Matrix Equation

A matrix equation represents a system of linear equations compactly using matrices. It is of the form AX = B, where A is the matrix of coefficients, X is the column vector of unknown variables, and B is the column vector of constants.

For example, for this system of linear equations: 
2x + 3y = 8
4x - y = 2

The matrix form is:
 

\(\[ \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 2 \end{bmatrix} \] \)

The matrix equation is AX=B
For a general system of n equations:
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2n + xn = b2
.
.
.
an1x1 + an2x2 + . . . + annxn = bn
The matrix equation AX = B, where,
A = coefficient matrix =aij
X = variable matrix = [x₁ , x₂,. . ., x_nT]
B = Constant matrix =]b₁, b₂, . . ., b_nT]

Let us take an example to understand how to write a matrix equation:

Consider the system: 
2x+3y-z=5
-x+4y+2z=6
3x-y+z=-2

Step 1: Check the order of variables to ensure all equations have the same order.
Here, all equations are of the order x, y, z.

Step 2: Make sure all the equations have variables on the left and constants on the right.

Step 3: Identify the coefficient matrix A, the variable matrix X, and the constant matrix B.

Here, 
Coefficient matrix A:

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

Variable matrix X:

\(\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]\)

Constant matrix B:

\(\[ B = \begin{bmatrix} 5 \\ 6 \\ -2 \end{bmatrix} \]\)

Step 4: Write in the form of a matrix equation AX = B

\(\[ AX = B \quad \Rightarrow \quad \begin{bmatrix} 2 & 3 & -1 \\ -1 & 4 & 2 \\ 3 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \\ -2 \end{bmatrix} \]\)

To solve a matrix equation of the form AX = B:
We multiply both sides by the inverse of A

A-1(AX) = A-1B

Then, we use the identity matrix property
A-1A = I (where I is the identity matrix)

So, IX = A-1B

Using the identity matrix rule, we know that IX = X
So, X = A-1B

This is the inverse matrix equation.
Let us solve an example using the inverse matrix method.

Question: Solve the following system of equations.

2x + 3y = 8
4x + y = 10

Writing as a matrix equation AX = B, where
 

\(\[ A = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 10 \end{bmatrix} \] \)

Now, we find the inverse A-1
Since A is a 2 × 2 matrix, we use the formula:
\( \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]\)

Here, a = 2, b = 3, c = 4, d = 1
Determinant A= (2)(1)-(3)(4)=2-12= -10

So, the inverse is:

\( \[ A^{-1} = \frac{1}{-10} \begin{bmatrix} 1 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -0.1 & 0.3 \\ 0.4 & -0.2 \end{bmatrix} \] \)
Now using the inverse matrix equation:
\(X = A^{-1}B = \begin{bmatrix} -0.1 & 0.3 \\ 0.4 & -0.2 \end{bmatrix} \begin{bmatrix} 8 \\ 10 \end{bmatrix} \)

First row: (-0.1)(8)+(0.3)(10)= -0.8 + 3 = 2.2
Second row:  (0.4)(8)+(0.3)(10)=3.2-2=1.2
So,
\(X = \begin{bmatrix} 2.2 \\ 1.2 \end{bmatrix} \)

X = 2.2 and y = 1.2
 

We can find the inverse of a matrix only when it is nonsingular, i.e., its determinant is not zero. The inverse A-1 only exists when det(A)  0. In such a case, the solution of the matrix equation AX = B is X = A-1B.
If det(A) = 0, we need to find adj(A)  B. Here, adj(A) is the adjoint of matrix A:
If adj(A)  B  0, then the system is inconsistent, meaning there is no solution to the equation AX = B.

If the product of adj(A)  B = 0, the system is either consistent with infinitely many solutions or is inconsistent.

Matrix equations simplify solving large systems of equations and allow matrix operations and techniques, including inverse, row reduction, and determinants. They are widely used in:

• Product pricing and production planning in business: Solving a matrix equation helps determine the number of units that can be produced under specific resource constraints.

• Rotating an image in Photoshop: Software programs use transformation matrices to rotate a photo perpendicularly. The matrix equation helps render the image correctly on screen.

• Solving electric circuits in engineering: Kirchhoff’s law is used for electrical circuits with 3 loops, and 3 equations are formed based on voltage and current. The matrix equation gives the current in each branch of the circuit accurately.

• Economics: The Leontief input-output model uses matrix equations to understand how much production India’s major economic sectors like agriculture and manufacturing need to meet the national demand.

• Robotics: Solving a matrix equation helps determine exact angles at which robotic movements occur efficiently.