3.3 Cramer's rule, volume, and linear transformations

Big idea

Determinants can solve a system outright. If the coefficient matrix is square and invertible, each unknown equals a ratio of two determinants: replace one column of the coefficient matrix by the right-hand side, take that determinant, and divide by the determinant of the original. That is Cramer’s rule, and it gives you one unknown without computing the others.

The same idea, applied one column at a time to the equation $A\mathbf{x} = \mathbf{e}_j$, produces a closed formula for the inverse. Every entry of $A^{-1}$ is a cofactor of $A$ divided by $\det A$, with the cofactors transposed into place. The matrix of transposed cofactors is called the adjugate.

The third strand is geometric. The absolute value of the determinant of a $2 \times 2$ matrix is the area of the parallelogram its columns span, and the $3 \times 3$ version gives the volume of a parallelepiped. That reading turns the determinant into a scaling factor: the transformation $\mathbf{x} \mapsto A\mathbf{x}$ multiplies the area or volume of any region by $|\det A|$. A determinant of zero means the transformation squashes space flat, which is the geometric face of “not invertible”.

Decoder

Each unknown is the determinant of the coefficient matrix with the corresponding column replaced by the constants, divided by the determinant of the coefficient matrix.

The phrase “the corresponding column” is the whole content. To find the third unknown, replace the third column. To find the first, replace the first. You never replace a row, and you never replace more than one column at once. Each unknown gets its own numerator; the denominator is the same for all of them.

The formula assumes the coefficient matrix is square and its determinant is nonzero. When the determinant is zero, the rule says nothing at all: the system might be inconsistent or might have infinitely many solutions, and you go back to row reduction to find out which.

Definitions and results

Notation for a replaced column. For a square matrix $A$ and a vector $\mathbf{b}$, write $A_i(\mathbf{b})$ for the matrix obtained from $A$ by replacing column $i$ with $\mathbf{b}$. Every other column stays exactly where it was.

Cramer’s rule. If $A$ is $n \times n$ and $\det A \ne 0$, the system $A\mathbf{x} = \mathbf{b}$ has the unique solution with entries

$$ x_i = \frac{\det A_i(\mathbf{b})}{\det A}, \qquad i = 1, 2, \dots, n $$

When to reach for it. Cramer’s rule is the right tool for a small system, for a system whose coefficients are symbols rather than numbers, and for the case where you need one unknown and not the rest. It is the wrong tool for anything large: it costs $n + 1$ determinants of size $n$, while row reduction solves the whole system for far less.

Adjugate. The adjugate of $A$, written $\text{adj}\,A$, is the transpose of the matrix of cofactors. Its entry in row $i$, column $j$ is $C_{ji}$, the cofactor of $A$ at row $j$, column $i$. The swap of indices is the transpose, and forgetting it is the most common error in the formula.

Inverse from the adjugate. If $\det A \ne 0$ then

$$ A^{-1} = \frac{1}{\det A}\,\text{adj}\,A $$

For a $2 \times 2$ matrix this reduces to the familiar rule: swap the diagonal entries, negate the off-diagonal ones, divide by the determinant. For larger matrices the formula is mainly a theoretical tool, since computing $n^2$ cofactors is slower than row reducing $[A \;\; I]$.

Area and volume. The absolute value of the determinant of a $2 \times 2$ matrix equals the area of the parallelogram whose sides are its columns. The absolute value of the determinant of a $3 \times 3$ matrix equals the volume of the parallelepiped whose edges are its columns. A degenerate case, where the columns lie on one line or in one plane, has area or volume zero and determinant zero.

Scaling by a transformation. Let $T(\mathbf{x}) = A\mathbf{x}$ with $A$ square. If $S$ is a region in the plane with finite area, the image $T(S)$ has area $|\det A|$ times the area of $S$. The same statement holds for volumes in three dimensions. The factor does not depend on the region, so one determinant describes what the transformation does to every shape at once.

Sign of the determinant. The absolute value carries the size; the sign carries orientation. A negative determinant means the transformation reverses orientation, turning a counterclockwise loop into a clockwise one.

Worked examples

Cramer’s rule on a $2 \times 2$ system

Solve

$$ 3x_1 + 2x_2 = 7 $$

$$ x_1 + 4x_2 = 9 $$

The coefficient matrix is $A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ with $\det A = 12 - 2 = 10$, which is not zero, so the rule applies.

Replace column one with the constants: $A_1(\mathbf{b}) = \begin{bmatrix} 7 & 2 \\ 9 & 4 \end{bmatrix}$, determinant $28 - 18 = 10$. So $x_1 = 10/10 = 1$.

Replace column two instead: $A_2(\mathbf{b}) = \begin{bmatrix} 3 & 7 \\ 1 & 9 \end{bmatrix}$, determinant $27 - 7 = 20$. So $x_2 = 20/10 = 2$.

Check both equations: $3(1) + 2(2) = 7$ and $1 + 4(2) = 9$. Both hold, so the solution is $(1, 2)$.

An inverse from cofactors

Take

$$ A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{bmatrix} $$

Expand along the first row: $1(-8 - 3) - 0(16 - 12) + 2(2 + 4) = -11 + 12 = 1$. So $\det A = 1$ and the inverse will have whole-number entries.

Now the nine cofactors, sign checkerboard included. Row one: $C_{11} = -8 - 3 = -11$, $C_{12} = -(16 - 12) = -4$, $C_{13} = 2 + 4 = 6$. Row two: $C_{21} = -(0 - 2) = 2$, $C_{22} = 8 - 8 = 0$, $C_{23} = -(1 - 0) = -1$. Row three: $C_{31} = 0 + 2 = 2$, $C_{32} = -(3 - 4) = 1$, $C_{33} = -1 - 0 = -1$.

Transpose that array to get the adjugate, then divide by $\det A = 1$:

$$ A^{-1} = \begin{bmatrix} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{bmatrix} $$

Check the first row of $A$ against the columns of the answer: $1(-11) + 0(-4) + 2(6) = 1$, then $1(2) + 0(0) + 2(-1) = 0$, then $1(2) + 0(1) + 2(-1) = 0$. That is the first row of the identity, and the other two rows work out the same way.

Area and how a transformation changes it

The columns $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$ span a parallelogram. Its area is

$$ \left| \det \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix} \right| = |12 - 1| = 11 $$

Now apply the transformation with matrix $M = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}$, whose determinant is $6$. The predicted area of the image is $6 \cdot 11 = 66$.

Verify by transforming the two sides. $M$ sends $(3, 1)$ to $(2 \cdot 3 + 1 \cdot 1,\; 3 \cdot 1) = (7, 3)$, and sends $(1, 4)$ to $(2 + 4,\; 12) = (6, 12)$. The image parallelogram has area

$$ \left| \det \begin{bmatrix} 7 & 6 \\ 3 & 12 \end{bmatrix} \right| = |84 - 18| = 66 $$

which is the prediction. Note that the factor $6$ would have applied to a disk or a triangle just as well, since it depends only on $M$.

Practice

First the solving rule: build the replaced-column matrices and divide.

Then the inverse formula: cofactors, transpose, divide by the determinant.

Videos

Watch the segment on what a determinant of zero does to a grid. Seeing the plane collapse onto a line is the clearest argument for why the ratio in Cramer’s rule needs a nonzero denominator.

Quiz

Five items on Cramer’s rule, the adjugate inverse, and area scaling.