# Matrix determinant

1. Jul 27, 2006

### Bob

Evaluate the following determinant. Write your answer as a polynomial in x.

$$\begin{array}{|lcr|}a-x&b&c\\1&-x&0\\0&1&-x\end{array}$$

2. Jul 28, 2006

### durt

Just find the determinant as usual. A cofactor expansion from one of the rows or columns containing a zero is probably the easiest way.

3. Jul 28, 2006

### HallsofIvy

Staff Emeritus
You do realize, don't you, that you are expected to show us what you have tried so we can suggest changes? What durt suggested is very general but without knowing where you are having trouble we can't be more specific. It won't help you for someone else to do it for you.

(I confess that I find the answer amusing!)

4. Jul 28, 2006

### Data

clever problem :rofl:

5. Jul 28, 2006

### Bob

The answer is $$-x^3+ax^2+(b-a)x-a+c$$

6. Jul 28, 2006

### Data

not quite

7. Jul 28, 2006

### interested_learner

Show us the work!!!! How did you get that wrong answer?

8. Jul 29, 2006

### Bob

$$-x^3+ax^2+bx+c$$

9. Jul 30, 2006

### HallsofIvy

Staff Emeritus
What part of "Show us the work!!!! How did you get that wrong answer?" did you not understand?

10. Jul 30, 2006

### Bob

I am sorry.

$$(a-x)(x*x-1)-1(-bx-c)+0\\=ax^2-a-x^3+x+bx+c\\=-x^3+ax^2-(a-b)x-a+c$$

11. Jul 30, 2006

### d_leet

Recheck your first term, (a-x)(x*x-1) isn't quite right.

12. Jul 31, 2006

### HallsofIvy

Staff Emeritus
Okay, you are expanding by the first column:
$$(a-x)\left|\begin{array}{cc}-x & 0 \\1 & -x\end{array}\right|- (1)\left|\begin{array}{cc}b & c \\ 1 & -x\end{array}\right|$$
As dleet said, check that first number. 0*1 is not 1!