Determinant properties, the adjoint and Cramer's rule

Last week we computed the determinant. This week we use it: telling with a single number whether a matrix is invertible, finding the determinant of products without multiplying anything, writing the inverse as a formula with the adjoint, and solving systems with Cramer’s rule. On the midterm, short questions of the type “det A = −2 is given, find these” almost certainly appear and are the cheapest points.

Golden rules:

1. $\det(cA) = c^n\det A$. Don’t forget the $n$.
2. $\det(AB) = \det A\det B$ but there is no rule for $\det(A + B)$.
3. Adjoint = the transpose of the cofactor matrix.
4. In Cramer’s rule $\mathbf{b}$ replaces a column.

Basic properties

$A$, $B$ are $n \times n$, $c$ is a number:

Example. $P = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, $Q = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}$: $\det P = -2$, $\det Q = 6$, $\det(PQ) = -12$. But $P + Q = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix}$, $\det = 13 \neq -2 + 6$.

PITFALL · det(2A) = 2 det A. For $3 \times 3$, $\det(2A) = 8\det A$; for $4 \times 4$, $16\det A$. A way to remember: in $2A$ every row is multiplied by 2, and one 2 comes out of each row.

Strategy for compound questions. In an expression like "$\det(2A^TB^{-1}A^2)$", first separate the constant ($2^n$), then write the determinant of each factor and multiply: $2^n \cdot \det A \cdot \frac{1}{\det B} \cdot (\det A)^2$.

Determinant and invertibility

$A \text{ is invertible} \iff \det A \neq 0.$

Why: reduce $A$ to RREF with row operations. Each operation multiplies the determinant by a nonzero number or leaves it unchanged. If the RREF is $I$, then $\det \neq 0$; if the RREF has a zero row, $\det = 0$.

So we added one more item to the invertible matrix theorem. For square $A$ the following are equivalent: $A$ is invertible, $\det A \neq 0$, $A\mathbf{x} = \mathbf{0}$ has only the trivial solution, $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$, $A \sim I$.

The question “For which $k$ is the system uniquely solvable?” is now one line for a square system: compute $\det A$ in terms of $k$ and find the values that make it zero (Q7). To tell in which $k$ the system has no solution and in which infinitely many, you need to go back to elimination; the determinant does not distinguish between the two.

The adjoint (adjugate)

Write all the cofactors of $A$ in their places: the cofactor matrix $[C_{ij}]$. Its transpose is the adjoint (adjugate):

$\operatorname{adj}A = [C_{ij}]^T.$

The basic identity:

$A \cdot \operatorname{adj}A = \operatorname{adj}A \cdot A = (\det A)I \qquad\Rightarrow\qquad A^{-1} = \frac{1}{\det A}\operatorname{adj}A.$

The diagonal entries come out as $\det A$ because this is the expansion of a row with its own cofactors. The off-diagonal entries come out 0 because multiplying a row by the cofactors of a different row is computing the determinant of a matrix with two equal rows.

Example. $A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 2 & 1 & 0 \end{bmatrix}$.

C11 = +(1·0 - 3·1) = -3     C12 = -(0·0 - 3·2) = 6      C13 = +(0·1 - 1·2) = -2
C21 = -(2·0 - 1·1) = 1      C22 = +(1·0 - 1·2) = -2     C23 = -(1·1 - 2·2) = 3
C31 = +(2·3 - 1·1) = 5      C32 = -(1·3 - 1·0) = -3     C33 = +(1·1 - 2·0) = 1
det A = a11 C11 + a12 C12 + a13 C13 = 1(-3) + 2(6) + 1(-2) = 7

$[C_{ij}] = \begin{bmatrix} -3 & 6 & -2 \\ 1 & -2 & 3 \\ 5 & -3 & 1 \end{bmatrix} \quad\Rightarrow\quad \operatorname{adj}A = \begin{bmatrix} -3 & 1 & 5 \\ 6 & -2 & -3 \\ -2 & 3 & 1 \end{bmatrix}, \qquad A^{-1} = \frac17\begin{bmatrix} -3 & 1 & 5 \\ 6 & -2 & -3 \\ -2 & 3 & 1 \end{bmatrix}.$

Check: $(A\operatorname{adj}A)_{11} = 1(-3) + 2(6) + 1(-2) = 7$, $(A\operatorname{adj}A)_{12} = 1(1) + 2(-2) + 1(3) = 0$.

For $2 \times 2$ the adjoint formula reduces to the recipe you know: $\operatorname{adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

Determinant of the adjoint. Take the determinant of $A\operatorname{adj}A = (\det A)I_n$: $\det A \cdot \det(\operatorname{adj}A) = (\det A)^n$, so $\det(\operatorname{adj}A) = (\det A)^{n-1}$.

PITFALL · forgetting the transpose. Writing the cofactor matrix directly as $\operatorname{adj}A$ is the most common way to lose points. For a non-symmetric matrix the result is wrong. Check: the off-diagonal entries of $A\operatorname{adj}A$ must come out 0.

Cramer’s rule

If $A$ is $n \times n$ and $\det A \neq 0$, the unique solution of $A\mathbf{x} = \mathbf{b}$ is

$x_i = \frac{\det A_i}{\det A}, \qquad A_i = \text{the matrix with } \mathbf{b} \text{ put in place of the } i\text{th column of } A.$

2×2 example. $2x + 5y = 1$, $x + 3y = 2$.

$\det A = \begin{vmatrix} 2 & 5 \\ 1 & 3 \end{vmatrix} = 1, \quad \det A_1 = \begin{vmatrix} 1 & 5 \\ 2 & 3 \end{vmatrix} = -7, \quad \det A_2 = \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 3 \quad\Rightarrow\quad x = -7,\ y = 3.$

3×3 example. With the $A$ above, $x + 2y + z = 1$, $y + 3z = 5$, $2x + y = 1$. $\det A = 7$.

$\det A_1 = \begin{vmatrix} 1 & 2 & 1 \\ 5 & 1 & 3 \\ 1 & 1 & 0 \end{vmatrix} = 7, \quad \det A_2 = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 5 & 3 \\ 2 & 1 & 0 \end{vmatrix} = -7, \quad \det A_3 = \begin{vmatrix} 1 & 2 & 1 \\ 0 & 1 & 5 \\ 2 & 1 & 1 \end{vmatrix} = 14.$

det A1 (row 1): 1(0 - 3) - 2(0 - 3) + 1(5 - 1)   = -3 + 6 + 4   = 7
det A2 (row 1): 1(0 - 3) - 1(0 - 6) + 1(0 - 10)  = -3 + 6 - 10  = -7
det A3 (row 1): 1(1 - 5) - 2(0 - 10) + 1(0 - 2)  = -4 + 20 - 2  = 14
x = 7/7 = 1,  y = -7/7 = -1,  z = 14/7 = 2

Check (3rd equation): $2(1) + (-1) = 1$.

When to use Cramer? For small systems (2×2, 3×3), especially when only one variable is asked. For large systems, Gaussian elimination needs far fewer operations. If $\det A = 0$ Cramer says nothing: the system has either no solution or infinitely many, and you decide with elimination.

Application: area

The area of the triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is

$\text{Area} = \frac12\left|\det\begin{bmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{bmatrix}\right|.$

If the determinant is 0 the three points lie on the same line (collinear). The same idea in $2 \times 2$: $\left|\det[\mathbf{u}\ \mathbf{v}]\right|$ is the area of the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$.

Common mistakes