Inversion of matrices

Vectors and matrices · Chapter 4.3 · Lecture notes (PDF) · Exercises: Inversion of matrices · Slides (PDF) · App: Inverse matrix

1. Properties of the inverse matrix

• Recall (see the chapter on matrices) that a square matrix \(\boldsymbol A \in \R^{n\times n}\) is invertible if there exists a matrix \(\boldsymbol A^{-1} \in \R^{n\times n}\) such that \(\boldsymbol A\boldsymbol A^{-1} = \boldsymbol A^{-1}\boldsymbol A = \boldsymbol I\). A square matrix which is not invertible is called singular.

• In this chapter, \(\boldsymbol A'\) denotes the transpose of \(\boldsymbol A\), as in the chapter on matrices.

Proposition 1: Uniqueness of the inverse

Let \(\boldsymbol A \in \R^{n\times n}\) be invertible. If \(\boldsymbol B, \boldsymbol C \in \R^{n\times n}\) are such that \(\boldsymbol A\boldsymbol B = \boldsymbol B\boldsymbol A = \boldsymbol I\) and \(\boldsymbol A\boldsymbol C = \boldsymbol C\boldsymbol A = \boldsymbol I\), then \(\boldsymbol B = \boldsymbol C\).

Proof

Using the associativity of the matrix product, we have:

\[ \boldsymbol B = \boldsymbol B\boldsymbol I = \boldsymbol B(\boldsymbol A\boldsymbol C) = (\boldsymbol B\boldsymbol A)\boldsymbol C = \boldsymbol I\boldsymbol C = \boldsymbol C. \]

□

Proposition 2: Properties of the inverse

Let \(\boldsymbol A, \boldsymbol B \in \R^{n\times n}\) be invertible matrices. Then:

1. \(\boldsymbol A^{-1}\) is invertible and \((\boldsymbol A^{-1})^{-1} = \boldsymbol A\);

2. \(\boldsymbol A\boldsymbol B\) is invertible and \((\boldsymbol A\boldsymbol B)^{-1} = \boldsymbol B^{-1}\boldsymbol A^{-1}\);

3. \(\boldsymbol A'\) is invertible and \((\boldsymbol A')^{-1} = (\boldsymbol A^{-1})'\);

4. \(\det(\boldsymbol A) \neq 0\) and \(\det(\boldsymbol A^{-1}) = \dfrac{1}{\det(\boldsymbol A)}\).

Proof

By the uniqueness of the inverse, in each case it is enough to exhibit a matrix that multiplied on the left and on the right gives \(\boldsymbol I\).

1. From \(\boldsymbol A^{-1}\boldsymbol A = \boldsymbol A\boldsymbol A^{-1} = \boldsymbol I\), the matrix \(\boldsymbol A\) is the inverse of \(\boldsymbol A^{-1}\).

2. Using associativity:

   \[ (\boldsymbol A\boldsymbol B)(\boldsymbol B^{-1}\boldsymbol A^{-1}) = \boldsymbol A(\boldsymbol B\boldsymbol B^{-1})\boldsymbol A^{-1} = \boldsymbol A\boldsymbol I\boldsymbol A^{-1} = \boldsymbol A\boldsymbol A^{-1} = \boldsymbol I, \]
   \[ (\boldsymbol B^{-1}\boldsymbol A^{-1})(\boldsymbol A\boldsymbol B) = \boldsymbol B^{-1}(\boldsymbol A^{-1}\boldsymbol A)\boldsymbol B = \boldsymbol B^{-1}\boldsymbol I\boldsymbol B = \boldsymbol B^{-1}\boldsymbol B = \boldsymbol I. \]

3. Recall that \((\boldsymbol C\boldsymbol D)' = \boldsymbol D'\boldsymbol C'\) for all \(\boldsymbol C, \boldsymbol D \in \R^{n\times n}\): indeed, the entry \((i,j)\) of \((\boldsymbol C\boldsymbol D)'\) is the entry \((j,i)\) of \(\boldsymbol C\boldsymbol D\), i.e., \(\sum_{k=1}^n c_{jk}d_{ki} = \sum_{k=1}^n d'_{ik}c'_{kj}\), which is the entry \((i,j)\) of \(\boldsymbol D'\boldsymbol C'\). Then, since \(\boldsymbol I' = \boldsymbol I\):

   \[ \boldsymbol A'(\boldsymbol A^{-1})' = (\boldsymbol A^{-1}\boldsymbol A)' = \boldsymbol I' = \boldsymbol I, \qquad (\boldsymbol A^{-1})'\boldsymbol A' = (\boldsymbol A\boldsymbol A^{-1})' = \boldsymbol I' = \boldsymbol I. \]

4. From the product rule of determinants:

   \[ \det(\boldsymbol A)\det(\boldsymbol A^{-1}) = \det(\boldsymbol A\boldsymbol A^{-1}) = \det(\boldsymbol I) = 1. \]

   Hence \(\det(\boldsymbol A)\neq 0\) and \(\det(\boldsymbol A^{-1}) = 1/\det(\boldsymbol A)\).

□

• Note the reversed order in \((\boldsymbol A\boldsymbol B)^{-1} = \boldsymbol B^{-1}\boldsymbol A^{-1}\): since the matrix product is not commutative, in general \((\boldsymbol A\boldsymbol B)^{-1} \neq \boldsymbol A^{-1}\boldsymbol B^{-1}\).

• Property 4 shows that a matrix with \(\det(\boldsymbol A) = 0\) cannot be invertible, i.e., it is singular.

2. Computation of the inverse

2.1 Gauss–Jordan method

Gauss–Jordan method (idea). To compute \( \boldsymbol A^{-1} \), we build the augmented matrix

\[ \left(\, \boldsymbol A \mid \boldsymbol I \,\right) \]

and we apply elementary row operations to transform the left block into the identity matrix. If we obtain

\[ \left(\, \boldsymbol I \mid \boldsymbol B \,\right), \]

then \( \boldsymbol B = \boldsymbol A^{-1} \).

Example 1: inverse matrix via Gauss–Jordan elimination

Let us consider

\[ \boldsymbol A= \begin{pmatrix} 2 & 1\\[0.5ex] 1 & 1 \end{pmatrix}. \]

We compute the inverse matrix \(\boldsymbol A^{-1}\) with the method of Gauss–Jordan.

\[\begin{align*} &\left( \begin{array}{cc|cc} 2 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 \end{array} \right) \\[2ex] &\left( \begin{array}{cc|cc} 1 & 1 & 0 & 1\\ 2 & 1 & 1 & 0 \end{array} \right) \hspace{1em}\text{(} R_1 \leftrightarrow R_2 \text{)} \\[2ex] &\left( \begin{array}{cc|cc} 1 & 1 & 0 & 1\\ 0 & -1 & 1 & -2 \end{array} \right) \hspace{1em}\text{(} R_2 \leftarrow R_2 - 2R_1 \text{)} \\[2ex] &\left( \begin{array}{cc|cc} 1 & 1 & 0 & 1\\ 0 & 1 & -1 & 2 \end{array} \right) \hspace{1em}\text{(} R_2 \leftarrow -R_2 \text{)} \\[2ex] &\left( \begin{array}{cc|cc} 1 & 0 & 1 & -1\\ 0 & 1 & -1 & 2 \end{array} \right) \hspace{1em}\text{(} R_1 \leftarrow R_1 - R_2 \text{)} \end{align*}\]

Since the left block is the identity matrix, we obtain:

\[ \boldsymbol A^{-1}= \begin{pmatrix} 1 & -1\\[0.5ex] -1 & 2 \end{pmatrix}. \]

We can verify:

\[ \boldsymbol A \boldsymbol A^{-1} = \begin{pmatrix} 2 & 1\\[0.5ex] 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1\\[0.5ex] -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0\\[0.5ex] 0 & 1 \end{pmatrix} = \boldsymbol I. \]

Example 2: inverse matrix via Gauss–Jordan elimination

Let us consider

\[ \boldsymbol A= \begin{pmatrix} 1 & 2 & 0\\[0.5ex] 0 & 1 & 1\\[0.5ex] 2 & 0 & 1 \end{pmatrix}. \]

We compute the inverse matrix \(\boldsymbol A^{-1}\) with the method of Gauss–Jordan.

\[\begin{align*} &\left( \begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 2 & 0 & 1 & 0 & 0 & 1 \\ \end{array} \right) \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & -4 & 1 & -2 & 0 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow R_3 - 2R_1 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & -2 & 1 & -2 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & -4 & 1 & -2 & 0 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_1 \leftarrow R_1 - 2R_2 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & -2 & 1 & -2 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 5 & -2 & 4 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow R_3 + 4R_2 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & -2 & 1 & -2 & 0 \\[1ex] 0 & 1 & 1 & 0 & 1 & 0 \\[1ex] 0 & 0 & 1 & -\tfrac{2}{5} & \tfrac{4}{5} & \tfrac{1}{5} \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow \tfrac{1}{5}R_3 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac{1}{5} & -\tfrac{2}{5} & \tfrac{2}{5} \\[1ex] 0 & 1 & 1 & 0 & 1 & 0 \\[1ex] 0 & 0 & 1 & -\tfrac{2}{5} & \tfrac{4}{5} & \tfrac{1}{5} \\ \end{array} \right) \hspace{1em} \text{(} R_1 \leftarrow R_1 + 2R_3 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac{1}{5} & -\tfrac{2}{5} & \tfrac{2}{5} \\[1ex] 0 & 1 & 0 & \tfrac{2}{5} & \tfrac{1}{5} & -\tfrac{1}{5} \\[1ex] 0 & 0 & 1 & -\tfrac{2}{5} & \tfrac{4}{5} & \tfrac{1}{5} \\ \end{array} \right) \hspace{1em} \text{(} R_2 \leftarrow R_2 - R_3 \text{)} \end{align*}\]

Since the left block is the identity matrix, we obtain:

\[ \boldsymbol A^{-1}= \begin{pmatrix} \tfrac{1}{5} & -\tfrac{2}{5} & \tfrac{2}{5}\\[0.8ex] \tfrac{2}{5} & \tfrac{1}{5} & -\tfrac{1}{5}\\[0.8ex] -\tfrac{2}{5} & \tfrac{4}{5} & \tfrac{1}{5} \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 1 & -2 & 2\\[0.5ex] 2 & 1 & -1\\[0.5ex] -2 & 4 & 1 \end{pmatrix} . \]

Example 3: inverse matrix via Gauss–Jordan elimination with a row swap

Let us consider

\[ \boldsymbol A= \begin{pmatrix} 0 & 1 & 2\\[0.5ex] 1 & 0 & 3\\[0.5ex] 4 & -3 & 8 \end{pmatrix}. \]

We compute the inverse matrix \(\boldsymbol A^{-1}\) with the method of Gauss–Jordan. Since \(a_{11}=0\), we start with a row swap.

\[\begin{align*} &\left( \begin{array}{ccc|ccc} 0 & 1 & 2 & 1 & 0 & 0 \\ 1 & 0 & 3 & 0 & 1 & 0 \\ 4 & -3 & 8 & 0 & 0 & 1 \\ \end{array} \right) \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 4 & -3 & 8 & 0 & 0 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_1 \leftrightarrow R_2 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & -3 & -4 & 0 & -4 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow R_3 - 4R_1 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 & -4 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow R_3 + 3R_2 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 3 & 0 & 1 & 0 \\[1ex] 0 & 1 & 2 & 1 & 0 & 0 \\[1ex] 0 & 0 & 1 & \tfrac{3}{2} & -2 & \tfrac{1}{2} \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow \tfrac{1}{2}R_3 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & -\tfrac{9}{2} & 7 & -\tfrac{3}{2} \\[1ex] 0 & 1 & 2 & 1 & 0 & 0 \\[1ex] 0 & 0 & 1 & \tfrac{3}{2} & -2 & \tfrac{1}{2} \\ \end{array} \right) \hspace{1em} \text{(} R_1 \leftarrow R_1 - 3R_3 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & -\tfrac{9}{2} & 7 & -\tfrac{3}{2} \\[1ex] 0 & 1 & 0 & -2 & 4 & -1 \\[1ex] 0 & 0 & 1 & \tfrac{3}{2} & -2 & \tfrac{1}{2} \\ \end{array} \right) \hspace{1em} \text{(} R_2 \leftarrow R_2 - 2R_3 \text{)} \end{align*}\]

Since the left block is the identity matrix, we obtain:

\[ \boldsymbol A^{-1}= \begin{pmatrix} -\tfrac{9}{2} & 7 & -\tfrac{3}{2}\\[0.8ex] -2 & 4 & -1\\[0.8ex] \tfrac{3}{2} & -2 & \tfrac{1}{2} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} -9 & 14 & -3\\[0.5ex] -4 & 8 & -2\\[0.5ex] 3 & -4 & 1 \end{pmatrix}. \]

Observation 1: Singular matrices and the Gauss–Jordan method

Let \(\boldsymbol A \in \R^{n\times n}\). If, while applying elementary row operations to \(\left(\boldsymbol A \mid \boldsymbol I\right)\), a zero row appears in the left block, then \(\det(\boldsymbol A) = 0\) and \(\boldsymbol A\) is singular: the method stops and \(\boldsymbol A^{-1}\) does not exist.

Proof

The left block is obtained from \(\boldsymbol A\) by elementary row operations. Each of them multiplies the determinant by a nonzero number (\(-1\) for a row swap, \(\lambda\neq 0\) for a row scaling, \(1\) for a row addition). A matrix with a zero row has zero determinant (Laplace expansion along that row), hence \(\det(\boldsymbol A) = 0\). By property 4 of Proposition Proposition 2, \(\boldsymbol A\) is not invertible. □

Example 4: Gauss–Jordan elimination on a singular matrix

Let us consider

\[ \boldsymbol A= \begin{pmatrix} 1 & 2 & 3\\[0.5ex] 2 & 5 & 7\\[0.5ex] 1 & 3 & 4 \end{pmatrix}. \]

We try to compute \(\boldsymbol A^{-1}\) with the method of Gauss–Jordan.

\[\begin{align*} &\left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 7 & 0 & 1 & 0 \\ 1 & 3 & 4 & 0 & 0 & 1 \\ \end{array} \right) \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 1 & 3 & 4 & 0 & 0 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_2 \leftarrow R_2 - 2R_1 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 0 & 1 & 1 & -1 & 0 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow R_3 - R_1 \text{)} \\[2ex] &\left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 1 \\ \end{array} \right) \hspace{1em} \text{(} R_3 \leftarrow R_3 - R_2 \text{)} \end{align*}\]

The third row of the left block is zero: the left block cannot be transformed into the identity matrix, and the method stops. Hence \(\boldsymbol A\) is singular. Indeed, the third row of \(\boldsymbol A\) is the difference of the second and the first row, and

\[ \det(\boldsymbol A) = 1\cdot(5\cdot 4-7\cdot 3) - 2\cdot(2\cdot 4 - 7\cdot 1) + 3\cdot(2\cdot 3 - 5\cdot 1) = -1 - 2 + 3 = 0. \]

2.2 Inverse via the adjugate matrix

• Recall (see the chapter on matrices) that, given \(\boldsymbol A \in \R^{n\times n}\), the cofactor of the entry \(a_{ij}\) is \(C_{ij} = (-1)^{i+j}\det(\boldsymbol A_{ij})\), where \(\boldsymbol A_{ij}\) is the minor matrix obtained from \(\boldsymbol A\) by deleting row \(i\) and column \(j\).

Definition 1: Cofactor matrix and adjugate matrix

Let \(\boldsymbol A \in \R^{n\times n}\) with \(n\ge 2\). The cofactor matrix of \(\boldsymbol A\) is the matrix \(\boldsymbol C = (C_{ij}) \in \R^{n\times n}\) of the cofactors of \(\boldsymbol A\). The adjugate matrix of \(\boldsymbol A\) is the transpose of the cofactor matrix:

\[ \mathrm{adj}(\boldsymbol A) = \boldsymbol C', \qquad \text{i.e.,} \qquad \big(\mathrm{adj}(\boldsymbol A)\big)_{ij} = C_{ji}. \]

Theorem 1: Adjugate formula

Let \(\boldsymbol A \in \R^{n\times n}\) with \(n\ge 2\). Then

\[ \boldsymbol A\,\mathrm{adj}(\boldsymbol A) = \mathrm{adj}(\boldsymbol A)\,\boldsymbol A = \det(\boldsymbol A)\,\boldsymbol I. \]

Consequently, if \(\det(\boldsymbol A)\neq 0\), then \(\boldsymbol A\) is invertible and

\[ \boldsymbol A^{-1} = \frac{1}{\det(\boldsymbol A)}\,\mathrm{adj}(\boldsymbol A). \]

Proof

The entry \((i,j)\) of \(\boldsymbol A\,\mathrm{adj}(\boldsymbol A)\) is

\[ \sum_{k=1}^{n} a_{ik}\,C_{jk}. \]

• If \(i=j\), this is the Laplace expansion of \(\det(\boldsymbol A)\) along row \(i\).

• If \(i\neq j\), this is the Laplace expansion along row \(j\) of the matrix \(\widetilde{\boldsymbol A}\) obtained from \(\boldsymbol A\) by replacing row \(j\) with row \(i\) (the cofactors \(C_{jk}\) do not depend on row \(j\)). The matrix \(\widetilde{\boldsymbol A}\) has two equal rows: swapping them leaves \(\widetilde{\boldsymbol A}\) unchanged and changes the sign of its determinant, so \(\det(\widetilde{\boldsymbol A}) = -\det(\widetilde{\boldsymbol A})\), i.e., \(\det(\widetilde{\boldsymbol A}) = 0\).

Hence \(\boldsymbol A\,\mathrm{adj}(\boldsymbol A) = \det(\boldsymbol A)\,\boldsymbol I\). The identity \(\mathrm{adj}(\boldsymbol A)\,\boldsymbol A = \det(\boldsymbol A)\,\boldsymbol I\) is obtained in the same way, using Laplace expansions along columns. If \(\det(\boldsymbol A) \neq 0\), dividing by \(\det(\boldsymbol A)\) shows that \(\frac{1}{\det(\boldsymbol A)}\mathrm{adj}(\boldsymbol A)\) is the inverse of \(\boldsymbol A\). □

Observation 2: Invertibility and determinant

A square matrix \(\boldsymbol A \in \R^{n\times n}\) is invertible if and only if \(\det(\boldsymbol A) \neq 0\).

Proof

If \(\boldsymbol A\) is invertible, then \(\det(\boldsymbol A)\neq 0\) by property 4 of Proposition Proposition 2. Conversely, if \(\det(\boldsymbol A)\neq 0\), then \(\boldsymbol A\) is invertible by Theorem Theorem 1 (for \(n=1\), \(\boldsymbol A = (a)\) with \(a\neq 0\) and \(\boldsymbol A^{-1} = (1/a)\)). □

Observation 3: Inverse of a \(2\times 2\) matrix

Let

\[ \boldsymbol A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in \R^{2\times 2} \quad \text{with} \quad \det(\boldsymbol A) = ad - bc \neq 0. \]

Then

\[ \boldsymbol A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b\\ -c & a \end{pmatrix}. \]

Proof

The minor matrices of \(\boldsymbol A\) are \(1\times 1\), so the cofactors are \(C_{11} = d\), \(C_{12} = -c\), \(C_{21} = -b\), \(C_{22} = a\). Hence

\[ \boldsymbol C = \begin{pmatrix} d & -c\\ -b & a \end{pmatrix}, \qquad \mathrm{adj}(\boldsymbol A) = \boldsymbol C' = \begin{pmatrix} d & -b\\ -c & a \end{pmatrix}, \]

and the formula follows from Theorem Theorem 1. □

Inverse of a \(2\times 2\) matrix (rule). Swap the two diagonal entries, change the sign of the two off-diagonal entries, and divide by the determinant.

Example 5: inverse of a \(2\times 2\) matrix with the formula

Let us consider again

\[ \boldsymbol A= \begin{pmatrix} 2 & 1\\[0.5ex] 1 & 1 \end{pmatrix}. \]

We have \(\det(\boldsymbol A) = 2\cdot 1 - 1\cdot 1 = 1 \neq 0\), hence

\[ \boldsymbol A^{-1} = \frac{1}{1} \begin{pmatrix} 1 & -1\\[0.5ex] -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -1\\[0.5ex] -1 & 2 \end{pmatrix}, \]

which is the matrix obtained with the method of Gauss–Jordan in Example Example 1.

Example 6: inverse of a \(3\times 3\) matrix with the adjugate formula

Let us consider again

\[ \boldsymbol A= \begin{pmatrix} 1 & 2 & 0\\[0.5ex] 0 & 1 & 1\\[0.5ex] 2 & 0 & 1 \end{pmatrix}. \]

By Laplace expansion along the first row:

\[ \det(\boldsymbol A) = 1\cdot(1\cdot 1 - 1\cdot 0) - 2\cdot(0\cdot 1 - 1\cdot 2) + 0 = 1 + 4 = 5 \neq 0. \]

The cofactors are:

\[\begin{align*} C_{11} &= +\det\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} = 1, & C_{12} &= -\det\begin{pmatrix} 0 & 1\\ 2 & 1\end{pmatrix} = 2, & C_{13} &= +\det\begin{pmatrix} 0 & 1\\ 2 & 0\end{pmatrix} = -2,\\[1ex] C_{21} &= -\det\begin{pmatrix} 2 & 0\\ 0 & 1\end{pmatrix} = -2, & C_{22} &= +\det\begin{pmatrix} 1 & 0\\ 2 & 1\end{pmatrix} = 1, & C_{23} &= -\det\begin{pmatrix} 1 & 2\\ 2 & 0\end{pmatrix} = 4,\\[1ex] C_{31} &= +\det\begin{pmatrix} 2 & 0\\ 1 & 1\end{pmatrix} = 2, & C_{32} &= -\det\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix} = -1, & C_{33} &= +\det\begin{pmatrix} 1 & 2\\ 0 & 1\end{pmatrix} = 1. \end{align*}\]

Hence

\[ \boldsymbol C = \begin{pmatrix} 1 & 2 & -2\\ -2 & 1 & 4\\ 2 & -1 & 1 \end{pmatrix}, \qquad \mathrm{adj}(\boldsymbol A) = \boldsymbol C' = \begin{pmatrix} 1 & -2 & 2\\ 2 & 1 & -1\\ -2 & 4 & 1 \end{pmatrix}, \]

and

\[ \boldsymbol A^{-1} = \frac{1}{5} \begin{pmatrix} 1 & -2 & 2\\ 2 & 1 & -1\\ -2 & 4 & 1 \end{pmatrix}, \]

which is the matrix obtained with the method of Gauss–Jordan in Example Example 2.

• For \(n=2\) (and often for \(n=3\)) the adjugate formula is convenient for computations by hand. For larger matrices, the method of Gauss–Jordan requires far fewer operations.

Try it in the lab

the same computation step by step: change the matrix and watch the steps change.

Exercises and lab

• Exercises · the exercise sheet of this chapter: 10 exercises with worked solutions
• Lab · Inverse matrix — Gauss–Jordan on \((\boldsymbol A \mid \boldsymbol I)\) or the cofactor formula.