Computing determinants

Shear a parallelogram, or slide one side along the other, and its area stays the same. Watch elimination turn it into a rectangle.

Three rules

The determinant as area gave the meaning: the factor by which a matrix scales area. To compute it for a big matrix, use elimination from Elimination, and three rules about rows.

1. Subtract a multiple of one row from another: the determinant does not change. The step is multiplication by an elimination matrix, which is a shear with determinant 1. In the picture at the top of the page, it sheared the parallelogram vertically: each vertical slice slid up or down at full length, so the area stayed 3.
2. Exchange two rows: the determinant changes sign. That is a mirror. After the clip, press the exchange button and watch the meter turn to $-3$.
3. Multiply a row by $c$: the determinant is multiplied by $c$. That stretches one direction.

The same three rules hold for columns, because $\det\mathbf{A}^\top=\det\mathbf{A}$. The second slide in the picture was a column step.

The pivots

Elimination turns $\mathbf{A}$ into an upper triangle $\mathbf{U}$ using only rule 1, and perhaps rule 2. A triangular matrix’s determinant is the product of its diagonal, the pivots: in the picture, the rectangle is 2 by 1.5. So

with a minus sign for each row exchange. For $\begin{bmatrix}2&-1\\1&1\end{bmatrix}$, the pivots are 2 and 1.5, and $2\cdot1.5=3$.

The $3\times3$ matrix of the elimination page, $\mathbf{A}=\begin{bmatrix}1&2&1\\3&8&1\\0&4&1\end{bmatrix}$, had pivots 1, 2 and 5, so $\det\mathbf{A}=10$. The matrix $\begin{bmatrix}1&1&0\\1&0&1\\0&1&1\end{bmatrix}$ has pivots 1, $-1$ and 2, so its determinant is $-2$, the box turned inside out from the area page.

With an exchange: $\begin{bmatrix}0&2\\3&1\end{bmatrix}$ has a zero first pivot. Exchange the rows to get $\begin{bmatrix}3&1\\0&2\end{bmatrix}$, with pivots 3 and 2, product 6. One exchange, so $\det=-6$. The $2\times2$ formula agrees: $0\cdot1-2\cdot3=-6$.

Cofactors

There is also a formula that needs no elimination. Pick a row $p$. For each entry $a_{pq}$ in it, cross out row $p$ and column $q$; the determinant of what is left is the minor $M_{pq}$. Give it the sign $(-1)^{p+q}$, a chessboard of signs starting with $+$ at the top left; the signed minor is the cofactor. Then

For the $3\times3$ matrix $\mathbf{A}$ above, along row 1:

Any row gives the same answer. The cofactors of $\mathbf{A}$ are $4,-3,12$ in row 1, $2,1,-4$ in row 2 and $-6,2,2$ in row 3. Along row 2: $3\cdot2+8\cdot1+1\cdot(-4)=10$. Along row 3: $0+4\cdot2+1\cdot2=10$.

The inverse and Cramer’s rule

Put the cofactors in a matrix and transpose it. That is the adjugate, $\mathrm{adj}\,\mathbf{A}$. Then

Check: row 1 of $\mathbf{A}$ times column 1 of the adjugate is $4-6+12=10$, and times column 2 is $2+2-4=0$. So $\mathbf{A}\,\mathrm{adj}\,\mathbf{A}=10\,\mathbf{I}$. For a $2\times2$ matrix this is the swap-and-change-signs formula of The inverse matrix.

Cramer’s rule solves $\mathbf{A}\mathbf{x}=\mathbf{b}$ with determinants alone. Replace column $q$ of $\mathbf{A}$ by $\mathbf{b}$ to get $\mathbf{A}_q$. Then

For the system $2x-y=4$, $x+y=5$ of Two pictures of a linear system, $\det\mathbf{A}=3$, and

Why programs eliminate

Expanding by cofactors again and again, down to $1\times1$ pieces, makes $n!$ products for an $n\times n$ matrix. Elimination needs about $n^3/3$ steps.

For a $3\times3$ matrix, cofactors are fine. For $n=20$, they need $20!=2.43\times10^{18}$ products. At a billion a second, that is about 77 years. Elimination needs about $20^3/3\approx2700$ steps.

Two warnings about determinants. They do not add: $\det(\mathbf{A}+\mathbf{B})$ is not $\det\mathbf{A}+\det\mathbf{B}$. For the $2\times2$ identity, $\det(\mathbf{I}+\mathbf{I})=\det(2\mathbf{I})=4$, not 2. And a tiny determinant does not mean a matrix is nearly singular. The $100\times100$ matrix $0.1\,\mathbf{I}$ has determinant $0.1^{100}=10^{-100}$, yet inverting it is perfectly safe: its inverse is $10\,\mathbf{I}$. Conditioning measures how close to singular a matrix really is.

Worked example

1. By pivots. $\begin{bmatrix}1&2&1\\3&8&1\\0&4&1\end{bmatrix}$: pivots 1, 2, 5, so $\det=10$.
2. By cofactors. Along row 1: $4-6+12=10$.
3. The inverse. $\mathbf{A}^{-1}=\tfrac1{10}\begin{bmatrix}4&2&-6\\-3&1&2\\12&-4&2\end{bmatrix}$.
4. With an exchange. $\det\begin{bmatrix}0&2\\3&1\end{bmatrix}=-6$.
5. Cramer. $2x-y=4$, $x+y=5$ gives $x=9/3=3$ and $y=6/3=2$.
6. Quick ones. Pivots 2, 3, $-4$ with no exchanges: $\det=-24$. Exchange two rows of a matrix with determinant 5: $-5$. $\det\begin{bmatrix}1&2\\3&4\end{bmatrix}=-2$. A $3\times3$ matrix with $\det\mathbf{A}=2$: $\det(3\mathbf{A})=3^3\cdot2=54$.

Where you’ll meet this

Cramer’s rule is a common way to code where two lines in a drawing meet: one $2\times2$ solve, written out. Programs that build meshes for 3-D models test which side of a plane a point lies on by the sign of a $3\times3$ determinant. Circuit simulators and SciPy never expand cofactors for large systems; they eliminate, as LU factorisation does, and read the determinant from the pivots if they need it.

Reference card