3.5. Matrix inversion

3.5.1. Motivation and definition

Gauss-Jordan elimination works well for solving Ax = b, but the process needs to be repeated for every new b. Is there an alternative if we need to solve Ax = b for many different b?

Matrix inverse

For a square (n x n) matrix A, the inverse \(\vv{A}^{-1}\) satisfies

(3.35)\[\begin{equation} \vv{A} \vv{A}^{-1} = \vv{A}^{-1} \vv{A} = \vv{I} \end{equation}\]

where I is the n x n identity matrix.

A matrix is called nonsingular or invertible if it has an inverse, but singular if it does not.

Invertible matrix theorem

A is invertible if and only if the determinant of A is nonzero.

(There are many more such conditions!)

If the inverse of A exists, it is unique and can be used to solve Ax = b.

(3.36)\[\begin{align} \vv{A} \vv{x} &= \vv{b} \\ \vv{A}^{-1} \vv{A} \vv{x} &= \vv{A}^{-1} \vv{b} \\ \vv{x} &= \vv{A}^{-1} \vv{b} \end{align}\]

Finding the inverse of A is usually hard. There is a general definition based on cofactors, as well as advanced numerical methods, that we will not cover. Instead, we focus on two options: a formula for 2 x 2 matrices, and use of Gauss-Jordan elimination for larger matrices.

3.5.2. Inverse of a 2 x 2 matrix

For a 2 x 2 matrix,

(3.37)\[\begin{equation} \vv{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \end{equation}\]

the matrix inverse is

(3.38)\[\begin{equation} \vv{A}^{-1} = \frac{1}{|\vv{A}|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \end{equation}\]

(Flip a and d, change the signs of b and c.)

Example: 2 x 2 inverse

To find the inverse of

(3.39)\[\begin{equation} \vv{A} = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} \end{equation}\]

First, compute its determinant:

(3.40)\[\begin{equation} |\vv{A}| = 3 \times 4 - 2 \times 1 = 12 - 2 = 10 \end{equation}\]

Then, compute its inverse

(3.41)\[\begin{equation} \vv{A}^{-1} = \frac{1}{10} \begin{bmatrix} 4 & -1 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{bmatrix} \end{equation}\]

3.5.3. Inverses using Gauss-Jordan elimination

For larger matrices, we can use Gauss–Jordan elimination to solve \(\vv{A} \vv{A}^{-1} = \vv{I}\) as a generalization of Ax = b.

• Check that \(|\vv{A}| \ne 0\) (i.e., A is invertible).

• Form the 2n x n augmented matrix \([ \vv{A} \, | \, \vv{I} ]\)

• Perform row operations to bring to \([ \vv{I} \, | \, \vv{A}^{-1} ]\).

3.5.4. Skill builder problems

1. Solve using matrix inversion or explain why this is not possible:

   (3.42)\[\begin{align} 5x_1 - 2x_2 &= 20.9 \\ -x_1 + 4x_2 &= -19.3 \end{align}\]
   Solution

   First, write in matrix form Ax = b with:

   (3.43)\[\begin{equation} \vv{A} = \begin{bmatrix} 5 & -2 \\ -1 & 4 \end{bmatrix} \qquad \vv{b} = \begin{bmatrix} 20.9 \\ -19.3 \end{bmatrix} \end{equation}\]

   Then, evaluate \(|\vv{A}|\) to check if an inverse exists:

   (3.44)\[\begin{equation} |\vv{A}| = (5 \cdot 4) - (-1 \cdot -2) = 18 \end{equation}\]

   \(|\vv{A}| \ne 0\), so an inverse can be found using the formula for a 2x2 matrix:

   (3.45)\[\begin{equation} \vv{A}^{-1} = \frac{1}{18} \begin{bmatrix} 5 & -2 \\ -1 & 4 \end{bmatrix} \end{equation}\]

   Last, solve for x:

   (3.46)\[\begin{align} \vv{x} = \vv{A}^{-1}\vv{b} &= \frac{1}{18} \begin{bmatrix} 5 & -2 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} 20.9 \\ -19.3 \end{bmatrix} \\ &= \frac{1}{18} \begin{bmatrix} 4 \cdot 20.9 + 2 \cdot -19.3 \\ 1 \cdot 20.9 + 5 \cdot -19.3 \end{bmatrix} \\ &= \begin{bmatrix} 2.5 \\ -4.2 \end{bmatrix} \end{align}\]

   Therefore, \(x_1 = 2.5\) and \(x_2 = -4.2\).

2. Solve using matrix inversion or explain why this is not possible:

   (3.47)\[\begin{align} x_1 + 4x_2 = 8 \\ 2x_1 + 8x_2 = 17 \end{align}\]
   Solution

   First, write in matrix form Ax = b with:

   (3.48)\[\begin{equation} \vv{A} = \begin{bmatrix} 1 & 4 \\ 2 & 8 \end{bmatrix} \qquad \vv{b} = \begin{bmatrix} 8 \\ 17 \end{bmatrix} \end{equation}\]

   Then, evalaute \(|\vv{A}|\) to check if an inverse exists:

   (3.49)\[\begin{equation} |\vv{A}| = (1 \cdot 8) - (2 \cdot 4) = 0 \end{equation}\]

   A is singular because \(|\vv{A}| = 0\), so these equations cannot be solved using an inverse.

3. Solve using matrix inversion or explain why this is not possible:

   (3.50)\[\begin{align} x_2 + x_3 = -2 \\ 4x_2 + 6x_3 = -12 \\ x_1 + x_2 + x_3 = 2 \end{align}\]
   Solution

   First, write in matrix form Ax = b with:

   (3.51)\[\begin{equation} \vv{A} = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \qquad \vv{b} = \begin{bmatrix} -2 \\ -12 \\ 2 \end{bmatrix} \end{equation}\]

   Then, evaluate \(|\vv{A}|\) to check if an inverse exists:

   (3.52)\[\begin{align} |\vv{A}| = 0 \cdot \begin{vmatrix} 4 & 6 \\ 1 & 1 \end{vmatrix} -1 \cdot \begin{vmatrix} 0 & 6 \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 4 \\ 1 & 1 \end{vmatrix} \end{align}\]

   (3.53)\[\begin{equation} |\vv{A}| = -1 \cdot (0-6) + 1 \cdot (0-4) = 2 \end{equation}\]

   Since \(|\vv{A}| \ne 0\), A is invertible. Use Gauss-Jordan elimination to find the inverse. Start with the augmented matrix \([\vv{A} | \vv{I}]\), then rearrange the rows

   (3.54)\[\begin{align} \begin{bmatrix} 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 4 & 6 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 4 & 6 & 0 & 1 & 0 \end{bmatrix} \begin{matrix} \vphantom{R_1} \\ \vphantom{R_1} \\ -4 R_1 \end{matrix} \\ &\to \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -4 & 1 & 0 \end{bmatrix} \begin{matrix} \vphantom{R_1} \\ \vphantom{R_1} \\ \div 2 \end{matrix} \\ &\to \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & -2 & 0.5 & 0 \end{bmatrix} \begin{matrix} -R_3 \\ -R_3 \\ \vphantom{R_1} \end{matrix} \\ &\to \begin{bmatrix} 1 & 1 & 0 & 2 & -0.5 & 1 \\ 0 & 1 & 0 & 3 & -0.5 & 0 \\ 0 & 0 & 1 & -2 & 0.5 & 0 \end{bmatrix} \begin{matrix} -R_2 \\ \vphantom{R_1} \\ \vphantom{R_1} \end{matrix} \\ &\to \begin{bmatrix} 1 & 0 & 0 & -1 & 0 & 1 \\ 0 & 1 & 0 & 3 & -0.5 & 0 \\ 0 & 0 & 1 & -2 & 0.5 & 0 \end{bmatrix} \end{align}\]

   Hence,

   (3.55)\[\begin{align} \vv{A}^{-1} = \begin{bmatrix} -1 & 0 & 1 \\ 3 & -0.5 & 0 \\ -2 & 0.5 & 0 \end{bmatrix} \end{align}\]

   and

   (3.56)\[\begin{align} \vv{x} = \vv{A}^{-1} \vv{b} &= \begin{bmatrix} -1 & 0 & 1 \\ 3 & -0.5 & 0 \\ -2 & 0.5 & 0 \end{bmatrix} \begin{bmatrix} -2 \\ -12 \\ 2 \end{bmatrix} \\ & = \begin{bmatrix} -1 \cdot -2 + 1 \cdot 2 \\ 3 \cdot -2 + -0.5 \cdot -12 \\ -2 \cdot -2 + 0.5 \cdot -12 \end{bmatrix} \\ &= \begin{bmatrix} 4 \\ 0 \\ -2 \end{bmatrix} \end{align}\]

   So, \(x_1 = 4\), \(x_2 = 0\), and \(x_3 = -2\).

4. Solve using matrix inversion or explain why this is not possible:

   (3.57)\[\begin{align} 4x_2 + 4x_3 = 24 \\ 3x_1 - 11x_2 - 2x_3 = -6 \\ 6x_1 - 17x_2 + x_3 = 18 \end{align}\]
   Solution

   First, write in matrix form Ax = b with:

   (3.58)\[\begin{equation} \vv{A} = \begin{bmatrix} 0 & 4 & 4 \\ 3 & -11 & -2 \\ 6 & -17 & 1 \end{bmatrix} \qquad \vv{b} = \begin{bmatrix} 24 \\ -6 \\ 18 \end{bmatrix} \end{equation}\]

   Then, evaluate \(|\vv{A}|\) to check if an inverse exists:

   (3.59)\[\begin{align} |\vv{A}| = 0 \cdot \begin{vmatrix} -11 & -2 \\ -17 & 1 \end{vmatrix} -4 \cdot \begin{vmatrix} 3 & -2 \\ 6 & 1 \end{vmatrix} + 4 \cdot \begin{vmatrix} 3 & -11 \\ 6 & -17 \end{vmatrix} \end{align}\]

   (3.60)\[\begin{align} |\vv{A}| &= -4 (3 \cdot 1 - 6 \cdot -2) + 4 (3 \cdot -17 - 6 \cdot -11) \\ &= -4 \cdot 15 + 4 \cdot 15 \\ &= 0 \end{align}\]

   Since \(|\vv{A}| = 0\), A is not invertible.

5. Solve using matrix inversion or explain why this is not possible:

   (3.61)\[\begin{align} 2x_1 - x_2 + 3x_3 = -1 \\ -4x_1 + 2x_2 - 6x_3 = 2 \end{align}\]
   Solution

   First, write in matrix form Ax = b with:

   (3.62)\[\begin{equation} \vv{A} = \begin{bmatrix} 2 & -1 & 3 \\ -4 & 2 & -6 \end{bmatrix} \qquad \vv{b} = \begin{bmatrix} -1 \\ 2 \end{bmatrix} \end{equation}\]

   Since A is not square, A is not invertible.