(B)} \text{adj}(B) $$
After full calculation the inverse of this matrix $B$ is:
$$ B^{-1} = \frac{1}{52} \begin{bmatrix} 15 & 5 & 2 \\ -22 & 10 & 4 \\ -3 & 1 & 10 \end{bmatrix} $$
You can verify by multiplying $B B^{-1}$: you’ll get the $3\times3$ identity matrix.
Inverse of a 3×3 Matrix Using the Adjugate Method
The adjugate method is the formal procedure for finding the inverse of a 3 by 3 matrix. Let’s walk through it in full detail with a new algebraic example.
Let $$C = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \\ 0 & 2 & -2 \end{bmatrix}$$
Step 1: Compute the determinant
Use cofactor expansion along row 1: $\det(C)=1(3\cdot(-2) – 1\cdot2) – 0( (-1)(-2) – 1\cdot0) + 2( (-1)\cdot2 – 3\cdot0) = 1(-6-2) + 2(-2-0) = -8 -4 = -12$
$\det(C) = -12 \neq 0$, so inverse exists.
Step 2: Matrix of minors
For each element, compute the 2×2 determinant of the submatrix after removing its row and column.
$$ M = \begin{bmatrix} \det\begin{bmatrix}3&1\\2&-2\end{bmatrix} & \det\begin{bmatrix}-1&1\\0&-2\end{bmatrix} & \det\begin{bmatrix}-1&3\\0&2\end{bmatrix} \\ \det\begin{bmatrix}0&2\\2&-2\end{bmatrix} & \det\begin{bmatrix}1&2\\0&-2\end{bmatrix} & \det\begin{bmatrix}1&0\\0&2\end{bmatrix} \\ \det\begin{bmatrix}0&2\\3&1\end{bmatrix} & \det\begin{bmatrix}1&2\\-1&1\end{bmatrix} & \det\begin{bmatrix}1&0\\-1&3\end{bmatrix} \end{bmatrix} $$
Compute each:
$$ M = \begin{bmatrix} (3)(-2)-(1)(2)=-6-2=-8 & (-1)(-2)-(1)(0)=2-0=2 & (-1)(2)-(3)(0)=-2-0=-2 \\ (0)(-2)-(2)(2)=0-4=-4 & (1)(-2)-(2)(0)=-2-0=-2 & (1)(2)-(0)(