Determinants

A single number attached to a square matrix, computable by cofactor expansion along any row or column, that vanishes exactly when the columns are dependent. It scales by the same factor a linear map scales volume, and it multiplies across products, which is what makes it useful rather than merely computable.

Definition

The determinant of a square matrix $A \in R^{n \times n}$ is written $det A$ or $| A |$ . For small sizes:

det (\begin{matrix} a \end{matrix}) = a, det (\begin{matrix} a & b \\ c & d \end{matrix}) = a d - b c .

Cofactor expansion. For $n \geq 2$ , let $M_{i j}$ be the $(n - 1) \times (n - 1)$ matrix obtained by deleting row $i$ and column $j$ , and let the cofactor be $C_{i j} = (- 1)^{i + j} det M_{i j}$ . Then for any fixed row $i$ ,

det A = \sum_{j = 1}^{n} a_{i j} C_{i j},

and for any fixed column $j$ , $det A = \sum_{i = 1}^{n} a_{i j} C_{i j}$ . Every choice of row or column gives the same value, which is a theorem rather than a convention.

Behaviour under row operations.

Operation	Effect on $det$
swap two rows	multiplied by $- 1$
scale a row by $k$	multiplied by $k$
add a multiple of one row to another	unchanged

The third is what makes elimination a practical method: reduce to triangular form without changing the determinant, then read it off.

Triangular matrices. If $A$ is upper or lower triangular, $det A$ is the product of the diagonal entries.

The two properties that matter.

det (A B) = det A \cdot det B, det A \neq 0 ⟺ A is invertible .

The second is the characterisation: $det A = 0$ exactly when the columns are linearly dependent, equivalently when the rank is less than $n$ , equivalently when the map $x \mapsto A x$ has a nontrivial kernel.

Also $det (A^{T}) = det A$ , which is why row and column expansions agree, and $det (A^{- 1}) = 1 / det A$ when the inverse exists.

Assumptions and scope

Only square matrices have determinants. A non-square matrix has a rank but no determinant, and asking for one usually signals a shape error upstream.
$det (A + B)$ is not $det A + det B$ . The determinant is multilinear in the rows separately, not additive in the matrix, and the identity matrices $I$ and $- I$ in dimension 2 already refute the additive version.
Vanishing determinant is not an error. It is the informative case: it says the columns are dependent and the map collapses a direction, which is what an eigenvalue computation looks for.
The sign of $det$ carries orientation, so $| det |$ is the volume factor while $det$ itself also says whether handedness is preserved.
Cofactor expansion is correct at any size and impractical beyond small ones. Choosing it over elimination for a large matrix is a complexity error rather than a mathematical one.

Worked material

Example

Determinants readable without computation

Several matrices give up their determinants on sight. Recognising them saves the arithmetic and, more usefully, says what the value means.

Matrix	$det$	Why
$I_{3}$	$1$	diagonal product; the identity changes no volume
$diag (2, - 3, 5)$	$- 30$	diagonal product $2 (- 3) (5)$
$(\begin{matrix} 3 & 7 & - 1 \\ 0 & - 2 & 5 \\ 0 & 0 & 4 \end{matrix})$	$- 24$	triangular: $3 (- 2) (4)$
$(\begin{matrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 4 & 5 & 6 \end{matrix})$	$0$	a zero row; expanding along it gives nothing
$(\begin{matrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 1 & 2 & 3 \end{matrix})$	$0$	two equal rows
$(\begin{matrix} 2 & 4 \\ 3 & 6 \end{matrix})$	$0$	second row is $1.5$ times the first
$(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$	$- 1$	a single row swap applied to $I$
$(\begin{matrix} 5 & 3 \\ 2 & 4 \end{matrix})$	$14$	$5 (4) - 3 (2)$

Why equal rows force zero. Swapping two rows negates the determinant. Swapping two equal rows leaves the matrix unchanged, so $det A = - det A$ , which over the reals forces $det A = 0$ . The same argument covers proportional rows, since scaling one row first turns the proportional case into the equal case.

Why a zero row forces zero. Expanding along it, every term carries a factor of zero. Geometrically the figure spanned by the rows has been flattened: one edge has length zero, so the volume is zero.

The permutation matrix. $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ swaps the two coordinates. It preserves area, the unit square maps to the unit square, so the magnitude is 1, and it reverses orientation, so the sign is negative. In general a permutation matrix has determinant $+ 1$ for an even permutation and $- 1$ for an odd one, which is the sign of the permutation.

Each value is read from structure rather than computed from entries: a diagonal, a dependency among rows, a single swap. Looking for that structure before starting an expansion is the cheapest step in any determinant calculation, and on a singular matrix it is the whole calculation.

Non-example

Four determinant errors

Adding determinants. " $det (A + B) = det A + det B$ ." Take $A = I$ and $B = - I$ in dimension 2: $det A = 1$ , $det B = 1$ , and $A + B$ is the zero matrix with determinant 0, not 2. The determinant is linear in each row separately with the others fixed, which is a different statement about a different object.

Scaling the matrix as though it were one row. " $det (3 A) = 3 det A$ ." Scaling the matrix scales all $n$ rows, so $det (3 A) = 3^{n} det A$ . For the worked example's $3 \times 3$ matrix that is $27 (- 21) = - 567$ , not $- 63$ .

Losing a sign in the expansion. Expanding along row 1 with entry $a_{12} = - 1$ and minor determinant 7, the learner writes $- 1 \times 7 = - 7$ , forgetting that the position also contributes $(- 1)^{1 + 2} = - 1$ . Both factors apply: $(- 1) (- 1) (7) = + 7$ . The chessboard sign and the entry's own sign are independent, and the error is invisible in the final number unless the computation is checked by a second expansion.

Treating a zero determinant as a failed calculation. " $det = 0$ , so I must have made a mistake." Zero is a legitimate and informative value: the columns are dependent, the rank is below $n$ , the kernel is nontrivial, and the map flattens space. Eigenvalue computations search for the values of $λ$ making $det (A - λ I) = 0$ , so the zero case is the one being hunted rather than an error to correct.

What separates these. The first two mistake which object the linearity applies to. The third is arithmetic with a conceptual cause, not knowing that two independent sign factors are in play. The fourth misreads an informative result as a broken one, and it is the most costly, because it leads to redoing correct work rather than acting on what it found.

Common errors

Common misconception

The determinant is a linear function of the matrix, so $det (A + B) = det A + det B$ and $det (k A) = k det A$ .

Determinants

Assumptions and scope

Worked material

Determinants readable without computation

Four determinant errors

Common errors

Related units

Requires

Connected

Learn this topic

Used in

Sources

Settings

Focus timer

Today

Last seven days

Where the time went

Recent sessions

Settings

Notifications

Your history