Given system of linear equations can be written in matrix form as shown below:
`[{:(2x-5y), (5x + 3y):}] = [(1),(18)]`
The LHS matrix can be further written as product of two matrices as shown below:
`[{:(2,-5), (5, 3):}][(x),(y)] = [(1),(18)]` or AX =B.
`"Here", A = [{:(2,-5), (5, 3):}]` is called coefficient matrix, X = `[(x),(y)]` is called variable matrix and B = `[(1),(18)]`
is called constant matrix. Now, we need to find value of x and y, i.e., the matrix X.
To find X, pre-multiplying both the sides of Eq. (1) with `A^(-1)`
`rArr A^(-1) (AX) = A^(-1) B`
`"or "(A^(-1)A) X = A^(-1) B ["since"A (BC) = (AB)C]`
`"or " I X = A^(-1) B, [because A^(-1) A= I] "or" X = A^(-1) B, [IX = X]`
`X = A^(-1) B`
So, to find X we have to find inverse of coefficient matrix (i.e., A) and multiply it with B.
`therefore X = [(x), (y)] = [{:(2, -5), (5, 3):}]^(-1)[(1), (18)]`
`= (1)/(2 xx 3-(-5) xx 5)[{:(3, 5), (-5, 2):}][(1), (18)]`
`= (1)/(31)[{:(3 xx 1 + 5 xx 18), (-5 xx 1 + 2 xx 18):}]`
`= (1)/(31)[(93), (31)] = [(3), (1)]`
Thus, X = `[(x), (y)] = [(3), (1)] rArr x = 3, y = 1.`
Thus, in general any system of linear equations px + qy =a, and rx + sy =b can be represented in matrix form (i.e., AX = B) as `[{:(p, q), (r, s):}] [(x), (y)] = [(a), (b)]`.
Here, A is coefficient matrix = `[{:(p, q), (r, s):}]m X " is variables matrix" = [(x), (y)] " and B" = [(a), (b)]` is constant matrix.
`X = A^(-1) B = [{:(p, q), (r, s):}]^(-1) [(a), (b)]`
