# Thread: Challenging MATRIX QUESTION

1. ## Challenging MATRIX QUESTION

Show working to sole for x in the equation

(x-3 1 -1)
(1 x-5 1 ) =0
( -1 1 x-3)

THSI IS A 3x3 MATRIX

2. ## Re: Challenging MATRIX QUESTION

Originally Posted by prittyboy
Show working to sole for x in the equation

(x-3 1 -1)
(1 x-5 1 ) =0
( -1 1 x-3)

THSI IS A 3x3 MATRIX
In which case this is not "challenging", it is simply non-sense. A matrix cannot be equal to a number. I suspect this is not a matrix question at all but is a deteriminant equation:
$\left|\begin{array}{ccc}x- 3 & 1 & -1 \\ 1 & x- 5 & 1 \\ -1 & 1 & x- 3\end{array}\right|= 0$

Expanding on the top row,
$\left|\begin{array}{ccc}x- 3 & 1 & -1 \\ 1 & x- 5 & 1 \\ -1 & 1 & x- 3\end{array}\right|= (x- 3)\left|\begin{array}{cc}x- 5 & 1 \\ 1 & x- 3\end{array}\right|- \left|\begin{array}{cc}1 & 1 \\ -1 & x- 3\end{array}\right|$ $- \left|\begin{array}{cc}1 & x- 5 \\ -1 & 1\end{array}\right|$
$= (x- 3)((x-5)(x-3)- 1)- (1(x-3)+ 1)- (1+ 1(x- 5))= (x- 3)(x^2- 8x+ 14)- x+ 2- x+ 4$
$= (x^3- 8x^2+14x)- (3x^2+ 24x- 42)-2x+ 6= x^3- 11x^2- 12x+ 48= 0$
a cubic equation. There is a general formula for solving cubic equations but it is a bit complicated. You might try checking for rational roots. The "rational root theorem" says that any rational number root of the polynomial equation $a_nx^n+ a_{n-1}x^{n-1}+\cdot\cdot\cdot+ a_1x+ a_0= 0$ must be of the form $\frac{m}{n}$ where the numerator, m, divides the constant term, $a_0$ and the denominator, n, divides the leading coefficent, $a_n$.

Since, here, the leading coefficient is 1, any rational number solution to the equation must be an integer that evenly divides 48. Try those to see if there is a rational root.