On this pageMotivation·Product Theorem·Invertibility Test·Transpose·Adjugate & Inverse·Cramer’s Rule·Exercises

MATH-120 · Linear Algebra · Lecture 1023 June 2026

Determinants and Matrix Inverses

The product theorem, the invertibility test, the adjugate formula for the inverse, and Cramer’s rule

§1A Short Motivation

So far we have learned how to compute a determinant. But the determinant is much more than a number you grind out by cofactor expansion — it is a detective. A single number tells you whether a matrix can be inverted, how two matrices interact when multiplied, and even lets you write down an exact formula for the solution of a linear system.

Here is the central question of this lecture. Suppose you multiply two matrices $A$ and $B$. The matrices themselves can be complicated, and computing $AB$ takes work. Yet their determinants behave in the simplest way you could hope for:

$$\det(AB) = (\det A)(\det B).$$

The determinant of a product is just the product of the determinants. This one clean rule unlocks almost everything else in this section: a test for invertibility, a formula for the inverse, and Cramer's rule for solving systems. Let us build it up step by step.

§2The Product Theorem

Theorem 3.2.1 — Product Theorem

If $A$ and $B$ are $n\times n$ matrices, then $$\det(AB) = (\det A)(\det B).$$

In words: to find the determinant of a product, you do not have to multiply the matrices first. Just take the determinant of each one and multiply those two numbers.

Example 1Checking the Product Theorem

Let $A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 5 \\ 2 & 1 \end{bmatrix}$.

First the easy way (each determinant separately):

$$\det A = (2)(4) - (1)(3) = 5, \qquad \det B = (1)(1) - (5)(2) = -9,$$

so the theorem predicts $\det(AB) = (5)(-9) = -45$.

Now check directly. Multiplying,

$$AB = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 5 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 11 \\ 11 & 19 \end{bmatrix}, \quad \det(AB) = (4)(19) - (11)(11) = 76 - 121 = -45. \;\checkmark$$

Both routes give $-45$, but the first one never required multiplying the matrices.

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A handy consequence

Applying the theorem to $A$ and itself, $\det(A^2) = (\det A)^2$, and more generally $\det(A^k) = (\det A)^k$ for any positive integer $k$. So powers of a matrix have easily predictable determinants.

§3The Invertibility Test

The Product Theorem immediately answers a question we have been circling around: which matrices have an inverse?

Theorem 3.2.2 — Determinant Test for Invertibility

An $n\times n$ matrix $A$ is invertible if and only if $\det A \neq 0$. When $A$ is invertible, $$\det(A^{-1}) = \frac{1}{\det A}.$$

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Why the formula for det(A⁻¹) is true

If $A$ is invertible then $AA^{-1} = I$. Take determinants of both sides and use the Product Theorem: $\det(A)\det(A^{-1}) = \det(I) = 1$, so $\det(A^{-1}) = \dfrac{1}{\det A}$. Notice this also forces $\det A \neq 0$: if $\det A$ were $0$, the left side would be $0$, never equal to $1$. So a matrix with zero determinant cannot possibly have an inverse.

Example 2Example 3.2.2 — For which c is A invertible?

For which values of $c$ does $A = \begin{bmatrix} 1 & 0 & -c \\ -1 & 3 & 1 \\ 0 & 2c & -4 \end{bmatrix}$ have an inverse?

§4Determinant of the Transpose

Theorem 3.2.3

If $A$ is any square matrix, then $\det(A^{\mathsf{T}}) = \det A$.

This is the reason that every property we proved for rows also holds for columns: transposing turns rows into columns without changing the determinant.

Example 3Transpose has the same determinant

Let $A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & 1 \\ 5 & 0 & -2 \end{bmatrix}$, $A^{\mathsf{T}} = \begin{bmatrix} 2 & 0 & 5 \\ 1 & 4 & 0 \\ 3 & 1 & -2 \end{bmatrix}$.

We computed $\det A = -81$ earlier (Lecture 9). Expanding $\det A^{\mathsf{T}}$ along its first column:

$$\det A^{\mathsf{T}} = 2\begin{vmatrix} 4 & 0 \\ 1 & -2 \end{vmatrix} - 1\begin{vmatrix} 0 & 5 \\ 1 & -2 \end{vmatrix} + 3\begin{vmatrix} 0 & 5 \\ 4 & 0 \end{vmatrix} = 2(-8) - 1(-5) + 3(-20) = -16 + 5 - 60 = -81.$$

So $\det A^{\mathsf{T}} = \det A = -81$. The point is that you never get a different answer by transposing.

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Application

Combined with the Product Theorem, Theorem 3.2.3 gives $\det(A^{\mathsf{T}}A) = \det(A^{\mathsf{T}})\det(A) = (\det A)^2 \geq 0$. So a matrix of the form $A^{\mathsf{T}}A$ always has a non-negative determinant — a fact used constantly in statistics and least-squares problems.

§5The Adjugate and a Formula for the Inverse

We now build a genuine formula for $A^{-1}$ — not an algorithm, but an explicit expression using determinants. To get there we need one new idea: the adjugate. Let us assemble it slowly.

$\textbf{Step 1 — recall the cofactor.}$ For a square matrix $A$, the $(i,j)$-cofactor $c_{ij}(A)$ is the signed minor we met in Lecture 9: delete row $i$ and column $j$, take the determinant of what remains, and attach the sign $(-1)^{i+j}$.

$\textbf{Step 2 — collect the cofactors into a matrix.}$ Replace every entry of $A$ by its cofactor. The result is the cofactor matrix, written $[c_{ij}(A)]$. It is the same size as $A$.

$\textbf{Step 3 — transpose it.}$ The adjugate of $A$ is the transpose of the cofactor matrix.

Definition 3.3 — Adjugate

The adjugate of $A$, written $\operatorname{adj}(A)$, is the transpose of the cofactor matrix: $$\operatorname{adj}(A) = \big[c_{ij}(A)\big]^{\mathsf{T}}.$$

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The transpose is easy to forget — here is how to keep it straight

The cofactor of position $(i,j)$ is computed at row $i$, column $j$, but in the adjugate it goes into row $j$, column $i$. In short: compute cofactors in the normal order, then flip across the main diagonal. For $2\times2$ matrices this gives the familiar swap-and-negate rule: $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \Rightarrow \operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. (Swap $a$ and $d$; put minus signs on $b$ and $c$.)

Example 4Warm-up — a 2×2 adjugate and inverse

Let $D = \begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix}$. Then $\det D = (4)(1) - (2)(3) = -2$, and by the swap-and-negate rule $\operatorname{adj}(D) = \begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix}$.

Quick check that $D\operatorname{adj}(D) = (\det D)I$:

$$\begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} = -2I. \;\checkmark$$

So $D^{-1} = \dfrac{1}{-2}\begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} -\tfrac12 & 1 \\ \tfrac32 & -2 \end{bmatrix}$.

★ 5Example 3.2.6 — computing an adjugate

Compute the adjugate of $A = \begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & 5 \\ -2 & -6 & 7 \end{bmatrix}$ and verify $A(\operatorname{adj}A) = (\operatorname{adj}A)A$.

Theorem 3.2.4 — Adjugate Formula

If $A$ is any square matrix, then $$A(\operatorname{adj}A) = (\det A)I = (\operatorname{adj}A)A.$$

In particular, if $\det A \neq 0$, the inverse of $A$ is $$A^{-1} = \frac{1}{\det A}\operatorname{adj}(A).$$

Example 6Finding an inverse with the adjugate formula

Let $C = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. First, $\det C$. Expand along the first column:

$$\det C = 2\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} - 1\begin{vmatrix} 1 & 0 \\ 1 & 2 \end{vmatrix} = 2(5) - 1(2) = 8.$$

Computing the nine cofactors (same method as Example 3.2.6) gives the cofactor matrix, whose transpose is

$$\operatorname{adj}(C) = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 5 \end{bmatrix}.$$

By the Adjugate Formula,

$$C^{-1} = \frac{1}{8}\begin{bmatrix} 5 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 5 \end{bmatrix} = \begin{bmatrix} \tfrac58 & -\tfrac14 & \tfrac18 \\ -\tfrac14 & \tfrac12 & -\tfrac14 \\ \tfrac18 & -\tfrac14 & \tfrac58 \end{bmatrix}.$$

⚠️ Is this a good way to compute inverses? No.

For a $10\times10$ matrix, the adjugate needs $10^2 = 100$ determinants of $9\times9$ matrices, which is a huge amount of work. Row reduction finds the inverse with far less effort. So Theorem 3.2.4 is not a practical numerical tool. Its value is theoretical: it gives an exact, closed-form expression for $A^{-1}$ that we can reason about in proofs.

§6Cramer's Rule

$\textbf{The need.}$ Suppose you have a square system $A\mathbf{x} = \mathbf{b}$ with $\det A \neq 0$. We already know it has a unique solution. Sometimes, though, you only want one of the unknowns — say $x_1$ — and you would like a direct formula for it rather than solving the whole system. Cramer's rule provides exactly that: a determinant formula for each variable, one at a time.

Theorem 3.2.5 — Cramer's Rule

Let $A\mathbf{x} = \mathbf{b}$ be a system of $n$ equations in $n$ unknowns with $\det A \neq 0$. Then the unique solution is given by $$x_i = \frac{\det A_i}{\det A}, \qquad i = 1, 2, \ldots, n,$$ where $A_i$ is the matrix obtained from $A$ by replacing its $i$-th column with the column of constants $\mathbf{b}$.

In plain language: to find $x_i$, swap the $i$-th column of $A$ for the right-hand side $\mathbf{b}$, take that determinant, and divide by $\det A$.

Example 7Example 3.2.9 — finding just one variable

Find $x_1$ for the system

$$5x_1 + x_2 - x_3 = 4, \quad 9x_1 + x_2 - x_3 = 1, \quad x_1 - x_2 + 5x_3 = 2.$$

Example 8Solving a 2×2 system completely

Solve $3x + 2y = 7$, $x - y = 1$. Here $A = \begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix}$, $\mathbf{b} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$, and $\det A = (3)(-1) - (2)(1) = -5$.

Replace column 1 by $\mathbf{b}$ for $x$, and column 2 by $\mathbf{b}$ for $y$:

$$x = \frac{\begin{vmatrix} 7 & 2 \\ 1 & -1 \end{vmatrix}}{-5} = \frac{-7-2}{-5} = \frac{-9}{-5} = \frac{9}{5}, \qquad y = \frac{\begin{vmatrix} 3 & 7 \\ 1 & 1 \end{vmatrix}}{-5} = \frac{3-7}{-5} = \frac{-4}{-5} = \frac{4}{5}.$$

So $(x, y) = \left(\tfrac{9}{5}, \tfrac{4}{5}\right)$.

★ 9A full 3×3 solution by Cramer's rule

Solve $2x + y + z = 7$, $x - y + z = 2$, $x + 2y - z = 2$.

⚠️ Is Cramer's rule a good way to solve systems? Again, no.

It looks attractive because it can give one variable without the others. But for a large system, computing even one of these determinants costs about as much work as solving the whole system by Gaussian elimination. Worse, Cramer's rule only works when the coefficient matrix is square and invertible, whereas elimination handles every case. Like the adjugate formula, Cramer's rule is a beautiful theoretical result, not a practical computational method.

Summary of Lecture 10

$\textbf{Product Theorem:}$ $\det(AB) = (\det A)(\det B)$. Consequence: $\det(A^k) = (\det A)^k$.
$\textbf{Invertibility test:}$ $A$ is invertible $\iff \det A \neq 0$, and then $\det(A^{-1}) = 1/\det A$.
$\textbf{Transpose:}$ $\det(A^{\mathsf{T}}) = \det A$ — this is why row and column rules match.
$\textbf{Adjugate:}$ $\operatorname{adj}(A) = [c_{ij}(A)]^{\mathsf{T}}$ (cofactor matrix, then transpose).
$\textbf{Adjugate Formula:}$ $A(\operatorname{adj}A) = (\det A)I$, so $A^{-1} = \tfrac{1}{\det A}\operatorname{adj}(A)$ when $\det A \neq 0$.
$\textbf{Cramer's Rule:}$ $x_i = \dfrac{\det A_i}{\det A}$, where $A_i$ replaces column $i$ of $A$ by $\mathbf{b}$.

§7Solutions to Section 3.2 Exercises

Exercise 3.2.1Find the adjugate of each matrix

$\textbf{Recipe.}$ For each entry compute its cofactor (signed minor), assemble the cofactor matrix, then transpose it to get the adjugate. The sign pattern is $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$.

Exercise 3.2.2Which real values of c make each matrix invertible?

A matrix is invertible exactly when its determinant is nonzero, so we compute the determinant as a function of $c$ and see where it is zero.

Exercise 3.2.3Given det A = −1, det B = 2, det C = 3, evaluate

$\textbf{Tools.}$ $\det(XY) = \det X\det Y$, $\det(X^{\mathsf{T}}) = \det X$, and $\det(X^{-1}) = 1/\det X$. Determinants are ordinary numbers, so we can freely rearrange the product.

Exercise 3.2.4Let A and B be invertible n×n matrices. Evaluate

Exercise 3.2.5A is 3×3 and det(2A⁻¹) = −4 = det(A³(B⁻¹)ᵀ). Find det A and det B

Exercise 3.2.6Let A with det A = 3. Compute det(2B⁻¹) and det(2C⁻¹)

Throughout, $A = \begin{bmatrix} a & b & c \\ p & q & r \\ u & v & w \end{bmatrix}$ with $\det A = 3$.

Exercise 3.2.7If det[[a,b],[c,d]] = −2, calculate a 3×3 determinant

Given $\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = -2$, calculate $\det\begin{bmatrix} 2 & -2 & 0 \\ c+1 & -1 & 2a \\ d-2 & 2 & 2b \end{bmatrix}$.

Exercise 3.2.8Solve each system by Cramer's rule

$\textbf{Reminder.}$ For $A\mathbf{x} = \mathbf{b}$, $x_i = \dfrac{\det A_i}{\det A}$, where $A_i$ replaces column $i$ of $A$ with $\mathbf{b}$.

Exercise 3.2.10What can be said about det A if…

$\textbf{Key idea.}$ Take determinants of both sides of the given equation and use $\det(XY) = \det X\det Y$, $\det(X^{\mathsf{T}}) = \det X$, and $\det(uX) = u^n\det X$. Let $t = \det A$; each part becomes a small equation in $t$.

Looking ahead

The determinant is now a full toolkit: it tests invertibility, multiplies cleanly, and hands you exact formulas for the inverse and for solutions.

With the product theorem, the adjugate, and Cramer's rule in hand, we are ready to leave the world of computation behind and start asking deeper structural questions — vector spaces, linear independence, and eventually eigenvalues, where the determinant returns as the star of the characteristic equation.

Lecture 10 — complete

MATH-120 · Shoaib Khan · LUMS · July 2026

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