Matrices and Determinants of Order 3

Lessons

Senior 5 · Mathematics

Matrices and Determinants of Order 3

5 lessons

Lesson 4 of 5

The inverse of a matrix of order 3

40 min

Undoing a transformation

A rotation through $90^\circ$ about the $z$ -axis is undone by a rotation through $-90^\circ$ ; an enlargement with scale factor 3 is undone by scale factor $\frac{1}{3}$ . The projection onto the $x y$ -plane cannot be undone: the points $(2, 1, 5)$ and $(2, 1, - 4)$ both land on $(2, 1, 0)$ , and nothing can tell them apart afterwards. For numbers, $3 \times \frac{1}{3} = 1$ and 0 has no reciprocal. For matrices the role of 1 is played by $I_{3}$ and the role of 0 by a zero determinant.

Definition

A square matrix $A$ is invertible (or non-singular) if there is a matrix $B$ with $A B = B A = I_{3}$ . Then $B$ is the inverse of $A$ , written $A^{- 1}$ . A matrix with no inverse is singular.

The inverse is unique: if $B$ and $B^{'}$ both work, then $B = B I_{3} = B (A B^{'}) = (B A) B^{'} = I_{3} B^{'} = B^{'}$ .

Definition

The adjugate (or adjoint) of $A$ , written $\operatorname{adj} A$ , is the transpose of the matrix of cofactors:

\operatorname{adj} A = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix}^T = \begin{pmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{pmatrix} .

Important

Theorem. For every $A$ of order 3, $A \operatorname{adj} A = (\operatorname{adj} A)A = (\det A)I_3$ . Hence $A$ is invertible if and only if $\det A \ne 0$ , and then

A^{-1} = \frac{1}{\det A}\operatorname{adj} A .

Reason. Entry $(i, i)$ of $A \operatorname{adj} A$ is row $i$ of $A$ times its own cofactors, which is the expansion of $\det A$ along row $i$ . Entry $(i, k)$ with $k \ne i$ is row $i$ times the cofactors of row $k$ : that is the expansion of a determinant whose rows $i$ and $k$ are equal, so it is 0. Dividing by $\det A \ne 0$ gives the formula. Conversely, if $A^{- 1}$ exists, $\det A \cdot \det A^{-1} = \det I_3 = 1$ , so $\det A \ne 0$ .

Worked example 1: the inverse by the adjugate

Find the inverse of $A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 0 \\ -1 & 1 & 3 \end{pmatrix}$ .

Step 1. $\det A = 1(3 - 0) - 2(6 - 0) + (-1)(2 + 1) = 3 - 12 - 3 = -12 \ne 0$ , so $A^{- 1}$ exists.

Steps 2 and 3. The cofactors, for example $C_{12} = -\begin{vmatrix} 2 & 0 \\ -1 & 3 \end{vmatrix} = -6$ and $C_{21} = -\begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = -7$ :

(C_{ij}) = \begin{pmatrix} 3 & -6 & 3 \\ -7 & 2 & -3 \\ 1 & -2 & -3 \end{pmatrix} .

Steps 4 and 5.

\operatorname{adj} A = \begin{pmatrix} 3 & -7 & 1 \\ -6 & 2 & -2 \\ 3 & -3 & -3 \end{pmatrix}, \qquad A^{-1} = -\frac{1}{12}\begin{pmatrix} 3 & -7 & 1 \\ -6 & 2 & -2 \\ 3 & -3 & -3 \end{pmatrix} = \frac{1}{12}\begin{pmatrix} -3 & 7 & -1 \\ 6 & -2 & 2 \\ -3 & 3 & 3 \end{pmatrix} .

Step 6. Row 1 of $A$ times column 1 of $\operatorname{adj} A$ : $3 - 12 - 3 = - 12$ ; row 1 times column 2: $- 7 + 4 + 3 = 0$ . Every product gives $A \operatorname{adj} A = -12I_3$ , so $AA^{-1} = I_3$ . ✓

Important

Properties of the inverse (for invertible $A, B$ of order 3 and $k \ne 0$ ):

$(A^{-1})^{-1} = A$ and $(kA)^{-1} = \frac{1}{k}A^{-1}$ ;
$(AB)^{-1} = B^{-1}A^{-1}$ , with the order reversed, because $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I_3$ ;
$(A^T)^{-1} = (A^{-1})^T$ , because $A^T(A^{-1})^T = (A^{-1}A)^T = I_3^T = I_3$ ;
$\det(A^{-1}) = \frac{1}{\det A}$ ;
if $A X = C$ then $X = A^{-1}C$ , and if $X A = C$ then $X = CA^{-1}$ . There is no "division" by a matrix, and the side matters.

Worked example 2: when does the inverse exist?

For which real $k$ is $M = \begin{pmatrix} 1 & k & 2 \\ k & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}$ invertible?

Solution.

\det M = 1(1 - 4) - k(k - 4) + 2(2k - 2) = -k^2 + 8k - 7 = -(k - 1)(k - 7) .

$M$ is invertible for every real $k$ except $k = 1$ and $k = 7$ . (For $k = 1$ , rows 1 and 2 are equal, which confirms the first root.)

Worked example 3: the inverse of a rotation

Find $R^{- 1}$ for the rotation $R = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ through $90^\circ$ about the $z$ -axis, and describe it.

Solution. $\det R = 0 - (-1)(1 - 0) + 0 = 1$ . The cofactors are $(C_{ij}) = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ , so

R^{-1} = \operatorname{adj} R = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = R^T ,

which is the rotation through $-90^\circ$ about the $z$ -axis, as expected. Every rotation matrix has $R^{-1} = R^T$ .

Worked example 4: an inverse from a matrix equation

Let $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ . Show that $A^{2} - 5 A + 4 I_{3} = O$ and use it to find $A^{- 1}$ .

Solution. $A^2 = \begin{pmatrix} 6 & 5 & 5 \\ 5 & 6 & 5 \\ 5 & 5 & 6 \end{pmatrix}$ . On the diagonal $6 - 10 + 4 = 0$ ; off the diagonal $5 - 5 + 0 = 0$ . So $A^{2} - 5 A + 4 I_{3} = O$ , that is $A (5 I_{3} - A) = 4 I_{3}$ . Therefore

A^{-1} = \frac{1}{4}(5I_3 - A) = \frac{1}{4}\begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix} ,

the same as $\frac{1}{\det A}\operatorname{adj} A$ with $\det A = 4$ .

Warning

Transpose the cofactors. $\operatorname{adj} A$ is the transpose of $(C_{ij})$ ; forgetting it gives a wrong inverse unless $A$ happens to be symmetric.
$(AB)^{-1} = B^{-1}A^{-1}$ , not $A^{-1}B^{-1}$ : to undo "put on socks, then shoes", take off the shoes first.
Check $\det A$ first. If it is 0, stop: computing cofactors cannot rescue a singular matrix.
Always check one or two entries of $AA^{-1}$ ; a single sign slip in a cofactor shows up at once.

Tip

Remember the formula as "adjugate over determinant": $A^{-1} = \frac{\operatorname{adj} A}{\det A}$ , just like the $2 \times 2$ case, where $\operatorname{adj}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .

3Blue1Brown: Inverse matrices, column space and null space, Chapter 7 of Essence of Linear Algebra (about 12 min). Watch why a transformation that squashes space onto a plane (determinant 0) has no inverse.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Find the inverse of $\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$ .
Find the inverse of $\begin{pmatrix} 3 & 1 & 2 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix}$ and check your answer.
Is $\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$ invertible? Give a reason.
For which real $k$ is $\begin{pmatrix} 1 & 2 & k \\ 0 & 1 & 2 \\ k & 0 & 1 \end{pmatrix}$ singular?
$A$ and $B$ are of order 3 with $\det A = 5$ and $\det B = -2$ . Find $\det(A^{-1})$ , $\det(AB^{-1})$ and $\det(2A^{-1})$ .
With $A$ the matrix of question 2, solve $AX = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ for $X$ .
Prove that if $A$ is invertible and symmetric, then $A^{- 1}$ is symmetric.

Show answersHide answers

$\det = 1$ ; inverse $\begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}$ .
$\det = 1$ ; inverse $\begin{pmatrix} 2 & -2 & -1 \\ -3 & 4 & 1 \\ -1 & 1 & 1 \end{pmatrix}$ ; the product with $A$ is $I_{3}$ .
No: its determinant is $1 (45 - 48) - 2 (36 - 42) + 3 (32 - 35) = - 3 + 12 - 9 = 0$ .
$\det = 1 + 4k - k^2 = 0$ , so $k = 2 \pm \sqrt{5}$ .
$\frac{1}{5}$ ; $5 \times \left(-\frac{1}{2}\right) = -\frac{5}{2}$ ; $2^3 \times \frac{1}{5} = \frac{8}{5}$ .
$X = A^{-1}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -4 & -3 \\ -3 & 8 & 3 \\ -1 & 2 & 3 \end{pmatrix}$ .
$(A^{-1})^T = (A^T)^{-1} = A^{-1}$ , using $A^{T} = A$ .

Sources and credits

Written by Ganzaa: the explanations, the flowchart, the worked examples, the practice items and the quiz questions. Every inverse was checked by multiplying back to the identity by computer.
Video: Inverse matrices, column space and null space | Chapter 7, Essence of linear algebra, 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1Undoing a transformation

Undoing a transformation