The Adjugate as a Practical Inverse Method
For a small square matrix, the adjugate formula turns determinant calculations into an explicit inverse. The steps are mechanical, but each has a place where an error can creep in: a minor must be calculated from the correct deleted row and column, its cofactor needs the correct sign, and the cofactor matrix must be transposed before it is used. A final multiplication is an essential check.
All matrices in this tutorial have real entries and are square of size \(2\times2\) or \(3\times3\). For a square matrix \(\mathbf{A}\), the minor \(M_{ij}\) is the determinant of the matrix left after deleting row \(i\) and column \(j\). The cofactor in position \((i,j)\) is \(C_{ij}=(-1)^{i+j}M_{ij}\). In particular, the signs follow a checkerboard pattern, beginning with a positive sign in the upper-left corner.
The cofactor matrix has entries \(C_{ij}\) in their original positions. The adjugate, written \(\operatorname{adj}\mathbf{A}\), is the transpose of that cofactor matrix: its entry in position \((i,j)\) is \(C_{ji}\). For a \(2\times2\) matrix, the minor is the single entry remaining after deletion, so these definitions give an especially compact formula.
The determinant condition matters: the formula requires division by \(\det\mathbf{A}\). If this number is zero, the displayed expression is undefined, and the matrix has no inverse. The next result states precisely when the method applies.
Proof. Suppose \(\det\mathbf{A}\ne0\). Define \(\mathbf{B}=\dfrac{1}{\det\mathbf{A}}\operatorname{adj}\mathbf{A}\). By the adjugate identity and scalar multiplication, \(\mathbf{A}\mathbf{B}=\mathbf{I}\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}\), so \(\mathbf{B}\) is an inverse of \(\mathbf{A}\). Conversely, suppose \(\mathbf{A}\) is invertible, with inverse \(\mathbf{B}\). The determinant product rule gives \(\det(\mathbf{A}\mathbf{B})=\det\mathbf{A}\det\mathbf{B}\). Since \(\mathbf{A}\mathbf{B}=\mathbf{I}\) and \(\det\mathbf{I}=1\), we have \(\det\mathbf{A}\det\mathbf{B}=1\), which forces \(\det\mathbf{A}\ne0\). Thus the condition is both necessary and sufficient, and in the nonzero case the inverse is the matrix \(\mathbf{B}\) already constructed. \(\square\)
The \(2\times2\) Formula
Write \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\). Its determinant is \(ad-bc\). The four signed cofactors, in order, are \(d,-c,-b,a\): for example, \(C_{12}=-c\), since deleting row 1 and column 2 leaves \(c\), and the cofactor sign is negative. Transposing the cofactor matrix gives the adjugate below.
Proof. Let \(\mathbf{B}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\). Multiplying in the first order gives \(\mathbf{A}\mathbf{B}=\dfrac{1}{ad-bc}\begin{bmatrix}ad-bc&-ab+ab\\cd-dc&-bc+ad\end{bmatrix}=\mathbf{I}_2\). Multiplying in the other order gives \(\mathbf{B}\mathbf{A}=\dfrac{1}{ad-bc}\begin{bmatrix}da-bc&db-bd\\-ca+ac&-cb+ad\end{bmatrix}=\mathbf{I}_2\). The denominator is nonzero by hypothesis, and both products are the identity, so \(\mathbf{B}\) is the inverse. \(\square\)
Worked Example: Inverting a \(2\times2\) Matrix
Find the inverse of \(\mathbf{A}=\begin{bmatrix}2&1\\5&3\end{bmatrix}\) by computing its cofactors and adjugate, then verify the result.
First calculate the determinant: \(\det\mathbf{A}=2(3)-1(5)=6-5=1\), which is nonzero. The minors are \(M_{11}=3\), \(M_{12}=5\), \(M_{21}=1\), and \(M_{22}=2\). Applying the signs \(+,-,-,+\), the cofactor matrix and its transpose are
Divide the adjugate by the determinant. Since the determinant is \(1\), the inverse is \(\mathbf{A}^{-1}=\begin{bmatrix}3&-1\\-5&2\end{bmatrix}\). To check, multiply by the original matrix:
The multiplication gives \(\mathbf{I}_2\), as required.
Computing a \(3\times3\) Adjugate
For a \(3\times3\) matrix, each minor is a \(2\times2\) determinant. Compute all nine signed cofactors in their original positions, then transpose the resulting matrix. The determinant can be found by cofactor expansion along any row or column; here we use the first row, so \(\det\mathbf{A}=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\). This is a convenient place to check the cofactor signs: the signs in the first row are \(+,-,+\).
Worked Example: Inverting a \(3\times3\) Matrix
Find and verify the inverse of \(\mathbf{A}=\begin{bmatrix}1&2&0\\0&1&1\\2&0&1\end{bmatrix}\).
Compute the minors and apply the checkerboard signs. The calculations by row are
For instance, \(M_{12}=\det\begin{bmatrix}0&1\\2&1\end{bmatrix}=0(1)-1(2)=-2\), and the negative sign for \(C_{12}\) makes \(C_{12}=2\). Applying all nine signs gives
Expand the determinant along the first row, using the first row of cofactors: \(\det\mathbf{A}=1(1)+2(2)+0(-2)=5\). Since \(5\ne0\), the adjugate formula applies, giving
Verify by multiplying \(\mathbf{A}\) by the adjugate first. Each entry is shown so the diagonal and off-diagonal results can be checked:
Dividing this product by \(5\) confirms \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_3\). This also checks the determinant and the computed adjugate together.
Why the Computation Works
The calculation is an algorithm, not just a list of manipulations. The adjugate identity ensures that the product of the original matrix and its adjugate is the determinant times the identity. If the determinant is nonzero, dividing by it changes that product to the identity on both sides. This explains why every stage matters: cofactors form the adjugate only after transposition, and the determinant is the scale factor that must be divided out.
The \(2\times2\) case shows the same principle by direct multiplication. For any \(a,b,c,d\) with \(ad-bc\ne0\), the two off-diagonal entries in either product cancel, while both diagonal entries equal \(ad-bc\) before division. The formula is therefore not a memorized pattern detached from the definition of inverse; it constructs a matrix whose products with \(\mathbf{A}\) are exactly the identity.
When the Determinant Is Zero
Worked Example: Detecting a Matrix with No Inverse
Consider \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\). Its determinant is \(1(4)-2(2)=0\). The cofactors are \(C_{11}=4\), \(C_{12}=-2\), \(C_{21}=-2\), and \(C_{22}=1\), so \(\operatorname{adj}\mathbf{A}=\begin{bmatrix}4&-2\\-2&1\end{bmatrix}\). The expression \((1/\det\mathbf{A})\operatorname{adj}\mathbf{A}\) is undefined: division by zero is not permitted.
To check directly that no inverse exists, take the nonzero vector \(\mathbf{x}=\begin{bmatrix}2\\-1\end{bmatrix}\). Then \(\mathbf{A}\mathbf{x}=\begin{bmatrix}2-2\\4-4\end{bmatrix}=\mathbf{0}\). If an inverse existed, multiplication by \(\mathbf{A}^{-1}\) would give \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{A}\mathbf{x}=\mathbf{A}^{-1}\mathbf{0}=\mathbf{0}\), contradicting \(\mathbf{x}\ne\mathbf{0}\). Thus this matrix genuinely has no inverse.
Common Mistake: Forgetting the Transpose
In the first worked example, the cofactor matrix is \(\begin{bmatrix}3&-5\\-1&2\end{bmatrix}\). A tempting wrong calculation is to use that matrix directly as the adjugate, without transposing it. That would give the candidate \(\begin{bmatrix}3&-5\\-1&2\end{bmatrix}\), since the determinant is \(1\). Multiplying this candidate by the original matrix gives
The correction is to transpose the cofactor matrix first, producing \(\operatorname{adj}\mathbf{A}=\begin{bmatrix}3&-1\\-5&2\end{bmatrix}\), and then divide by the determinant. A second useful check is to confirm that the computed product equals \((\det\mathbf{A})\mathbf{I}\) before dividing. If it does not, revisit the minors, signs, and transpose.
For an exact small-matrix calculation, use this order:
If it is zero, stop: the matrix has no inverse. If it is nonzero, continue.
Delete the indexed row and column to form each minor, then apply the sign \((-1)^{i+j}\).
The resulting matrix is \(\operatorname{adj}\mathbf{A}\).
Multiply the original matrix by the proposed inverse and check that the result is the identity.
Check Your Understanding
Use exact arithmetic, keep the cofactors in their original positions until the transpose step, and verify each proposed inverse by multiplication.
- For \(\begin{bmatrix}3&1\\2&1\end{bmatrix}\), calculate the cofactors, adjugate, determinant, and inverse.
- Compute the inverse of \(\begin{bmatrix}1&0&1\\0&1&1\\1&1&0\end{bmatrix}\) using cofactors and the adjugate.
- For \(\begin{bmatrix}2&4\\1&2\end{bmatrix}\), calculate the determinant and explain why the adjugate formula cannot produce an inverse.
- In a \(3\times3\) cofactor matrix, what operation produces the adjugate, and what goes wrong if that operation is omitted?
- State the identity used to verify the adjugate formula when \(\det\mathbf{A}\ne0\).
