Inverse of a Matrix using Adjoint

MATThe Inverse of a Matrix

The inverse of a square matrix is found by $A^{-1} = \dfrac{1}{|A|}\,\text{adj}\,A$ , which exists only when $A$ is non-singular ( $|A| \ne 0$ ). The adjoint is the transpose of the cofactor matrix, so the method threads together minors, cofactors, the adjoint and the determinant.

ISC tests this directly for $2\times2$ and $3\times3$ matrices, often asking you to verify $A A^{-1} = I$ as a check, and it underpins the matrix method for solving systems.

Inverse via adjoint

A^{-1} = \dfrac{1}{|A|}\,\text{adj}\,A,\qquad |A| \ne 0

A

is an

n\times n

matrix;

\text{adj}\,A

is the transpose of the cofactor matrix and

|A|

its determinant.

Fundamental adjoint identity

A\,(\text{adj}\,A) = (\text{adj}\,A)\,A = |A|\,I

I

is the identity matrix of the same order; dividing by

|A|

gives the inverse formula.

2x2 inverse shortcut

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

Valid when

|A| = ad - bc \ne 0

; swap the diagonal entries, negate the off-diagonal, divide by

|A|

Defining property of the inverse

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

Used to verify a computed inverse; the product with

A

must return the identity matrix.

Step 1: compute $|A|$ . If $|A| = 0$ the matrix is singular and has no inverse.
Step 2: find every cofactor $C_{ij} = (-1)^{i+j}M_{ij}$ to build the cofactor matrix.
Step 3: take the transpose of the cofactor matrix to get $\text{adj}\,A$ .
Step 4: divide by the determinant: $A^{-1} = \dfrac{1}{|A|}\text{adj}\,A$ .
For a $2\times2$ matrix use the direct shortcut; for $3\times3$ you must compute all nine cofactors.
Verify by checking $A A^{-1} = I$ ; this catches sign and transpose errors instantly.
Useful properties: $(A^{-1})^{-1} = A$ , $(AB)^{-1} = B^{-1}A^{-1}$ , and $|A^{-1}| = \dfrac{1}{|A|}$ .

Where the marks go

Forgetting to transpose the cofactor matrix, i.e. using the cofactor matrix itself as the adjoint.
Sign errors in cofactors from mishandling $(-1)^{i+j}$ , especially at positions $(1,2),(2,1),(2,3),(3,2)$ .
Computing $A^{-1}$ when $|A| = 0$ ; a singular matrix is non-invertible and the method must stop at Step 1.
Dividing each cofactor by $|A|$ before transposing, or only dividing some entries, giving a wrong inverse.

How the board asks it

Numericalthe four-step adjoint method $A^{-1} = \dfrac{1}{|A|}\,\text{adj}\,A$
Find the inverse of the matrix $A = \begin{bmatrix} 2 & 3 & 1 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$ using the adjoint method.
Numericalcofactors and the transpose of the cofactor matrix
For the matrix $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$ , find all the cofactors and hence write down $\text{adj}\,A$ .
Derive / provethe adjoint identity $A(\text{adj}\,A) = |A|\,I$
If $A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$ , verify that $A(\text{adj}\,A) = (\text{adj}\,A)A = |A|\,I$ and hence obtain $A^{-1}$ .
Numericalthe non-singularity condition $|A| \ne 0$
For what value of $k$ is the matrix $A = \begin{bmatrix} k & 2 \\ 3 & 4 \end{bmatrix}$ not invertible? Give reasons for your answer.
Multiple choicethe determinant result $|\text{adj}\,A| = |A|^{n-1}$
If $A$ is a $3 \times 3$ matrix with $|A| = 4$ , then $|\text{adj}\,A|$ equals: (a) $4$ , (b) $16$ , (c) $12$ , (d) $64$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.