Inverse of a Square Matrix by Adjoint Method

The elementary-transformation method of section 2.2.1 works but is elaborate, needing a series of transformations tracked carefully. This section develops a second, direct route: the adjoint method. It relies on the definitions of a minor, a cofactor, and the adjoint of a matrix.

Recalling minor and cofactor. The minor of an element $a_{ij}$ of a determinant is the determinant obtained by deleting the $i$-th row and $j$-th column in which $a_{ij}$ lies, denoted $M_{ij}$. The cofactor of $a_{ij}$ is $A_{ij} = (-1)^{i+j}M_{ij}$.

Worked illustration. For $A=\begin{bmatrix}1&-2&3\\4&5&-6\\7&-8&9\end{bmatrix}$, the minor of the element $4$ (which sits in row 2, column 1) is $M_{21}=\begin{vmatrix}-2&3\\-8&9\end{vmatrix}=-18-(-24)=6$. The corresponding cofactor is $A_{21}=(-1)^{2+1}M_{21}=(-1)(6)=-6$.

Definition of the adjoint. For a square matrix $A=[a_{ij}]_{m\times m}$, the adjoint, written $\text{adj}\,A$, is defined as the transpose of the matrix of cofactors $[A_{ij}]_{m\times m}$, where $A_{ij}$ is the cofactor of $a_{ij}$, for every $i,j=1,2,\dots,m$. For a $3\times3$ matrix, the cofactor matrix is $\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}$ and the adjoint -- its transpose -- is $\text{adj}\,A=\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}$. The transpose step is easy to skip by mistake, but it is exactly what turns the "cofactor matrix" into the "adjoint".

Worked Examples -- cofactors and adjoint. For $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$: $A_{11}=(-1)^{1+1}(4)=4$, $A_{12}=(-1)^{1+2}(3)=-3$, $A_{21}=(-1)^{2+1}(2)=-2$, $A_{22}=(-1)^{2+2}(1)=1$; the required cofactors are $4,-3,-2,1$. For $A=\begin{bmatrix}2&-3\\4&1\end{bmatrix}$: $A_{11}=1$, $A_{12}=-4$, $A_{21}=3$, $A_{22}=2$, so the cofactor matrix is $\begin{bmatrix}1&-4\\3&2\end{bmatrix}$ and $\text{adj}\,A=\begin{bmatrix}1&3\\-4&2\end{bmatrix}$. For $A=\begin{bmatrix}2&0&-1\\3&1&2\\-1&1&2\end{bmatrix}$, computing all nine cofactors gives the cofactor matrix $\begin{bmatrix}0&-8&4\\-1&3&-2\\1&-7&2\end{bmatrix}$, so $\text{adj}\,A=\begin{bmatrix}0&-1&1\\-8&3&-7\\4&-2&2\end{bmatrix}$.

Why $A(\text{adj}\,A)=|A|I$. A determinant can be expanded along any row using its own cofactors: e.g. $a_{21}A_{21}+a_{22}A_{22}+\dots+a_{2n}A_{2n}=|A|$. But if a row's entries are combined with a different row's cofactors, the sum is always $0$: e.g. $a_{21}A_{31}+a_{22}A_{32}+\dots+a_{2n}A_{3n}=0$. Multiplying these two facts out across every row/column pair shows $A\cdot\text{adj}\,A$ is a matrix with $|A|$ on every diagonal entry and $0$ everywhere else -- i.e. $A\cdot\text{adj}\,A = |A|\cdot I$. Dividing both sides by the scalar $|A|$ (valid whenever $|A|\neq0$) gives the key formula:

So if $A=[a_{ij}]_{m\times m}$ is non-singular, its inverse exists and is given directly by this formula. (Think about why $A^{-1}$ cannot exist when $A$ is singular: with $|A|=0$, dividing by $|A|$ is undefined, and indeed $A\cdot\text{adj}\,A=0\cdot I$ is the zero matrix, not $I$, whenever $A$ is singular.)

Worked Example -- adjoint method, $2\times2$. For $A=\begin{bmatrix}2&-2\\4&3\end{bmatrix}$: $M_{11}=3,\ A_{11}=3$; $M_{12}=4,\ A_{12}=-4$; $M_{21}=-2,\ A_{21}=2$; $M_{22}=2,\ A_{22}=2$. So $\text{adj}\,A=\begin{bmatrix}3&2\\-4&2\end{bmatrix}$, and $|A|=2(3)-4(-2)=6+8=14\neq0$. Hence $A^{-1}=\dfrac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}$.

Worked Example -- adjoint method, $3\times3$. For $A=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$: computing all nine cofactors gives $\text{adj}\,A=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$, and $|A|=2(4-1)+1(-2+1)+1(1-2)=6-1-1=4$. Hence $A^{-1}=\dfrac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$. …