Determinants and Inverse Matrices

With ordinary numbers, you solve $5x = 15$ by dividing by $5$, which is the same as multiplying by $5^{-1}$. Matrices have no division, but many square matrices have an inverse that does the same job. The determinant, a single number worked out from a square matrix, tells you whether that inverse exists. Together they let you solve whole systems of equations in one step, and they’re the key to codes, transformations and much more.

Key ideas

The determinant of a 2 × 2 matrix

For a $2 \times 2$ matrix, the determinant is “leading diagonal minus the other diagonal”:

$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \qquad \Rightarrow \qquad \det A = |A| = ad - bc$$

For example, $\det \begin{pmatrix} 4 & 3 \\ 2 & 5 \end{pmatrix} = 4(5) - 3(2) = 14$.

Only square matrices have determinants. For $3 \times 3$ and larger matrices, use your GDC’s $\det$ function: in this course you find those with technology, not by hand.

The inverse of a matrix

The inverse of a square matrix $A$ is the matrix $A^{-1}$ that undoes it:

$$AA^{-1} = A^{-1}A = I$$

For a $2 \times 2$ matrix: swap the entries on the leading diagonal, change the signs of the other two, and divide by the determinant.

$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \qquad \Rightarrow \qquad A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

For larger matrices, enter $A$ in the GDC and use $A^{-1}$ (the $x^{-1}$ key). Many GDCs can show the result as fractions.

Singular matrices

If $\det A = 0$, the formula would divide by zero, and $A$ has no inverse. Such a matrix is called singular. A matrix with a non-zero determinant is non-singular (or invertible). For a $2 \times 2$ matrix, $\det A = 0$ means one row is a multiple of the other: for example, $\begin{pmatrix} 6 & -4 \\ 3 & -2 \end{pmatrix}$ has determinant $-12 - (-12) = 0$.

Systems of equations as AX = B

Any system of linear equations can be written as one matrix equation. For example,

$$\begin{aligned} 4x + 3y &= 1 \\ 2x + 5y &= 11 \end{aligned} \qquad \Longleftrightarrow \qquad \underbrace{\begin{pmatrix} 4 & 3 \\ 2 & 5 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x \\ y \end{pmatrix}}_{X} = \underbrace{\begin{pmatrix} 1 \\ 11 \end{pmatrix}}_{B}$$

$A$ is the coefficient matrix. If $A$ is invertible, multiply both sides on the left by $A^{-1}$:

$$A^{-1}AX = A^{-1}B \quad\Rightarrow\quad IX = A^{-1}B \quad\Rightarrow\quad X = A^{-1}B$$

Order matters. $A^{-1}$ must go on the left of both sides, so the answer is $A^{-1}B$, never $BA^{-1}$. (If instead the equation is $XA = B$, multiply on the right: $X = BA^{-1}$.) In IB exams, $A$ will always be invertible in these questions, so each system has a unique solution.

Coding and decoding

One classic use: turn letters into numbers ($A = 1$, $B = 2$, …, $Z = 26$), arrange them in columns of a matrix $M$, and multiply by a key matrix $K$ to get the coded matrix $C = KM$. Anyone who knows $K$ decodes with $M = K^{-1}C$. The key must be invertible, or the message could never be recovered.

Worked examples

Example 1: An inverse by hand

Let $A = \begin{pmatrix} 4 & 3 \\ 2 & 5 \end{pmatrix}$. Find $A^{-1}$ and check your answer.

Solution. $\det A = 4(5) - 3(2) = 14 \ne 0$, so the inverse exists. Swap $4$ and $5$, negate $3$ and $2$, and divide by $14$:

$$A^{-1} = \frac{1}{14}\begin{pmatrix} 5 & -3 \\ -2 & 4 \end{pmatrix}$$

Check:

$$AA^{-1} = \frac{1}{14}\begin{pmatrix} 4(5) + 3(-2) & 4(-3) + 3(4) \\ 2(5) + 5(-2) & 2(-3) + 5(4) \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix} = I \;\checkmark$$

Leaving the $\dfrac{1}{14}$ outside keeps the fractions out of the way.

Example 2: When is a matrix singular?

Find the values of $k$ for which $\begin{pmatrix} k & 6 \\ 2 & k - 1 \end{pmatrix}$ is singular.

Solution. Set the determinant equal to $0$:

$$\begin{aligned} k(k - 1) - 6(2) &= 0 \\ k^2 - k - 12 &= 0 \\ (k - 4)(k + 3) &= 0 \end{aligned}$$

So $k = 4$ or $k = -3$. For any other value of $k$, the matrix has an inverse.

Check $k = 4$: $\begin{pmatrix} 4 & 6 \\ 2 & 3 \end{pmatrix}$, and the first row is twice the second. ✓

Example 3: Solving a 3 × 3 system with an inverse

Write this system as $AX = B$ and solve it using the inverse matrix.

$$\begin{aligned} x + y + z &= 9 \\ 2x - y + 3z &= 13 \\ 3x + 2y - 4z &= -4 \end{aligned}$$

Solution.

$$\begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -4 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 9 \\ 13 \\ -4 \end{pmatrix}$$

On the GDC, $\det A = 22$, which isn’t $0$, so $A^{-1}$ exists:

$$A^{-1} = \frac{1}{22}\begin{pmatrix} -2 & 6 & 4 \\ 17 & -7 & -1 \\ 7 & 1 & -3 \end{pmatrix}$$

Then $X = A^{-1}B$ (calculate $A^{-1}B$ directly on the GDC):

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1}\begin{pmatrix} 9 \\ 13 \\ -4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$$

So $x = 2$, $y = 3$, $z = 4$. Check the third equation: $3(2) + 2(3) - 4(4) = 6 + 6 - 16 = -4$ ✓.

Example 4: Decoding a message

A message is coded by writing letters as numbers ($A = 1$, …, $Z = 26$) in the columns of a $2 \times 2$ matrix $M$, then finding $C = KM$ with key $K = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$. The coded matrix received is $C = \begin{pmatrix} 27 & 48 \\ 41 & 76 \end{pmatrix}$. Decode it. (Read the message down the first column, then down the second.)

Solution. $\det K = 4 - 3 = 1$, so

$$K^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}$$

Since $C = KM$, we get $M = K^{-1}C$:

$$M = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}\begin{pmatrix} 27 & 48 \\ 41 & 76 \end{pmatrix} = \begin{pmatrix} 54 - 41 & 96 - 76 \\ -81 + 82 & -144 + 152 \end{pmatrix} = \begin{pmatrix} 13 & 20 \\ 1 & 8 \end{pmatrix}$$

Reading down the columns: $13, 1, 20, 8$, which is $M, A, T, H$. The message is MATH.

Common mistakes

Forgetting to divide by the determinant. $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ on its own is not the inverse unless $\det A = 1$. Always multiply by $\dfrac{1}{ad - bc}$.

Swapping and negating the wrong entries. The leading-diagonal entries ($a$ and $d$) swap places; the other two ($b$ and $c$) stay where they are but change sign. Check by multiplying: $AA^{-1}$ should be $I$.

Writing BA⁻¹ instead of A⁻¹B. From $AX = B$, the inverse goes on the left: $X = A^{-1}B$. Usually $BA^{-1}$ isn’t even defined (a $2 \times 1$ times a $2 \times 2$), and when it is, it’s a different answer.

Trying to invert a singular matrix. If $\det A = 0$ there is no inverse; your GDC will say “singular matrix”. In a system of equations, that means there’s no unique solution.

Finding the determinant of a non-square matrix. Only square matrices have determinants and inverses. A $2 \times 3$ matrix has neither.

Entering a system with the variables out of order. Each column of $A$ belongs to one variable. If an equation is written $3z + x = 5$, its row is $\begin{pmatrix} 1 & 0 & 3 \end{pmatrix}$ when the variables are in the order $x, y, z$.

Practice

1. (Warm-up) Find the determinant of each matrix.

• (a) $\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$
• (b) $\begin{pmatrix} -2 & 5 \\ 3 & -1 \end{pmatrix}$
• (c) $\begin{pmatrix} 6 & -4 \\ 3 & -2 \end{pmatrix}$

Solution

(a) $3(4) - 2(1) = 12 - 2 = 10$

(b) $(-2)(-1) - 5(3) = 2 - 15 = -13$

(c) $6(-2) - (-4)(3) = -12 + 12 = 0$ (this matrix is singular)

2. (Warm-up) Find the inverse of $B = \begin{pmatrix} 3 & -1 \\ 5 & -2 \end{pmatrix}$.

Solution

$\det B = 3(-2) - (-1)(5) = -6 + 5 = -1$.

$$B^{-1} = \frac{1}{-1}\begin{pmatrix} -2 & 1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 5 & -3 \end{pmatrix}$$

Check: $BB^{-1} = \begin{pmatrix} 6 - 5 & -3 + 3 \\ 10 - 10 & -5 + 6 \end{pmatrix} = I$ ✓.

3. (Warm-up) Which of these matrices are singular?

$$P = \begin{pmatrix} 2 & 8 \\ 1 & 4 \end{pmatrix}, \qquad Q = \begin{pmatrix} 3 & -6 \\ -1 & 2 \end{pmatrix}, \qquad R = \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}$$

Solution

$\det P = 8 - 8 = 0$, $\det Q = 6 - 6 = 0$, $\det R = 5 - 4 = 1$.

$P$ and $Q$ are singular. $R$ is not (its inverse is $\begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix}$).

4. (Core) By hand, use an inverse matrix to solve:

$$\begin{aligned} 3x + 2y &= 4 \\ 5x + 4y &= 10 \end{aligned}$$

Solution

$A = \begin{pmatrix} 3 & 2 \\ 5 & 4 \end{pmatrix}$, $\det A = 12 - 10 = 2$, so

$$A^{-1} = \frac{1}{2}\begin{pmatrix} 4 & -2 \\ -5 & 3 \end{pmatrix}$$$$X = A^{-1}B = \frac{1}{2}\begin{pmatrix} 4 & -2 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 4 \\ 10 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 16 - 20 \\ -20 + 30 \end{pmatrix} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$$

So $x = -2$, $y = 5$. Check: $3(-2) + 2(5) = 4$ ✓ and $5(-2) + 4(5) = 10$ ✓.

5. (Core) A school places three supply orders:

• $2$ binders, $1$ pack of pens and $1$ ruler: $11.00
• $1$ binder, $3$ packs of pens and $2$ rulers: $13.75
• $1$ binder, $1$ pack of pens and $4$ rulers: $12.75

Write this as a matrix equation and use an inverse matrix to find the price of each item.

Solution

Let $b$, $p$, $r$ be the prices in dollars.

$$\begin{pmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 1 & 4 \end{pmatrix}\begin{pmatrix} b \\ p \\ r \end{pmatrix} = \begin{pmatrix} 11.00 \\ 13.75 \\ 12.75 \end{pmatrix}$$

GDC: $\det A = 16 \ne 0$, and

$$\begin{pmatrix} b \\ p \\ r \end{pmatrix} = A^{-1}\begin{pmatrix} 11.00 \\ 13.75 \\ 12.75 \end{pmatrix} = \begin{pmatrix} 3.50 \\ 2.25 \\ 1.75 \end{pmatrix}$$

A binder costs $3.50, a pack of pens $2.25 and a ruler $1.75. Check the first order: $7.00 + 2.25 + 1.75 = 11.00$ ✓.

6. (Core) Let $A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & 3 \\ 1 & -2 \end{pmatrix}$. Find the matrix $X$ such that $XA = B$.

Solution

Multiply both sides on the right by $A^{-1}$: $XAA^{-1} = BA^{-1}$, so $X = BA^{-1}$.

$\det A = 6 - 5 = 1$, so $A^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$.

$$X = \begin{pmatrix} 4 & 3 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 12 - 15 & -4 + 6 \\ 3 + 10 & -1 - 4 \end{pmatrix} = \begin{pmatrix} -3 & 2 \\ 13 & -5 \end{pmatrix}$$

Check: $XA = \begin{pmatrix} -6 + 10 & -3 + 6 \\ 26 - 25 & 13 - 15 \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ 1 & -2 \end{pmatrix} = B$ ✓.

7. (Core) A four-letter word was coded as in Example 4 ($A = 1$, …, $Z = 26$, letters in the columns of $M$, $C = KM$), but with key $K = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$. The coded matrix is $C = \begin{pmatrix} 24 & 17 \\ 45 & 30 \end{pmatrix}$. Find the word.

Solution

$\det K = 6 - 5 = 1$, so $K^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$.

$$M = K^{-1}C = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 24 & 17 \\ 45 & 30 \end{pmatrix} = \begin{pmatrix} 48 - 45 & 34 - 30 \\ -120 + 135 & -85 + 90 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 15 & 5 \end{pmatrix}$$

Down the columns: $3, 15, 4, 5$, which is $C, O, D, E$. The word is CODE.

8. (Challenge) Let $P = \begin{pmatrix} k & 2 \\ 3 & k + 1 \end{pmatrix}$.

• (a) Find the values of $k$ for which $P$ has no inverse.
• (b) Find $P^{-1}$ in terms of $k$, for all other values of $k$.
• (c) Use your answer to find $P^{-1}$ when $k = 3$.

Solution

(a) $\det P = k(k + 1) - 6 = k^2 + k - 6 = (k + 3)(k - 2)$. This is $0$ when $k = -3$ or $k = 2$.

(b)

$$P^{-1} = \frac{1}{(k + 3)(k - 2)}\begin{pmatrix} k + 1 & -2 \\ -3 & k \end{pmatrix}, \qquad k \ne -3,\ k \ne 2$$

(c) For $k = 3$: $(k + 3)(k - 2) = 6 \times 1 = 6$, so

$$P^{-1} = \frac{1}{6}\begin{pmatrix} 4 & -2 \\ -3 & 3 \end{pmatrix}$$

Check: $P = \begin{pmatrix} 3 & 2 \\ 3 & 4 \end{pmatrix}$, and $\dfrac{1}{6}\begin{pmatrix} 3 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 4 & -2 \\ -3 & 3 \end{pmatrix} = \dfrac{1}{6}\begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix} = I$ ✓.

9. (Challenge) Let $A = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$.

• (a) Show that $A^2 = 3A - 2I$.
• (b) Multiply both sides of this equation by $A^{-1}$ to show that $A^{-1} = \dfrac{1}{2}(3I - A)$.
• (c) Use (b) to find $A^{-1}$, and check it with the $2 \times 2$ inverse formula.

Solution

(a)

$$A^2 = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 3 & 1 \end{pmatrix}, \qquad 3A - 2I = \begin{pmatrix} 6 - 2 & 0 \\ 3 & 3 - 2 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 3 & 1 \end{pmatrix}$$

They’re equal. ✓

(b) $\det A = 2 \ne 0$, so $A^{-1}$ exists. Multiply $A^2 = 3A - 2I$ on the left by $A^{-1}$:

$$A^{-1}AA = 3A^{-1}A - 2A^{-1}I \quad\Rightarrow\quad A = 3I - 2A^{-1}$$

Rearrange: $2A^{-1} = 3I - A$, so $A^{-1} = \dfrac{1}{2}(3I - A)$.

(c)

$$A^{-1} = \frac{1}{2}\begin{pmatrix} 3 - 2 & 0 \\ -1 & 3 - 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix}$$

Formula: $\dfrac{1}{2}\begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix}$, the same. ✓