0
$\begingroup$

Let $\beta \in \mathbb{R}$ and the following matrices:

$$A = \begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}, B = \begin{pmatrix} 2 & 2\\ 1 & 1 \end{pmatrix}, C = \begin{pmatrix} 2 & 2\\ 2 & 1 \end{pmatrix}, D = \begin{pmatrix} 2\beta & 2\beta\\ 2\beta & 2\beta \end{pmatrix}$$

Find all real quadruples $(x_1, x_2, x_3, x_4)$ so that $x_1A + x_2B + x_3C + x_4D = \begin{pmatrix} 8 & 7\\ 6 & 5 \end{pmatrix}$.

My approach is the following:

I created a system of equations using the given formula:

$$(S): \left\{\begin{matrix} 2x_1 + 2x_2 + 2x_3 + 2\beta x_4 = 8\\ x_1 + 2x_2 + 2x_3 + 2\beta x_4 = 7\\ x_1 + x_2 + 2x_3 + 2\beta x_4 = 6\\ x_1 + x_2 + x_3 + 2\beta x_4 = 5 \end{matrix}\right.$$

Now this needs to be solved as a regular system of equations.

I wrote its associated matrices:

$$A = \begin{pmatrix} 2 & 2 & 2 & 2\beta\\ 1 & 2 & 2 & 2\beta\\ 1 & 1 & 2 & 2\beta\\ 1 & 1 & 1 & 2\beta \end{pmatrix}, B = \begin{pmatrix} 8\\ 7\\ 6\\ 5 \end{pmatrix}$$

Now, I calculated $\det A$ and got $-2\beta$. For $\beta = 0$ the system has no solutions. For $\beta \neq 0$ I know the system has unique solution, but I don't know how to find them, because using Kramer's rule would be too complicated in this case.

Thank you in advance!

$\endgroup$
1
  • $\begingroup$ Kramer's rule comes in handy because the determinants have the property that is one column has a multiplier of $\beta$, the determinants will be the multiplier times the determinant such that column divides $\beta$. $\endgroup$
    – Zau
    Aug 19, 2016 at 11:36

1 Answer 1

1
$\begingroup$

You can calculate the quadruple by using the gauss method:

$\left( \begin{array}{ccc} 2 & 2 & 2 & 2\beta & | & 8\\ 1 & 2 & 2 & 2\beta & | & 7\\ 1 & 1 & 2 & 2\beta & | & 6\\ 1 & 1 & 1 & 2\beta& | & 5\end{array} \right)\sim \left( \begin{array}{ccc} 0 & 0 & 0 & -2\beta & | & -2\\ 0 & 1 & 1 & 0 & | & 2\\ 0 & 0 & 1 & 0 & | & 1\\ 1 & 1 & 1 & 2\beta& | & 5\end{array} \right)$

And from that transformation of the linear equations you can obtain:

$x_4=\frac{1}{\beta}$

$x_3=1$

$x_2 = 2-x_3=1$

$x_1=5-1-1-2=1$

And as you stated, there's no solution for the system with $\beta=0$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .