# Determinants homework help

1. Jul 25, 2009

### dracolnyte

1. The problem statement, all variables and given/known data
detA = 0
Matrix A =

| (x+5) 4 4 |
| -4 (x-3) -4 |
| -4 -4 (x-3)|

3. The attempt at a solution

I know that if one row or column is equal to another, then detA = 0, so using the last 2 rows, i can find out that x has to be -1 for row2 and row3 to be equal for detA = 0.

but the answer at the back says x = -1 or 3, how can i solve the 3? I have tried to reduce it to the triangular form, but it got way too messy to be correct. I also tried using the 3x3 matrix trick where you copy the first 2 rows and make a 4th and 5th row out of them and solve for the determinant, also got pretty messy.

Is there some rule that i missed out that can make my life easier on solving this question?

2. Jul 25, 2009

### Office_Shredder

Staff Emeritus
Re: Determinants

Do you know the expansion by rows trick?

Definitely the best way to do this is to calculate the determinant and then solve the polynomial... trying to find linear dependence conditions is too error prone and you can miss solutions

3. Jul 25, 2009

### dracolnyte

Re: Determinants

you mean the one the one where you the multiply the diagonals and the subtract it by the other diagonals? in other words
(x+5) 4 4
-4 (x-3) -4
-4 -4 (x-3)
------------
(x+5) 4 4
-4 (x-3) -4

4. Jul 25, 2009

### dracolnyte

Re: Determinants

a11a22a33 - a11a23a32 - a12a21a33 + a12a23a31 + a13a21a32 - a13a22a31 = 0
it gets really messy with like = x^3 - 11x^2 + 55x - 93

Last edited: Jul 25, 2009
5. Jul 25, 2009

### Staff: Mentor

Re: Determinants

Either your formula isn't right, or you have made in error in calculation.
I worked it out and got a different polynomial, which when factored and set to zero, had roots equal to 3 and -1.

6. Jul 25, 2009

### dracolnyte

Re: Determinants

ya sorry, i realized and i did it again, i got x = 3 and -1. my bad, it must be getting late