Problem of the week - Find the nonsingular matrix

This problem of the week is about the inverse of a square matrix, where you should need to know some basic properties and how to calculate it. Before exposing the problem, let's introduce some concepts and applications related to the inverse of a matrix.

The inverse of a square matrix A is a matrix A¯¹ such that AA¯¹=I, where I is the identity matrix. A is called invertible or nonsingular, if there exists its inverse.

It can be proved that a square matrix A is invertible if and only if its determinant is not zero. The existence of a matrix inverse is associated with a number of other equivalent properties in Linear Algebra.

Inverse matrices are useful for a variety of fields, such as:

• Cryptography, to decrypt a message that was encrypted by using an invertible matrix.

• Image processing, where several transformations can be expressed as matrix multiplication, and to reverse those transformation, you just need multiply by the inverse.

• Solving consistent independent systems of linear equations.

Find a 3×3 nonsingular matrix A satisfying 4A - BA + I = 0, where:

    ┌            ┐
    │  5  -3  -1 │
B = │  1   0  -1 │
    │ -1   3   6 │
    └            ┘

Assume that A is nonsingular. Then the inverse matrix A¯¹ exists.

Multiplying the given equality by A¯¹ on the right, we obtain:

4AA¯¹ - BAA¯¹ + A¯¹ = 0
4I - B + A¯¹ = 0

solving for A¯¹, we have

A¯¹ = B - 4I

that is

      ┌            ┐
      │  1  -3  -1 │
A¯¹ = │  1  -4  -1 │
      │ -1   3   2 │
      └            ┘

Since (A¯¹)¯¹ = A, to find A, we need to find the inverse of A¯¹.

We will use the linear row reduction method, but it can also be solved using the adjugate matrix method.

We reduce the augmented matrix [A¯¹ | I] as follows:

┌                      ┐
│  1  -3  -1 | 1  0  0 │
│  1  -4  -1 | 0  1  0 │
│ -1   3   2 | 0  0  1 │
└                      ┘

r2 <———> r2 - r1
r3 <———> r3 + r1
┌                      ┐
│ 1  -3  -1 |  1  0  0 │
│ 0  -1   0 | -1  1  0 │
│ 0   0   1 |  1  0  1 │
└                      ┘

r1 <———> r1 + r3
┌                     ┐
│ 1  -3  0 |  2  0  1 │
│ 0  -1  0 | -1  1  0 │
│ 0   0  1 |  1  0  1 │
└                     ┘
r1 <———> r1 - 3•r2
┌                      ┐
│ 1   0  0 |  5  -3  1 │
│ 0  -1  0 | -1   1  0 │
│ 0   0  1 |  1   0  1 │
└                      ┘

r2 <———> -r2
┌                     ┐
│  1  0  0 | 5  -3  1 │
│  0  1  0 | 1  -1  0 │
│  0  0  1 | 1   0  1 │
└                     ┘

So, the matrix is given by the right 3 by 3 part:

    ┌          ┐
    │ 5  -3  1 │
A = │ 1  -1  0 │
    │ 1   0  1 │
    └          ┘