Question

Find the Square matrix M order of two such that M square of 2 - 2M= (-1 0) 2x2
6 3

Answer 1

Firstly, let x₁, x₂, x₃ and x₄ be the corresponding elements of matrix, so that:

$\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix}$ * $\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix}$ - 2 * $\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix}$ = $\begin{pmatrix} -1 & 0 \\ 6 & 3 \end{pmatrix}$

When we multiply matrix by itself and multiply it by two as well, we get:

$\begin{pmatrix} x_1^2 + x_2x_3 & x_1x_2 + x_2x_4 \\ x_1x_3 + x_3x_4 & x_2x_3 + x_4^2 \end{pmatrix}$ - $\begin{pmatrix} 2x_1 & 2x_2 \\ 2x_3 & 2x_4 \end{pmatrix}$ = $\begin{pmatrix} -1 & 0 \\ 6 & 3 \end{pmatrix}$

After matrix subtraction, we get equations (one per each element):

$1) x_1^2 + x_2x_3 - 2x_1 = -1$

$2) x_1x_2 + x_2x_4 - 2x_2 = 0$

$3) x_1x_3 + x_3x_4 - 2x_3 = 6$

$4) x_2x_3 + x_4^2 - 2x_4 = 3$

Let's group the second and the third equations:

$2) x_2(x_1 + x_4 - 2) = 0$

$3) x_3(x_1 + x_4 - 2) = 6$

From the second equation we understand that x₂ = 0 or x₁ + x₄ - 2 = 0. The second assumption is false because if it were true, then the third equation would be:

$3) x_3 * 0 = 6$

which is impossible. Thus, $x_2=0$ .

Let's group the first and the fourth equations:

$1) x_2x_3 = -1 + 2x_1 - x_1^2$

$4) x_2x_3 = 3 + 2x_4 - x_4^2$

As x₂ = 0, then x₂x₃ = 0. That means:

$1) x_1^2 - 2x_1 + 1 = 0$

$4) x_4^2 - 2x_4 - 3 = 0$

Let's solve the first equation:

$x_1^2 - 2x_1 + 1 = 0$

$D = b^2 - 4*a*c = 4 - 4 = 0.$

That means there is only one solution for this equation:

$x_1 = \frac{-b}{2*a} = \frac{2}{2} = 1$

Thus, $x_1=1$.

Let's solve the fourth equation:

$x_4^2 - 2x_4 - 3 = 0$

$D = b^2 - 4*a*c = 4 + 12 = 16; \sqrt{D} = 4;$

$x_{4;1} = \frac{ -b - \sqrt{D}}{2*a} = \frac{2-4}{2} = -1$

$x_{4;2} = \frac{ -b + \sqrt{D}}{2*a} = \frac{2+4}{2} = 3$

So, x₄ has two solutions.

Let's get back to the third equation again:

$x_3(x_1 + x_4 - 2) = 6$

$x_3 = \frac{6}{x_1 + x_4 - 2} = \frac{6}{x_4 - 1}$

If x₄ = -1, then:

$x_3 = \frac{6}{-2} = -3$

If x₄ = 3, then:

$x_3 = \frac{6}{2} = 3$

Thus, (x₃;x₄) is (-3;-1) or (3;3).

So, there are two solutions, the matrices are as follows:

1) $\begin{pmatrix} 1 & 0 \\ -3 &-1 \end{pmatrix}$

2) $\begin{pmatrix} 1 & 0 \\ 3 &3 \end{pmatrix}$