Adjoint and inverse of a matrix

What you will be able to do

• say what the inverse of a matrix is and when it exists;
• find the adjoint and the inverse of a $2 \times 2$ and a $3 \times 3$ matrix;
• tell a singular matrix from a non-singular one;
• use the properties of the inverse and of the adjoint.

Undoing a matrix

A matrix $A$ moves points of the plane. Is there a matrix that moves them back? Such a matrix is the inverse of $A$, written $A^{-1}$. It is defined by

Only a square matrix can have an inverse, and if it has one, it has only one. A matrix that has an inverse is called non-singular, or invertible. A matrix with no inverse is singular.

The picture shows why some matrices have no inverse. A matrix with determinant $0$ squashes the unit square onto a line. Many points land on the same spot, so nothing can send them back.

The adjoint

Recall the cofactor $C_{ij} = (-1)^{i+j}M_{ij}$. The cofactor matrix has $C_{ij}$ in row $i$ and column $j$. The adjoint (also called the adjugate) of $A$, written $\text{adj}\,A$, is the transpose of the cofactor matrix.

The key fact comes from the lesson on determinants. A row of $A$ times its own cofactors gives $|A|$. A row times the cofactors of a different row gives $0$. Put these together in the product $A\,\text{adj}\,A$: the diagonal entries are $|A|$ and all the others are $0$. So

The inverse

If $|A| \ne 0$, divide by $|A|$:

If $|A| = 0$, then $A\,\text{adj}\,A = O$, and $A$ has no inverse. So a square matrix has an inverse if and only if its determinant is not zero.

For a $2 \times 2$ matrix the adjoint is easy to write: swap the two diagonal entries and change the signs of the other two:

For $A =

$,$|A| = 5$and$A^{-1} = \dfrac{1}{5}

$. Check: the entry in row$1$, column$1$of$AA^{-1}$is$\tfrac{1}{5}(8 - 3) = 1$.

For a $3 \times 3$ matrix take $A =

$.

1. The determinant. Along row $1$: $1(6 - 9) - 2(4 - 3) + 2(6 - 3) = -3 - 2 + 6 = 1$.
2. The cofactors. $C_{11} = -3$, $C_{12} = -(4 - 3) = -1$, $C_{13} = 6 - 3 = 3$. Next $C_{21} = -(4 - 6) = 2$, $C_{22} = 2 - 2 = 0$, $C_{23} = -(3 - 2) = -1$. Last $C_{31} = 6 - 6 = 0$, $C_{32} = -(3 - 4) = 1$, $C_{33} = 3 - 4 = -1$.
3. The adjoint is the transpose of the cofactor matrix: $\text{adj}\,A =

$.

4. The inverse. Since $|A| = 1$, $A^{-1} = \text{adj}\,A$. Check the first row of $A$ with the first column of $\text{adj}\,A$: $1(-3) + 2(-1) + 2(3) = 1$.

Properties

Let $A$ and $B$ be invertible matrices of order $n$.

• $(A^{-1})^{-1} = A$.
• $(AB)^{-1} = B^{-1}A^{-1}$, with the order reversed, just as for the transpose.
• $(A^T)^{-1} = (A^{-1})^T$.
• $|A^{-1}| = \dfrac{1}{|A|}$, because $|A||A^{-1}| = |I| = 1$.
• $|\text{adj}\,A| = |A|^{n-1}$, because $A\,\text{adj}\,A = |A|I$ and taking determinants gives $|A||\text{adj}\,A| = |A|^n$.
• $\text{adj}(AB) = (\text{adj}\,B)(\text{adj}\,A)$.

A worked example

Find the inverse of $A =

$, and check it.

1. The determinant. $|A| = 3 \cdot 1 - (-2)(4) = 11$. It is not $0$, so the inverse exists.
2. The adjoint. Swap $3$ and $1$, and change the signs of $-2$ and $4$: $\text{adj}\,A =

$.

3. The inverse. $A^{-1} = \dfrac{1}{11}

$.

4. Check. Row $2$ of $A$ times column $2$ of $\text{adj}\,A$ is $4 \cdot 2 + 1 \cdot 3 = 11$, and dividing by $11$ gives $1$, as it should for a diagonal entry of $I$.

Practise with new numbers

Try it in Python

The program finds the cofactors of the $3 \times 3$ matrix of the lesson, forms the adjoint as their transpose, and checks that $A\,\text{adj}\,A$ is $|A|I$. Then it divides to get the inverse.

Common mistakes

• Forgetting to transpose. The adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself.
• Forgetting to divide by $|A|$. $\text{adj}\,A$ is the inverse only when $|A| = 1$.
• Writing $(AB)^{-1} = A^{-1}B^{-1}$. The order reverses.
• Looking for the inverse of a singular matrix. Check $|A| \ne 0$ first.
• Sign slips in the cofactors. Use the chessboard pattern $+\ -\ +$.

Remember these

• $AA^{-1} = A^{-1}A = I$, and $A^{-1}$ exists exactly when $|A| \ne 0$.
• $A\,\text{adj}\,A = |A|I$, so $A^{-1} = \dfrac{\text{adj}\,A}{|A|}$.
• For $2 \times 2$, swap the diagonal and negate the other two entries.
• $(AB)^{-1} = B^{-1}A^{-1}$, $|A^{-1}| = \dfrac{1}{|A|}$ and $|\text{adj}\,A| = |A|^{n-1}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the inverse of $

$.

$.

2. Does $

$ have an inverse?

3. A $3 \times 3$ matrix $A$ has $|A| = 3$. Find $|\text{adj}\,A|$.

4. A $2 \times 2$ matrix has $|A| = 5$. What is $|A^{-1}|$?

5. Write $\text{adj}\,A$ for $A =

$.

$.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 4 (Determinants), the section on the adjoint and inverse of a square matrix.
• OpenStax, Precalculus 2e, Section 9.7 (Solving Systems with Inverses).