What does it mean if the determinant equals zero?

A zero determinant means the matrix is singular and not invertible; the columns are linearly dependent. Geometrically, the transformation collapses 3D space onto a lower-dimensional subspace. Any system using this matrix will have either no solution or infinitely many solutions instead of a unique solution.

Why does the formula have alternating plus and minus signs?

The alternating signs come from the cofactor expansion pattern, where each minor is multiplied by (-1) raised to the power of (row index plus column index). For the first row—positions (1,1), (1,2), (1,3)—this gives (-1)^{2} = +1, (-1)^{3} = -1, and (-1)^{4} = +1 respectively.

Can I compute the determinant by expanding along a different row or column?

Yes, the determinant is the same regardless of which row or column you expand along. The formula shown here expands along the first row, which is standard, but you can use any row or column with the same alternating sign pattern adjusted for the position of each entry.

How does the determinant relate to whether a matrix has an inverse?

If \det A \neq 0, the matrix has an inverse A^{-1}, and the determinant appears explicitly in the inverse formula: each entry of A^{-1} is divided by \det A. If \det A = 0, the inverse does not exist.

Determinant of a 3x3 Matrix

The determinant of a 3x3 matrix reveals whether it is invertible and how the transformation scales volume in three-dimensional space.

\det A = a(ei - fh) - b(di - fg) + c(dh - eg)

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What each symbol means

What Determinant of a 3x3 Matrix takes

Symbol	Meaning
$A$	The 3×3 matrix, a rectangular array of three rows and three columns where each of the nine entries occupies one position; if you misread any entry or confuse rows with columns, the entire calculation will fail.

When to use it

Use the determinant when you need to solve a system of linear equations, check if a matrix is invertible, or verify that vectors are linearly independent.

Level

Usually taught in: Pre-Calculus

Worked examples

1. Find the determinant of a 3×3 matrix with small integers

Problem

Find

\det A

where

A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{pmatrix}

$ei - fh = 1 \cdot 1 - 2 \cdot 0 = 1 - 0 = 1$
Compute the first $2 \times 2$ minor by taking elements $e = 1$ , $f = 2$ , $h = 0$ , $i = 1$ and calculating $ei - fh$ .
$di - fg = 0 \cdot 1 - 2 \cdot 1 = 0 - 2 = -2$
Compute the second minor with $d = 0$ , $i = 1$ , $f = 2$ , $g = 1$ to get $di - fg = -2$ .
$dh - eg = 0 \cdot 0 - 1 \cdot 1 = 0 - 1 = -1$
Compute the third minor: $d = 0$ , $h = 0$ , $e = 1$ , $g = 1$ gives $dh - eg = -1$ .
$\det A = a(ei - fh) - b(di - fg) + c(dh - eg) = 1(1) - 2(-2) + 3(-1)$
Substitute $a = 1$ , $b = 2$ , $c = 3$ and the three minors into the determinant formula with the $-$ before $b$ .
$\det A = 1 + 4 - 3 = 2$
Evaluate each term: $1 \times 1 = 1$ , then $-2 \times (-2) = +4$ (negative times negative is positive), then $3 \times (-1) = -3$ . Sum: $1 + 4 - 3 = 2$ .

Answer: $\det A = 2$

Since the determinant is positive and nonzero, the matrix is invertible and the linear transformation preserves orientation while scaling volume by a factor of 2. A system of equations using this coefficient matrix has a unique solution.

2. Find the determinant when the matrix has negative entries and zeros

Problem

Calculate

\det A

for

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

$ei - fh = (-2) \cdot (-1) - 3 \cdot 0 = 2 - 0 = 2$
For the first minor, use $e = -2$ , $i = -1$ , $f = 3$ , $h = 0$ . The product of two negatives gives a positive result: $(-2)(-1) = 2$ .
$di - fg = 1 \cdot (-1) - 3 \cdot 2 = -1 - 6 = -7$
For the second minor with $d = 1$ , $i = -1$ , $f = 3$ , $g = 2$ , compute the difference $-1 - 6$ carefully to get $-7$ .
$dh - eg = 1 \cdot 0 - (-2) \cdot 2 = 0 - (-4) = 0 + 4 = 4$
For the third minor with $d = 1$ , $h = 0$ , $e = -2$ , $g = 2$ : subtracting a negative value $-(-4)$ equals addition, so $0 + 4 = 4$ .
$\det A = (-1)(2) - 2(-7) + 0(4) = -2 - (-14) + 0$
Substitute $a = -1$ , $b = 2$ , $c = 0$ and the minors. The coefficient $-b = -2$ applied to the second minor gives $-2(-7) = +14$ .
$\det A = -2 + 14 + 0 = 12$
Simplify: $-2 - (-14) = -2 + 14 = 12$ , and the third term vanishes because $0 \cdot 4 = 0$ . Final answer is $12$ .

Answer: $\det A = 12$

Negative entries and zeros frequently appear in real systems. Even with negative numbers, the determinant is positive and nonzero, confirming invertibility. Whenever a coefficient like $c = 0$ appears, the entire corresponding term drops out, simplifying the calculation.

3. Use the determinant in a road-trip scheduling problem

Problem

A delivery truck travels on a road trip with two highway legs, carrying cargo distributed among three weight classes. A system of three equations relates the distances, speeds, and payload weights using the coefficient matrix

A = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 1 \\ -1 & 1 & 2 \end{pmatrix}

. Find

\det A

to verify the system has a unique solution for the three unknowns.

$ei - fh = 2 \cdot 2 - 1 \cdot 1 = 4 - 1 = 3$
The first minor comes from $e = 2$ , $f = 1$ , $h = 1$ , $i = 2$ , which is the $2 \times 2$ determinant of the lower-right block of $A$ .
$di - fg = 1 \cdot 2 - 1 \cdot (-1) = 2 - (-1) = 2 + 1 = 3$
The second minor uses $d = 1$ , $i = 2$ , $f = 1$ , $g = -1$ . Subtracting a negative $-(-1)$ becomes addition, so $2 + 1 = 3$ .
$dh - eg = 1 \cdot 1 - 2 \cdot (-1) = 1 - (-2) = 1 + 2 = 3$
The third minor uses $d = 1$ , $h = 1$ , $e = 2$ , $g = -1$ . Again subtracting a negative: $1 - (-2) = 3$ .
$\det A = 2(3) - (-1)(3) + 1(3) = 6 - (-3) + 3$
Substitute $a = 2$ , $b = -1$ , $c = 1$ and the three minors. The second term's coefficient is $-b = -(-1) = +1$ , so we compute $-(-1)(3) = +3$ .
$\det A = 6 + 3 + 3 = 12$
Simplify: $6 - (-3) = 6 + 3 = 9$ , then add the third term: $9 + 3 = 12$ . All three minors happened to equal 3, making the calculation elegant.

Answer: $\det A = 12$

A nonzero determinant guarantees that this system of three equations in three unknowns has a unique solution, meaning the truck's weight distribution, speed profile, and distance adjustments across the two highway legs are uniquely determined by the constraints.

Common mistakes

Where Determinant of a 3x3 Matrix usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\det A = a(ei + fh) - b(di + fg) + c(dh + eg)$ with addition signs inside the minors instead of subtraction.	Each $2 \times 2$ minor must use subtraction of the cross products, not addition; using $+$ gives cofactors that do not correspond to the standard matrix structure and produces an incorrect determinant.	Always use subtraction in the minors: $\det A = a(ei - fh) - b(di - fg) + c(dh - eg)$ .
Writing $\det A = a(ei - fh) + b(di - fg) + c(dh - eg)$ where all three terms are added instead of using alternating signs.	The second term must be subtracted, not added; the formula requires alternating signs $(+\ -\ +)$ along the first row, and changing the minus to a plus gives the wrong answer.	Use the correct alternating pattern: $\det A = a(ei - fh) - b(di - fg) + c(dh - eg)$ , with the $b$ term subtracted.
Computing the wrong $2 \times 2$ minors, such as using $af - be$ instead of $ei - fh$ for the first term with coefficient $a$ .	When expanding along coefficient $a$ in position $(1,1)$ , the minor must exclude all entries in row 1 and column 1, leaving only the bottom-right $2 \times 2$ block with entries $e, f, h, i$ ; using entries from $a$ 's row or column produces an incorrect minor.	For the term with coefficient $a = \text{position}(1,1)$ , the minor is always $\begin{vmatrix} e & f \\ h & i \end{vmatrix} = ei - fh$ . Remember: exclude $a$ 's entire row and column.

Tips and when to use something else

Remember the alternating sign pattern along the first row: plus, minus, plus ( $+\ -\ +$ ). This is essential and easy to forget.
If $\det A = 0$ , the matrix is singular (not invertible), and the system has either no solution or infinitely many solutions; switch to Gaussian Elimination to analyze the system further.
Master the $2 \times 2$ determinant formula $ei - fh$ first, since the entire $3 \times 3$ method is built from three of these small determinants.
The absolute value $|\det A|$ tells you by what factor the transformation scales volumes in 3D space; this makes determinants invaluable in physics, engineering, and computer graphics.

Related formulas

Determinant of a 2x2 MatrixA single number computed from a 2x2 matrix that tells whether the matrix is invertible and helps solve systems of linear equations.Matrix MultiplicationMatrix multiplication combines rows from the first matrix with columns of the second to produce a new matrix, useful for solving systems and transformations.Cramer's RuleCramer's Rule uses determinants to solve a square system of linear equations by replacing columns of the coefficient matrix with the constants.Gaussian EliminationGaussian Elimination transforms a system of linear equations into row echelon form to solve for unknown variables systematically.Inverse of a 2x2 MatrixCompute the unique 2×2 matrix that reverses multiplication by another 2×2 matrix, essential for solving systems of linear equations.

Frequently asked questions

What does it mean if the determinant equals zero?: A zero determinant means the matrix is singular and not invertible; the columns are linearly dependent. Geometrically, the transformation collapses 3D space onto a lower-dimensional subspace. Any system using this matrix will have either no solution or infinitely many solutions instead of a unique solution.
Why does the formula have alternating plus and minus signs?: The alternating signs come from the cofactor expansion pattern, where each minor is multiplied by $(-1)$ raised to the power of (row index plus column index). For the first row—positions $(1,1)$ , $(1,2)$ , $(1,3)$ —this gives $(-1)^{2} = +1$ , $(-1)^{3} = -1$ , and $(-1)^{4} = +1$ respectively.
Can I compute the determinant by expanding along a different row or column?: Yes, the determinant is the same regardless of which row or column you expand along. The formula shown here expands along the first row, which is standard, but you can use any row or column with the same alternating sign pattern adjusted for the position of each entry.
How does the determinant relate to whether a matrix has an inverse?: If $\det A \neq 0$ , the matrix has an inverse $A^{-1}$ , and the determinant appears explicitly in the inverse formula: each entry of $A^{-1}$ is divided by $\det A$ . If $\det A = 0$ , the inverse does not exist.

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Reviewed 2026-09-18