How do I find the inverse of a 3×3 matrix using the adjoint method?

In one line

If \(|A|\ne0\), then \(A^{-1}=\dfrac{1}{|A|}\,\mathrm{adj}\,A\), where \(\mathrm{adj}\,A\) is the transpose of the matrix of cofactors.

Explained simply

The inverse of a number 5 is \(\frac15\) because \(5\times\frac15=1\). The inverse of a matrix \(A\) is the matrix \(A^{-1}\) with \(AA^{-1}=A^{-1}A=I\). Just as 0 has no reciprocal, a matrix with determinant 0 (a singular matrix) has no inverse.

Technically, the cofactor of entry \(a_{ij}\) is \(C_{ij}=(-1)^{i+j}M_{ij}\), where \(M_{ij}\) is the minor (determinant left after deleting row \(i\) and column \(j\)). The adjoint is \(\mathrm{adj}\,A=[C_{ij}]^T\). The key identity \(A\,(\mathrm{adj}\,A)=|A|\,I\) gives the formula. This is a pre-requisite topic, needed for unitary/orthogonal matrices and for coding theory.

Why it matters: a system of equations \(AX=B\) can be solved as \(X=A^{-1}B\) when \(A\) is invertible, which is how circuit equations from KVL and KCL are often written. For large systems, numerical methods (Module 6) are faster, but for \(3\times3\) systems the adjoint method is exact and straightforward.

How it works

1. Compute \(|A|\). If it is 0, stop: no inverse.
2. Find all nine cofactors \(C_{ij}\), remembering the sign pattern \(\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}\).
3. Write the cofactor matrix and transpose it to get \(\mathrm{adj}\,A\).
4. Divide by \(|A|\).
5. Check: \(AA^{-1}=I\).

Key formula

\(A^{-1}=\dfrac{\mathrm{adj}\,A}{|A|}\), \(\ C_{ij}=(-1)^{i+j}M_{ij}\). Properties: \((AB)^{-1}=B^{-1}A^{-1}\), \((A^T)^{-1}=(A^{-1})^T\).

Quick example

\(A=\begin{bmatrix}1&2&3\\0&1&4\\5&6&0\end{bmatrix}\).

\(|A|=1(0-24)-2(0-20)+3(0-5)=-24+40-15=1\).

Cofactors: row 1: \(-24,\ 20,\ -5\); row 2: \(18,\ -15,\ 4\); row 3: \(5,\ -4,\ 1\).

\(\mathrm{adj}\,A=\begin{bmatrix}-24&18&5\\20&-15&-4\\-5&4&1\end{bmatrix}\), so \(A^{-1}=\mathrm{adj}\,A\) (since \(|A|=1\)).

Check first row of \(AA^{-1}\): \(1(-24)+2(20)+3(-5)=1\), \(1(18)+2(-15)+3(4)=0\), \(1(5)+2(-4)+3(1)=0\). Correct.

Common mistakes

• Forgetting to transpose the cofactor matrix.
• Sign errors in cofactors; use the checkerboard pattern.
• Not checking \(|A|\ne0\) first.

How MU asks it

A pre-requisite, so it rarely carries marks on its own, but it is needed inside questions like "Show that \(A\) is orthogonal and hence find \(A^{-1}\)" or coding theory problems. In tutorials you may be asked to find \(A^{-1}\) and verify \(AA^{-1}=I\). Show the determinant, cofactors, adjoint, and the final matrix.