A = (2x₁ − 3x₂ + x₃ = 5, x₁ + x₂ − 3x₃ = 7, 5x₁ − x₂ + 6x₃ = 1

To solve this system of equations, we can use the method of substitution or elimination. Let's use the elimination method.

First, let's rewrite the system of equations in matrix form:

[2 -3 1 | 5] [1 1 -3 | 7] [5 -1 6 | 1]

Now, let's perform row operations to simplify the matrix:

R2 = R2 - 0.5R1 R3 = R3 - 2.5R1

This gives us:

[2 -3 1 | 5] [0 2 -4 | 4] [0 5 3.5 | -11.5]

Next, let's perform row operations to further simplify the matrix:

R3 = R3 - 2.5*R2

This gives us:

[2 -3 1 | 5] [0 2 -4 | 4] [0 0 13 | -21]

Now, let's back-substitute to find the values of x₁, x₂, and x₃:

From the third equation, we have:

13x₃ = -21 x₃ = -21/13

Substitute x₃ back into the second equation:

2x₂ - 4(-21/13) = 4 2x₂ + 84/13 = 4 2x₂ = 4 - 84/13 2x₂ = 52/13 - 84/13 2x₂ = -32/13 x₂ = -16/13

Substitute x₃ and x₂ back into the first equation:

2x₁ - 3(-16/13) + (-21/13) = 5 2x₁ + 48/13 - 21/13 = 5 2x₁ = 65/13 - 27/13 2x₁ = 38/13 x₁ = 19/13

Therefore, the solution to the system of equations is: x₁ = 19/13 x₂ = -16/13 x₃ = -21/13