Dove's LibraryDove's Library

4.11 - The Inverse and Determinant of a Matrix

Overview

In Topic 4.10 you learned how to multiply matrices. This topic adds three tools that make matrices useful. The identity matrix acts like the number 1. The inverse of a matrix acts like a reciprocal, so it lets you undo a matrix multiplication. The determinant is a single number that tells you whether an inverse exists, and it also has a picture behind it: it measures the area of a parallelogram built from two vectors.

The identity matrix

The identity matrix, written II, is a square matrix with 1s on the main diagonal (from the top left to the bottom right) and 0s everywhere else. Here are the 2×22 \times 2 and 3×33 \times 3 identity matrices:
I2=[1001]I3=[100010001]I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
Multiplying a square matrix by the identity matrix of the same size gives back the original matrix, no matter which side II is on:
AI=AandIA=AAI = A \quad \text{and} \quad IA = A
You can see why by multiplying it out. Each entry of the product is a dot product, and the 0s in II wipe out everything except the one entry you started with:
[abcd][1001]=[a(1)+b(0)a(0)+b(1)c(1)+d(0)c(0)+d(1)]=[abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a(1) + b(0) & a(0) + b(1) \\ c(1) + d(0) & c(0) + d(1) \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}
That's why II plays the role of 1. For numbers, 5⋅1=55 \cdot 1 = 5. For matrices, AI=AAI = A. Matrix multiplication usually depends on order (ABAB and BABA are often different), but the identity matrix gives the same answer on either side.

The inverse of a matrix

For a nonzero number, the reciprocal undoes multiplication: 5⋅15=15 \cdot \frac{1}{5} = 1. The matrix version is the inverse. The inverse of a square matrix AA is written A−1A^{-1}, and it's the matrix that multiplies with AA to give the identity matrix of the same size:
AA−1=IandA−1A=IAA^{-1} = I \quad \text{and} \quad A^{-1}A = I
A few things to know about inverses:
  • Only square matrices have inverses. A 2×32 \times 3 matrix doesn't have one.
  • Not every square matrix has an inverse, just like the number 0 has no reciprocal. The determinant (below) tells you which ones do.
  • There's no matrix division. You never write 1A\frac{1}{A} or divide by a matrix. You multiply by A−1A^{-1} instead.
  • To check an inverse, multiply. If AA times your answer gives II, your answer is right.
Inverses are how you undo a matrix multiplication. If AX=BAX = B and you know AA and BB, multiply both sides on the left by A−1A^{-1}:
A−1AX=A−1B⇒IX=A−1B⇒X=A−1BA^{-1}AX = A^{-1}B \quad \Rightarrow \quad IX = A^{-1}B \quad \Rightarrow \quad X = A^{-1}B
Because order matters, A−1A^{-1} has to go on the left of both sides. You'll use exactly this move in Topic 4.13 to undo a transformation and in Topic 4.14 to go backward in time.

The determinant

Every square matrix has a number attached to it called its determinant. For a 2×22 \times 2 matrix, it's the product of the main diagonal minus the product of the other diagonal:
A=[abcd]⇒det⁡(A)=ad−bcA = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \quad \Rightarrow \quad \det(A) = ad - bc
The determinant is written det⁡(A)\det(A), and some books use vertical bars, ∣A∣|A|. It can be positive, negative, or zero. For example:
det⁡[4723]=(4)(3)−(7)(2)=12−14=−2\det \begin{bmatrix} 4 & 7 \\ 2 & 3 \end{bmatrix} = (4)(3) - (7)(2) = 12 - 14 = -2
You can find determinants with or without technology. On a TI-84, press 2nd and then x−1x^{-1} to open the MATRIX menu, go to EDIT, and enter the matrix as [A]. Back on the home screen, MATRIX, then MATH, then det( gives det([A]).

Finding the inverse of a 2×22 \times 2 matrix

When the inverse of a 2×22 \times 2 matrix exists, this formula gives it:
A=[abcd]⇒A−1=1ad−bc[d−b−ca]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \quad \Rightarrow \quad A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
In words:
  1. Find det⁡(A)=ad−bc\det(A) = ad - bc. If it's 0, stop: there's no inverse.
  1. Swap the two entries on the main diagonal (aa and dd).
  1. Change the signs of the other two entries (bb and cc), but leave them where they are.
  1. Multiply every entry by 1det⁡(A)\frac{1}{\det(A)}.
Why it works. Multiply AA by the swapped-and-negated matrix and see what comes out:
[abcd][d−b−ca]=[ad−bc−ab+bacd−dc−cb+da]=[ad−bc00ad−bc]\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} ad - bc & -ab + ba \\ cd - dc & -cb + da \end{bmatrix} = \begin{bmatrix} ad - bc & 0 \\ 0 & ad - bc \end{bmatrix}
That's (ad−bc)(ad - bc) times the identity matrix. Dividing every entry by ad−bcad - bc turns it into exactly II, which is what an inverse has to produce.
With technology, type [A] on the home screen and press the x−1x^{-1} key. MATH, then Frac, shows the entries as fractions.

Example 1: Finding an inverse

Find the inverse of A=[4723]A = \begin{bmatrix} 4 & 7 \\ 2 & 3 \end{bmatrix} and check your answer.
Step 1: Find the determinant.
det⁡(A)=(4)(3)−(7)(2)=−2\det(A) = (4)(3) - (7)(2) = -2
It isn't 0, so the inverse exists.
Step 2: Swap the diagonal entries and change the signs of the other two.
[3−7−24]\begin{bmatrix} 3 & -7 \\ -2 & 4 \end{bmatrix}
Step 3: Multiply every entry by 1−2\frac{1}{-2}.
A−1=−12[3−7−24]=[−32721−2]A^{-1} = -\frac{1}{2}\begin{bmatrix} 3 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -\frac{3}{2} & \frac{7}{2} \\ 1 & -2 \end{bmatrix}
Step 4: Check by multiplying.
AA−1=[4723][−32721−2]=[−6+714−14−3+37−6]=[1001]AA^{-1} = \begin{bmatrix} 4 & 7 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} -\frac{3}{2} & \frac{7}{2} \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} -6 + 7 & 14 - 14 \\ -3 + 3 & 7 - 6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
Answer: A−1=[−32721−2]A^{-1} = \begin{bmatrix} -\frac{3}{2} & \frac{7}{2} \\ 1 & -2 \end{bmatrix}, and AA−1=IAA^{-1} = I confirms it.

When does an inverse exist?

A square matrix AA has an inverse if and only if det⁡(A)≠0\det(A) \neq 0. A matrix with an inverse is called invertible. A matrix with determinant 0 has no inverse (it's sometimes called singular).
The formula shows part of the reason: when det⁡(A)≠0\det(A) \neq 0, the formula builds an inverse, and when det⁡(A)=0\det(A) = 0 it would divide by zero. But it's more than a formula problem. When the determinant is 0, no matrix at all can work as an inverse. Take this matrix:
B=[2412],det⁡(B)=(2)(2)−(4)(1)=0B = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}, \qquad \det(B) = (2)(2) - (4)(1) = 0
The first row of BB is exactly twice the second row. Each row of a product BCBC comes from dotting that row of BB with the columns of CC, so the first row of BCBC is always twice the second row of BCBC, whatever CC is. In II, the first row is [10]\begin{bmatrix} 1 & 0 \end{bmatrix} and the second is [01]\begin{bmatrix} 0 & 1 \end{bmatrix}, and the first isn't twice the second. So BCBC can never equal II, and BB has no inverse.

Example 2: When there's no inverse

For what values of kk does M=[k41k]M = \begin{bmatrix} k & 4 \\ 1 & k \end{bmatrix} have no inverse?
MM has no inverse exactly when its determinant is 0.
det⁡(M)=(k)(k)−(4)(1)=k2−4\det(M) = (k)(k) - (4)(1) = k^2 - 4k2−4=0⇒k2=4⇒k=2 or k=−2k^2 - 4 = 0 \quad \Rightarrow \quad k^2 = 4 \quad \Rightarrow \quad k = 2 \text{ or } k = -2
Answer: MM has no inverse when k=2k = 2 or k=−2k = -2. For every other value of kk, MM is invertible. (At k=2k = 2, MM is the matrix BB from above.)

The determinant as area

Put two vectors from R2\mathbb{R}^2 into a 2×22 \times 2 matrix, either as its columns or as its rows. The absolute value of the determinant is the area of the parallelogram spanned by the vectors: the parallelogram that has the two vectors as adjacent sides, starting from the same point.
For example, u⃗=⟨3,1⟩\vec{u} = \langle 3, 1 \rangle and v⃗=⟨2,3⟩\vec{v} = \langle 2, 3 \rangle as columns give
det⁡[3213]=(3)(3)−(2)(1)=7\det \begin{bmatrix} 3 & 2 \\ 1 & 3 \end{bmatrix} = (3)(3) - (2)(1) = 7
so the parallelogram they span has area 7. Using them as rows instead gives det⁡[3123]=9−2=7\det \begin{bmatrix} 3 & 1 \\ 2 & 3 \end{bmatrix} = 9 - 2 = 7, the same area.
4.11 - The Inverse and Determinant of a Matrix
Why it works. Draw a box around the parallelogram. The box is 5 by 4, so its area is 20. The parts of the box outside the parallelogram are two triangles with area 1.5, two rectangles with area 2, and two triangles with area 3, for a total of 13. That leaves 20−13=720 - 13 = 7 for the parallelogram, which matches the determinant.
4.11 - The Inverse and Determinant of a Matrix
The same subtraction works with letters. For columns ⟨a,c⟩\langle a, c \rangle and ⟨b,d⟩\langle b, d \rangle, the box is (a+b)(a + b) by (c+d)(c + d), and the pieces outside add up to ac+bd+2bcac + bd + 2bc:
(a+b)(c+d)−ac−bd−2bc=ad−bc(a + b)(c + d) - ac - bd - 2bc = ad - bc
The determinant can come out negative (that just depends on which vector you list first), so take the absolute value to get the area.

A determinant of 0 means parallel vectors

If the determinant is 0, the two vectors are parallel: they lie along the same line, and one is a scalar multiple of the other. The parallelogram gets flattened into a segment with zero area. In the right-hand graph above, ⟨4,2⟩=2⟨2,1⟩\langle 4, 2 \rangle = 2\langle 2, 1 \rangle, and
det⁡[2412]=(2)(2)−(4)(1)=0\det \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} = (2)(2) - (4)(1) = 0
That's the matrix BB from earlier, which had no inverse. The two ideas fit together: a 2×22 \times 2 matrix is invertible exactly when its columns are not parallel.

Example 3: Area and parallel vectors

(a) Find the area of the parallelogram spanned by ⟨−2,5⟩\langle -2, 5 \rangle and ⟨3,1⟩\langle 3, 1 \rangle. (b) Are ⟨−2,6⟩\langle -2, 6 \rangle and ⟨1,−3⟩\langle 1, -3 \rangle parallel?
(a) Put the vectors in as columns and find the determinant.
det⁡[−2351]=(−2)(1)−(3)(5)=−2−15=−17\det \begin{bmatrix} -2 & 3 \\ 5 & 1 \end{bmatrix} = (-2)(1) - (3)(5) = -2 - 15 = -17
The area is the absolute value: ∣−17∣=17|-17| = 17 square units.
(b) Do the same with the second pair.
det⁡[−216−3]=(−2)(−3)−(1)(6)=6−6=0\det \begin{bmatrix} -2 & 1 \\ 6 & -3 \end{bmatrix} = (-2)(-3) - (1)(6) = 6 - 6 = 0
The determinant is 0, so the vectors are parallel. You can confirm it directly: ⟨−2,6⟩=−2⟨1,−3⟩\langle -2, 6 \rangle = -2\langle 1, -3 \rangle.

Example 4: Putting it together

Let A=[2−15k]A = \begin{bmatrix} 2 & -1 \\ 5 & k \end{bmatrix}.
  • (a) Find the value of kk for which AA has no inverse.
  • (b) Let k=1k = 1. Find A−1A^{-1} and verify that AA−1=IAA^{-1} = I.
  • (c) Let k=1k = 1. Find the area of the parallelogram spanned by the columns of AA.
  • (d) For the value of kk from part (a), what is true about the columns of AA?
(a) det⁡(A)=(2)(k)−(−1)(5)=2k+5\det(A) = (2)(k) - (-1)(5) = 2k + 5. Set it equal to 0:
2k+5=0⇒k=−522k + 5 = 0 \quad \Rightarrow \quad k = -\frac{5}{2}
AA has no inverse when k=−52k = -\frac{5}{2}.
(b) With k=1k = 1, det⁡(A)=2(1)+5=7\det(A) = 2(1) + 5 = 7. Swap, change signs, and divide by 7:
A−1=17[11−52]=[1717−5727]A^{-1} = \frac{1}{7}\begin{bmatrix} 1 & 1 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & \frac{1}{7} \\ -\frac{5}{7} & \frac{2}{7} \end{bmatrix}
Verify:
AA−1=17[2−151][11−52]=17[2+52−25−55+2]=17[7007]=[1001]AA^{-1} = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 5 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -5 & 2 \end{bmatrix} = \frac{1}{7}\begin{bmatrix} 2 + 5 & 2 - 2 \\ 5 - 5 & 5 + 2 \end{bmatrix} = \frac{1}{7}\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
(c) The area is ∣det⁡(A)∣=∣7∣=7|\det(A)| = |7| = 7 square units.
(d) At k=−52k = -\frac{5}{2}, the columns are ⟨2,5⟩\langle 2, 5 \rangle and ⟨−1,−52⟩\langle -1, -\frac{5}{2} \rangle. The second is −12-\frac{1}{2} times the first, so the columns are parallel and span a parallelogram with area 0. That's the geometric reason AA has no inverse there.

Common mistakes

  • Mixing up the determinant formula. It's ad−bcad - bc: main diagonal first, then subtract the other diagonal. Not ad+bcad + bc, and not bc−adbc - ad.
  • Moving the wrong entries. In the inverse formula, aa and dd trade places, while bb and cc stay put and only change sign.
  • Forgetting to divide by the determinant, or multiplying by it instead.
  • Inverting a matrix with determinant 0. Check the determinant first. If it's 0, the answer is "no inverse."
  • Reporting a negative area. The area of the parallelogram is ∣det⁡(A)∣|\det(A)|.
  • Dividing by a matrix. There's no 1A\frac{1}{A}. Multiply by A−1A^{-1} on the correct side: AX=BAX = B gives X=A−1BX = A^{-1}B, not BA−1BA^{-1}.
  • Skipping the check. Multiplying AA by your inverse takes a minute and catches almost every sign error.

AP® is a registered trademark of the College Board, which is not affiliated with Dove's Library and was not involved in the production of, and does not endorse, this website.

Providing free, high-quality AP notes to help students succeed, earn college credit, and reach their educational goals, ensuring everyone has access to the resources they need.

© 2026 Standard Learning Group, Inc., which operates Dove's Library. All rights reserved.