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Using matrices for solving systems of linear algebraic equations

Let’s note, that if ∆A = 0, then the inverse matrix doesn’t exist, that is the inverse matrix A −1
exists only for square and nonsingular matrices.

2.4. Using matrices for solving systems of linear algebraic equations

Let’s consider a system of three linear equations with three unknowns:


 a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 ,
a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 , (2.2)

a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 .
     
a 11 a 12 a 13 x 1 b 1
Let’s mark the following matrices: A =  a 21 a 22 a 23  , X =   , B =   . Then, using
b 2
x 2
a 31 a 32 a 33 x 3 b 3
the rule of multiplying matrices, the system (2.2) can be presented in an equivalent view:
A · X = B (2.3)

where A — the given matrix which consists of digits near the unknowns (the major matrix); B
— the matrix - column consisting of free terms; X — the unknown matrix - column. Matrix
X is considered to be the solution of the matrix equation (2.3), when it tranforms that equation
into equality.
Let’s assume, that ∆A ̸= 0. Then, according to Cramer’s rule, system (2.2) has the only
solution, which can be solved by Cramer’s formulas. And now let’s find the solution by a
matrix method — by means of the inverse matrix. Multiplying the equation (2.3) on its left side
−1
on matrix A , which exists, because ∆A ̸= 0, we will have:
A −1 · A · X = A −1 · B. (2.4)

Because of A · A −1 = E, E · X = E, then according to (2.4), we will get:

X = A −1 · B. (2.5)

Formula (2.5) is called the formula for the matrix method for solving the system (2.2).

Example 2.1. Let’s solve the below system using the matrix method:


x 1 + 3x 2 + x 3 = 7,

2x 1 − x 2 + 5x 3 = −19,

3x 1 − 3x 2 + 2x 3 = −11.


     
1 3 1 7 x 1
Solution. Let’s write matrices: A =  2 −1 5  , B =  −19  , X =   and calculate the
x 2
3 −3 2 −11 x 3
determinant of matrix A :

1 3 1

∆A = 2 −1 5 = −2 + 45 − 6 + 3 + 15 − 12 = 43 ⇒

3 −3 2
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