Problem of the week - Find the idempotent matrix

Every Wednesday Nibcode Solutions will bring you a new Linear Algebra problem, which answer will be published on Tuesdays following week.

Problems will cover the most important Linear Algebra topics, ranging from simple problems to some a little more complex, but all of them aimed to provide examples for you to understand the primary concepts and problem solving methods of Linear Algebra.

This problem of the week involves multiplication of matrices, which is considered the most important matrix operation, and although is not quite as straightforward as addition, the method is not difficult to grasp.

Find all values of x, y and z, if any, such that matrix A is idempotent?

    ┌               ┐
    │  x  -x-3  x+1 │
A = │ 25     z   30 │
    │  y  -y-8  y+5 │
    └               ┘

From definition, if matrix A is idempotent, then:

┌               ┐┌               ┐   ┌               ┐
│  x  -x-3  x+1 ││  x  -x-3  x+1 │   │  x  -x-3  x+1 │
│ 25     z   30 ││ 25     z   30 │ = │ 25     z   30 │
│  y  -y-8  y+5 ││  y  -y-8  y+5 │   │  y  -y-8  y+5 │
└               ┘└               ┘   └               ┘

To multiply two matrices A and B, we multiply every row of A by every column of B. To find the entry associated to row i and column j in the result matrix, multiply the i-th row of A by the j-th column of B, which is the sum of all products of corresponding elements. In this problem, every combination leads to an equation, resulting in the following 9 equations:

1. x²       + 25(-x-3) + y(x+1)      = x
2. x(-x-3)  +  z(-x-3) + (x+1)(-y-8) = -x-3
3. x(x+1)   + 30(-x-3) + (x+1)(y+5)  = x+1

4. 25x      + 25z      + 30y         = 25
5. 25(-x-3) + z²       + 30(-y-8)    = z
6. 25(x+1)  + 30z      + 30(y+5)     = 30

7. yx       + 25(-y-8) + y(y+5)      = y
8. y(-x-3)  + z(-y-8)  + (y+5)(-y-8) = -y-8
9. y(x+1)   + 30(-y-8) + (y+5)(y+5)  = y+5

Subtracting equation 6 from 4 we get:

z = -34

Substituting the value of z in equation 4:

5x + 6y = 175 
y = (175 - 5x)/6

Substituting the expression of y in the first equation, and multiplying each side of the resulting equation by 6:

6x² + 150(-x-3) + (175-5x)(x+1) = 6x
6x² - 150x - 450 + 175x + 175 - 5x² - 5x - 6x = 0
x² + 14x - 275 = 0
(x + 25)(x - 11) = 0

We get two possible solutions for the previous equation, leading to the possible solutions for the original problem:

x = -25 => y = 50

and

x =  11 => y = 20

To check which answers are correct, we need to put the values of x, y and z back into the original equations. When checking both solutions we get that only the solution x = 11, y= 20, z = -34 is valid, obtaining the matrix:

┌             ┐
│ 11  -14  12 │
│ 25  -34  30 │
│ 20  -28  25 │
└             ┘