Solving equations using matrices — KCSE Mathematics

To solve the matrix equation AX = B, where A is a square matrix and B is a matrix of constants, we use the formula X = A⁻¹B. This approach is applicable only when the inverse of A exists.

Steps to solve:

1. Calculate the inverse of A (denoted as A⁻¹).
2. Multiply A⁻¹ by B to find X.

Example:
Given the matrix equation:
[ A = \begin{pmatrix} 2 & 1 \ 1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 5 \ 3 \end{pmatrix} ]

1. Find A⁻¹:
   • Determinant of A = (2)(1) - (1)(1) = 1
   • A⁻¹ = [ \frac{1}{1} \begin{pmatrix} 1 & -1 \ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -1 \ -1 & 2 \end{pmatrix} ]
2. Calculate X:
   • X = A⁻¹B = [ \begin{pmatrix} 1 & -1 \ -1 & 2 \end{pmatrix} \begin{pmatrix} 5 \ 3 \end{pmatrix} = \begin{pmatrix} 1(5) + (-1)(3) \ -1(5) + 2(3) \end{pmatrix} = \begin{pmatrix} 2 \ 1 \end{pmatrix} ]
     Therefore, X = \begin{pmatrix} 2 \ 1 \end{pmatrix}.

To solve equations using matrices, we express the system of equations in matrix form. For example, consider the equations:

1. 2x + 3y = 8
2. 4x - y = 2

We can represent this system as:
[ A = \begin{bmatrix} 2 & 3 \ 4 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 8 \ 2 \end{bmatrix} ]

Thus, the matrix equation is:
[ AX = B ]

To find ( X ), we need to compute ( A^{-1} ):
[ X = A^{-1}B ]

Steps to solve:

1. Calculate the inverse of matrix A.
2. Multiply the inverse by matrix B to find X.

This method is efficient for larger systems of equations, as it allows us to handle multiple variables systematically.