Adjoint and inverse

Minors and cofactors, restated. For a square matrix $A=[a_{ij}]$, the minor $M_{ij}$ of the entry $a_{ij}$ is the determinant of the matrix left after deleting row $i$ and column $j$. Attaching the checkerboard sign gives the cofactor:

where: $i$ is the row index, $j$ is the column index, and the sign alternates $+,-,+,-,\dots$ starting from $C_{11}=+M_{11}$.

Building the adjoint. Collect every cofactor into a matrix of the same shape as $A$, then transpose it:

Transposing matters: the $(i,j)$ entry of $\operatorname{Adj}A$ is $C_{ji}$, not $C_{ij}$. The two coincide only when $A$ is symmetric, which is why "forgetting to transpose" is the single most common slip on this topic.

Singular versus non-singular. A square matrix $A$ is non-singular if $|A|\neq0$ and singular if $|A|=0$. Only a non-singular matrix has an inverse:

where: $|A|$ is the determinant of $A$ and $\operatorname{Adj}A$ is the adjoint constructed above. If $|A|=0$, no inverse exists at all — dividing by $|A|$ is simply impossible, not merely inconvenient.

Why the formula works. The identity $A\cdot\operatorname{Adj}A=\operatorname{Adj}A\cdot A=|A|\,I$ holds for every square matrix, singular or not (it is proved below). Dividing through by $|A|$, valid only when $|A|\neq0$, gives $A\left(\tfrac{1}{|A|}\operatorname{Adj}A\right)=I$, so $\tfrac{1}{|A|}\operatorname{Adj}A$ is exactly the two-sided inverse of $A$.

A useful $2\times2$ shortcut. For $A=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$, the cofactors work out to $C_{11}=d,\ C_{12}=-c,\ C_{21}=-b,\ C_{22}=a$, and transposing gives

— swap the diagonal entries, negate the off-diagonal entries. This is exact, not approximate, and it saves real time whenever the matrix is $2\times2$; for $3\times3$ and larger, the full minor-and-cofactor process below is required.

Uniqueness of the inverse. The syllabus explicitly asks students to prove that an inverse, when it exists, is unique. Suppose $B$ and $C$ are both inverses of $A$, so $AB=BA=I$ and $AC=CA=I$. Then $B = BI = B(AC) = (BA)C = IC = C$, so $B=C$. There can never be two different inverses of the same matrix.

The claim that $A\left(\tfrac{1}{|A|}\operatorname{Adj}A\right)=I$ deserves a full argument, since exam questions sometimes ask for exactly this justification, and it is what makes the whole formula trustworthy rather than a rule to memorise blindly. The goal is to show that $A\cdot\operatorname{Adj}A=|A|\,I$ for every square matrix — an identity with $n^2$ entries to check, each one either a determinant expansion or the determinant of a matrix with a repeated row.

Step 1. Look at a diagonal entry of the product $A\cdot\operatorname{Adj}A$. Since $\operatorname{Adj}A=[C_{ij}]^T$, its $(k,i)$ entry is $C_{ik}$, so the $(i,i)$ entry of the product is $\sum_k a_{ik}C_{ik}$. This sum is precisely the cofactor expansion of $|A|$ along row $i$ — by definition of a determinant expansion. So every diagonal entry of $A\cdot\operatorname{Adj}A$ equals $|A|$.

Step 2. Look at an off-diagonal entry, say $(i,j)$ with $i\neq j$. By the same rule for matrix multiplication with the transposed cofactor matrix, this entry is $\sum_k a_{ik}C_{jk}$. That sum is the cofactor expansion (along row $j$) of the determinant of a different matrix: take $A$ and replace its row $j$ with a second copy of row $i$.

Step 3. That constructed matrix has two identical rows (row $i$ appears twice), and a standard determinant property says any determinant with two identical rows is zero. So the sum $\sum_k a_{ik}C_{jk}=0$, which means every off-diagonal entry of $A\cdot\operatorname{Adj}A$ vanishes.

Step 4. Combine Steps 1 and 3: $A\cdot\operatorname{Adj}A$ has $|A|$ on every diagonal entry and $0$ everywhere else, which is exactly $|A|\,I$. The identical argument run along columns instead of rows gives $\operatorname{Adj}A\cdot A=|A|\,I$ too. Dividing this identity by $|A|$ is the one and only place the calculation can fail — precisely when $|A|=0$, the singular case — which is why the inverse formula carries that condition and no other.

Find $\operatorname{Adj}A$ and $A^{-1}$ for $A=\begin{pmatrix}2 & -1\\ 4 & 3\end{pmatrix}$, then verify the result.

Step 1. Check non-singularity by computing the determinant: $|A| = 2(3)-(-1)(4) = 6+4 = 10 \neq 0$. Since $|A|\neq0$, $A^{-1}$ exists.

Step 2. Build the adjoint using the $2\times2$ shortcut — swap the diagonal entries, negate the off-diagonal entries:

Step 3. Apply the inverse formula:

Step 4. Verify by multiplying $A$ and $A^{-1}$ together — this is the fastest arithmetic check available:

The inverse is $A^{-1}=\begin{pmatrix}0.3 & 0.1\\ -0.4 & 0.2\end{pmatrix}$, confirmed by $AA^{-1}=I$.

Competitive stretch

For a $3\times3$ matrix $A=\begin{pmatrix}1 & 0 & 2\\ 2 & 1 & 0\\ 0 & 2 & 1\end{pmatrix}$, expanding along the first row gives $|A|=1(1)-0+2(4)=9\neq0$. Computing all nine cofactors and transposing gives

The pattern is identical to the $2\times2$ case, just with nine cofactors instead of four — the $2\times2$ shortcut of Step 2 no longer applies once the matrix grows beyond order $2$.

• Forgetting to transpose the cofactor matrix — the adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself; the two agree only when $A$ is symmetric.
• Sign errors in cofactors — the checkerboard pattern $(-1)^{i+j}$ must be applied to every entry; one missed sign ruins the whole inverse.
• Trying to invert a singular matrix — if $|A|=0$, $A^{-1}$ simply does not exist; there is nothing to "divide by zero" in the formula.
• Skipping the verification step $AA^{-1}=I$ — in a multi-step $3\times3$ cofactor calculation this is the quickest way to catch an arithmetic slip before it costs marks.

Ask the tutor: if a $3\times3$ cofactor calculation feels error-prone, ask the AI tutor to walk through the adjoint one cofactor at a time, or to check your worked answer for a sign mistake.