Invertible Matrices

Invertible Matrices#

Table of Contents#

Definition of a Square Matrix#

A square matrix is a matrix having the same number of rows and columns.

Definition of an Invertible Matrix#

A square \(n \times n\) matrix \(\mathbf{A}\) is said to be invertible if there exists an \(n \times n\) matrix \(\mathbf{C}\) such that

\(\mathbf{CA = I_n = AC}\)

where \(\mathbf{I_n}\) denotes the \(n \times n\) identity matrix.

If it exists this matrix \(\mathbf{C}\) is the inverse of \(\mathbf{A}\) and is usually written as \(\mathbf{A^{-1}}\)

\(\mathbf{A^{-1}A = I_n = AA^{-1}}\)

Note the following.

• Not all square matrices are invertible.

• If a square matrix is invertible then its inverse is unique.

• Matrices that are not invertible are called singular and matrices that are invertible are called non singular.

Theorem#

Let \(\mathbf{A}\) be an \(n \times n\) matrix.

\(\mathbf{A}\) is invertible iff for each vector \(\mathbf{b} \in \mathbb{R}^n\) the equation \(\mathbf{Ax = b}\) admits a single solution.

Proof of if \(\mathbf{A}\) is invertible then for each vector \(\mathbf{b} \in \mathbb{R}^n\) the equation \(\mathbf{Ax = b}\) admits a single solution.

\( \begin{aligned} \mathbf{Ax} &= \mathbf{b} \\ \mathbf{A^{-1}Ax} &= \mathbf{A^{-1}b} \\ \mathbf{x} &= \mathbf{A^{-1}b} \\ \mathbf{Ax} &= \mathbf{AA^{-1}b} \\ \mathbf{Ax} &= \mathbf{b} \\ \end {aligned} \)

Theorem#

Let \(\mathbf{A}\) be an invertible \(n \times n\) matrix.

We know that

• For each vector \(\mathbf{b} \in \mathbb{R}^n\) the equation \(\mathbf{Ax = b}\) has at least one solution \(\iff\) the linear transformation \(\mathbf{x \mapsto Ax}\) is onto \(\iff\) \(\mathbf{A}\) is row-equivalent to a row echelon matrix with a pivot in each row.

• For each vector \(\mathbf{b} \in \mathbb{R}^n\) the equation \(\mathbf{Ax = b}\) has at most one solution \(\iff\) the linear transformation \(\mathbf{x \mapsto Ax}\) is one-to-one \(\iff\) \(\mathbf{A}\) is row-equivalent to a row echelon matrix with a pivot in each column.

• The only reduced row echelon matrices having a pivot in each row and a pivot in each column are the identity matrices.

A (square) matrix is invertible iff the linear transformation \(\mathbf{x \mapsto Ax}\) is both onto and one-to-one.

A (square) matrix is invertible iff it is row-equivalent to the identity matrix of the same size.

Example#

PROBLEM STATEMENT

Determine whether the following matrix is invertible.

ROW REDUCE

\( \begin{bmatrix*}[r] \boxed{1} & -3 & -6 \\ 0 & 4 & 3 \\ -3 & 6 & 0 \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + 3r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{1} & -3 & -6 \\ 0 & \boxed{ 4} & 3 \\ 0 & -3 & -18 \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + \frac{3}{4} r_2}{\rightarrow} \underbrace{ \begin{bmatrix*}[r] \boxed{1} & -3 & -6 \\ 0 & \boxed{ 4} & 3 \\ 0 & 0 & \boxed{-\frac{63}{4}} \\ \end {bmatrix*} }_{\textbf{row echelon}} \)

The matrix is invertible because it is both onto (there is a pivot in each row) and one-to-one (there is a pivot in each column).

(The \(\textcolor{#0096FF}{3} \times \textcolor{#0096FF}{3}\) matrix has \(\textcolor{#0096FF}{3}\) pivots!)

The determinant: computing the inverse of an invertible \(2 \times 2\) matrix#

Let \(\mathbf{A} = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}\).

The determinant of \(\mathbf{A}\) is the number \(\text{det}(\mathbf{A}) = ad - bc\).

If \(ad - bc = 0\) then the matrix \(\mathbf{A}\) is not invertible.

If \(ad - bc \ne 0\) then the matrix \(\mathbf{A}\) is invertible and its inverse is

\(\mathbf{A^{-1}} = \textcolor{#0096FF}{\frac{1}{ad - bc}} \begin{bmatrix*}[r] \textcolor{green}{d} & \textcolor{purple}{-b} \\ \textcolor{purple}{-c} & \textcolor{green}{a} \\ \end{bmatrix*}\)

1. Swap the diagonal entries.

2. Change the signs of the off-diagonal entries.

3. Divide by the determinant.

\( \mathbf{A^{-1}A} = \frac{1}{ad - bc} \begin{bmatrix*}[r] d & -b \\ -c & a \\ \end {bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \\ \end {bmatrix*} = \frac{1}{ad - bc} \begin{bmatrix*}[r] ad - bc & bd - bd \\ ac - ac & ad - bc \\ \end {bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \\ \end {bmatrix*} = \mathbf{I_2} \)

Example#

Determine the inverse of the matrix

\( \mathbf{A} = \begin{bmatrix} 3 & 2 \\ 8 & 5 \\ \end {bmatrix} \)

\(\text{det}(\mathbf{A}) = 3 \cdot 5 - 2 \cdot 8 = -1\)

\( \mathbf{A^{-1}} = -1 \begin{bmatrix*}[r] 5 & -2 \\ -8 & 3 \\ \end {bmatrix*} = \begin{bmatrix*}[r] -5 & 2 \\ 8 & -3 \\ \end {bmatrix*} \)

Computing the inverse of an invertible \(n \times n\) matrix#

STEP 1 - Row reduce until a row echelon matrix is obtained

\(\mathbf{A \to I_n}\)

STEP 2 - Check: a square matrix is invertible iff it is row equivalent to the identity matrix of the same size

STEP 3 - Apply the same elementary row operations to the identity matrix in the same order

\(\mathbf{I_n \to A^\top}\)

Example#

PROBLEM STATEMENT

Determine the inverse of the matrix

\( \mathbf{A} = \begin{bmatrix*}[r] 0 & 1 & 1 \\ 1 & 2 & 2 \\ -1 & 4 & 7 \\ \end {bmatrix*} \)

ROW REDUCE

\( \begin{bmatrix*}[r] \boxed{0} & 1 & 1 \\ 1 & 2 & 2 \\ -1 & 4 & 7 \\ \end {bmatrix*} \underset{r_1 \leftrightarrow\,\,\, r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{-1} & 4 & 7 \\ 1 & 2 & 2 \\ 0 & 1 & 1 \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, r_2 + r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{-1} & 4 & 7 \\ 0 & \boxed{6} & 9 \\ 0 & 1 & 1 \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{-1} & 4 & 7 \\ 0 & \boxed{6} & 9 \\ 0 & 1 & 1 \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + (-\frac{1}{6})r_2}{\rightarrow} \underbrace{ \begin{bmatrix*}[r] \boxed{-1} & 4 & 7 \\ 0 & \boxed{6} & 9 \\ 0 & 0 & \boxed{-\frac{1}{2}} \\ \end {bmatrix*} }_{\textbf{row echelon}} \underset{r_3 \leftarrow\,\,\, (-2)r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{-1} & 4 & 7 \\ 0 & \boxed{6} & 9 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{-1} & 4 & 7 \\ 0 & \boxed{6} & 9 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, r_2 + (-9)r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{-1} & 4 & 7 \\ 0 & \boxed{6} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, r_1 + (-7)r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{-1} & 4 & 0 \\ 0 & \boxed{6} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{-1} & 4 & 0 \\ 0 & \boxed{6} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, \frac{1}{6} r_2}{\rightarrow} \begin{bmatrix*}[r] \boxed{-1} & 4 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, r_1 + (-4)r_2}{\rightarrow} \begin{bmatrix*}[r] \boxed{-1} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{-1} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, (-1)r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{1} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \)

INVERSE MATRIX

\( \begin{bmatrix*}[r] \boxed{1} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_1 \leftrightarrow\,\,\, r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{0} & 0 & 1 \\ 0 & \boxed{1} & 0 \\ 1 & 0 & \boxed{0} \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, r_2 + r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{0} & 0 & 1 \\ 0 & \boxed{1} & 1 \\ 1 & 0 & \boxed{0} \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{0} & 0 & 1 \\ 0 & \boxed{1} & 1 \\ 1 & 0 & \boxed{0} \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + (-\frac{1}{6})r_2}{\rightarrow} \begin{bmatrix*}[r] \boxed{0} & 0 & 1 \\ 0 & \boxed{1} & 1 \\ 1 & -\frac{1}{6} & \boxed{-\frac{1}{6}} \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, (-2)r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{0} & 0 & 1 \\ 0 & \boxed{1} & 1 \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{0} & 0 & 1 \\ 0 & \boxed{1} & 1 \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, r_2 + (-9)r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{0} & 0 & 1 \\ 18 & \boxed{-2} & -2 \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, r_1 + (-7)r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{14} & -\frac{7}{3} & \frac{4}{3} \\ 18 & \boxed{-2} & -2 \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{14} & -\frac{7}{3} & -\frac{4}{3} \\ 18 & \boxed{-2} & -2 \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, \frac{1}{6} r_2}{\rightarrow} \begin{bmatrix*}[r] \boxed{14} & -\frac{7}{3} & -\frac{4}{3} \\ 3 & \boxed{-\frac{1}{3}} & -\frac{1}{3} \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, r_1 + (-4)r_2}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & -1 & 0 \\ 3 & \boxed{-\frac{1}{3}} & -\frac{1}{3} \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{2} & -1 & 0 \\ 3 & \boxed{-\frac{1}{3}} & -\frac{1}{3} \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, (-1)r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{-2} & 1 & 0 \\ 3 & \boxed{-\frac{1}{3}} & -\frac{1}{3} \\ -2 & \frac{1}{3} & \boxed{\frac{1}{3}} \\ \end {bmatrix*} \)

VERIFY

\( \mathbf{A^{-1}A} = \begin{bmatrix*}[r] -2 & 1 & 0 \\ 3 & -\frac{1}{3} & -\frac{1}{3} \\ -2 & \frac{1}{3} & \frac{1}{3} \\ \end {bmatrix*} \begin{bmatrix*}[r] 0 & 1 & 1 \\ 1 & 2 & 2 \\ -1 & 4 & 7 \\ \end {bmatrix*} = \begin{bmatrix*}[r] 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix*} \)

Example#

PROBLEM STATEMENT

Determine the inverse of the matrix

\( \mathbf{A} = \begin{bmatrix*}[r] 1 & 2 & -1 \\ -4 & -7 & 3 \\ -2 & -6 & 4 \\ \end {bmatrix*} \)

ROW REDUCE

\( \begin{bmatrix*}[r] \boxed{1} & 2 & -1 \\ -4 & -7 & 3 \\ -2 & -6 & 4 \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, r_2 + 4r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{1} & 2 & -1 \\ 0 & 1 & -1 \\ -2 & -6 & 4 \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + 2r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{1} & 2 & -1 \\ 0 & \boxed{1} & -1 \\ 0 & -2 & 2 \\ \end {bmatrix*} \)

\( \begin{bmatrix*}[r] \boxed{1} & 2 & -1 \\ 0 & \boxed{1} & -1 \\ 0 & -2 & 2 \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + 2r_2}{\rightarrow} \underbrace{ \begin{bmatrix*}[r] \boxed{1} & 2 & -1 \\ 0 & \boxed{1} & -1 \\ 0 & 0 & 0 \\ \end {bmatrix*} }_{\textbf{row echelon}} \)

The matrix is not invertible.

Inverse Matrix Theorem#

Let \(\mathbf{A}\) be a square \(n \times n\) matrix.

The following statements are equivalent.

• The matrix \(\mathbf{A}\) is invertible.

• The matrix \(\mathbf{A}\) is row equivalent to the \(n \times n\) identity matrix.

• The matrix \(\mathbf{A}\) has \(n\) pivots.

• The equation \(\mathbf{Ax = 0}\) only admits the trivial solution.

• The columns of the matrix \(\mathbf{A}\) form a linearly independent set.

• The linear transformation \(T(\mathbf{x}) = \mathbf{Ax}\) is one-to-one because there is a pivot in each column.

• For each vector \(\mathbf{b} \in \mathbb{R}^n\) the equation \(\mathbf{Ax = b}\) has at least one solution.

• The columns of the matrix \(\mathbf{A}\) span \(\mathbb{R}^n\).

• The linear transformation \(T(\mathbf{x}) = \mathbf{Ax}\) is onto because there is a pivot in each row.

• There exists an \(n \times n\) matrix \(\mathbf{C}\) such that \(\mathbf{CA = I_n}\).

• There exists an \(n \times n\) matrix \(\mathbf{D}\) such that \(\mathbf{AD = I_n}\).

• The matrix \(\mathbf{A^\top}\) is invertible.

Example#

PROBLEM STATEMENT

Determine whether the matrix is invertible.

\( \begin{bmatrix*}[r] 1 & -2 & 0 & 3 \\ 0 & 4 & 0 & 7 \\ 3 & 1 & 0 & 9 \\ 8 & 4 & 0 & 5 \\ \end {bmatrix*} \)

The columns of the matrix are linearly dependent and so the matrix is not invertible.

Example#

PROBLEM STATEMENT

Determine whether the matrix is invertible.

\( \begin{bmatrix*}[r] -5 & 1 & 4 \\ 6 & 3 & 5 \\ 1 & 4 & 9 \\ \end {bmatrix*} \)

The transpose of the matrix is

\( \begin{bmatrix*}[r] -5 & 6 & 1 \\ 1 & 3 & 4 \\ 4 & 5 & 9 \\ \end {bmatrix*} \)

Its third column is the sum of its first two columns and so its columns are linearly dependent. Therefore neither the matrix or its transpose are invertible.

Properties of matrix inversion#

If \(\mathbf{A}\) is invertible then \(\mathbf{A^{-1}}\) is invertible and \(\mathbf{(A^{-1})^{-1} = A}\).

If \(\mathbf{A}\) and \(\mathbf{B}\) are both invertible then \(\mathbf{AB}\) is invertible and \(\mathbf{(AB)^{-1} = B^{-1} A^{-1}}\).

A matrix \(\mathbf{A}\) is invertible iff its transpose \(\mathbf{A^\top}\) is invertible.

\(\mathbf{(A^\top)^{-1} = (A^{-1})^\top}\)

Example#

Let \(\mathbf{A, B, C}\) be invertible \(n \times n\) matrices.

Simplify \(\mathbf{(AB)^\top(CA^\top)^{-1}}\).

\( \mathbf{(AB)^\top(CA^\top)^{-1}} = \mathbf{B^\top A^\top (A^\top)^{-1} C^{-1}} = \mathbf{B^\top C^{-1}} \)

Example#

Let \(\mathbf{A, B, C, X}\) be \(n \times n\) matrices with \(\mathbf{C}\) invertible.

Solve the equation \(\mathbf{B(X + A)^{-1} = C}\) for the matrix \(\mathbf{X}\).

\( \begin{aligned} \mathbf{B(X + A)^{-1}} &= \mathbf{C} \\ \mathbf{B(X + A)^{-1}(X + A)} &= \mathbf{C(X + A)} \\ \mathbf{B} &= \mathbf{C(X + A)} \\ \mathbf{C^{-1}B} &= \mathbf{C^{-1}C(X + A)} \\ \mathbf{C^{-1}B} &= \mathbf{X + A} \\ \mathbf{C^{-1}B - A} &= \mathbf{X} \\ \end {aligned} \)

Invertible Linear Transformation#

Let \(T : \mathbb{R}^n \to \mathbb{R}^n\) be a linear transformation and let \(\mathbf{A}\) be the square matrix associated with \(T\) so that \(T(\mathbf{x}) = \mathbf{Ax}\).

The linear transformation is invertible iff the matrix \(\mathbf{A}\) is invertible.

If \(T\) is invertible then \(T^{-1}(\mathbf{x}) = \mathbf{A^{-1}x}\).

Example#

Given the linear transformation

\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] 2x_1 & - & 8x_2 \\ -2x_1 & + & 7x_2 \\ \end {bmatrix*} \)

find

\(T^{-1} \left( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right)\)

\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] 2x_1 & - & 8x_2 \\ -2x_1 & + & 7x_2 \\ \end {bmatrix*} = \begin{bmatrix*}[r] 2 & -8 \\ -2 & 7 \\ \end {bmatrix*} \begin{bmatrix*}[r] x_1 \\ x_2 \\ \end {bmatrix*} \)

\( \mathbf{A^{-1}} = \frac{1}{2 \cdot 7 - (-8) \cdot (-2)} \begin{bmatrix*}[r] 7 & 8 \\ 2 & 2 \\ \end {bmatrix*} = -\frac{1}{2} \begin{bmatrix*}[r] 7 & 8 \\ 2 & 2 \\ \end {bmatrix*} = \begin{bmatrix*}[r] -\frac{7}{2} & -4 \\ -1 & -1 \\ \end {bmatrix*} \)

\( T^{-1} \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] -\frac{7}{2} & -4 \\ -1 & -1 \\ \end {bmatrix*} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix*}[r] -\frac{7}{2}x_1 & - & 4x_2 \\ -x_1 & - & x_2 \\ \end {bmatrix*} \)

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