The cofactors of a matrix, arranged the right way, almost give its inverse. Put them in a matrix, transpose it, and divide by the determinant. The same idea solves a square system one unknown at a time, each unknown a ratio of two determinants.

A recipe can be written out in full, every quantity spelled out, even when a cook would take a shortcut. The adjugate writes the inverse out in full, entry by entry; elimination, later, is the shortcut.

The adjugate of an $n\times n$ matrix $A$ is the transpose of its matrix of cofactors: the entry in row $i$ and column $j$ of $\operatorname{adj}A$ is $C_{ji}$. Laplace's expansion shows that $A\operatorname{adj}A=(\det A)I$, so when $\det A\neq0$ the inverse is the adjugate divided by the determinant. For $n=2$ this is exactly the formula of the lesson on $2\times2$ inverses.

• A square matrix is invertible exactly when $\det A\neq0$. The last chapter proved one direction; the adjugate proves the other by writing the inverse down.
• Cramer's rule: when $\det A\neq0$, the solution of $A\mathbf{x}=\mathbf{b}$ has $x_i=\det A_i/\det A$, where $A_i$ is $A$ with column $i$ replaced by $\mathbf{b}$.
• Both formulas are exact, and good for small matrices and for reasoning. For large matrices elimination is far quicker, because every cofactor is a determinant.
• Each entry of $A^{-1}$ is a ratio of determinants: a cofactor over $\det A$.

Entry $(i,j)$ of the product pairs row $i$ of $A$ with column $j$ of $\operatorname{adj}A$, which holds the cofactors of row $j$. When $i=j$ this is Laplace's expansion along row $i$. When $i\neq j$ it pairs a row with the cofactors of a different row, and that gives $0$. So the product has $\det A$ on the diagonal and zeros elsewhere.

Row $i$ of $A$ against column $j$ of $\operatorname{adj}A$, whose entries are the cofactors of row $j$ of $A$.

On the diagonal this is Laplace's expansion along row $i$.

Off the diagonal a row meets the cofactors of another row, which is the expansion of a matrix with two equal rows.

$\det A$ on the diagonal and $0$ everywhere else is $\det A$ times the identity.

Divide by the determinant when it is not zero. The product in the other order works the same way with columns.

Form the matrix of cofactors, $C_{ij}$ in position $(i,j)$, and then transpose it. The transpose is easy to forget, and forgetting it gives the wrong inverse whenever the matrix is not symmetric.

To find $x_i$, replace column $i$ of $A$ by $\mathbf{b}$ and divide that determinant by $\det A$. It works only when $\det A\neq0$.

For $2x+y=5$, $x+3y=5$: $\det A=5$, and replacing column 1 by $(5,5)$ gives $\det A_1=\begin{vmatrix}5&1\\5&3\end{vmatrix}=10$, so $x=2$.

• $\operatorname{adj}A$ is the transpose of the cofactor matrix: $(\operatorname{adj}A)_{ij}=C_{ji}$.
• $A\operatorname{adj}A=(\det A)I$, so $A^{-1}=\tfrac{1}{\det A}\operatorname{adj}A$ when .
• A square matrix is invertible exactly when $\det A\neq0$.
• Cramer's rule: $x_i=\det A_i/\det A$, with column $i$ replaced by .

Great work! You now know how to:

• build the adjugate from the cofactors, transposed;
• write $A^{-1}$ as $\tfrac{1}{\det A}\operatorname{adj}A$;
• state why $\det A\neq0$ is exactly invertibility;
• solve a small square system by Cramer’s rule.