## Matrix determinant

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

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

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 Just find the determinant as usual. A cofactor expansion from one of the rows or columns containing a zero is probably the easiest way.
 Recognitions: Gold Member Science Advisor 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!)

## Matrix determinant

clever problem
 The answer is $$-x^3+ax^2+(b-a)x-a+c$$
 not quite
 Show us the work!!!! How did you get that wrong answer?

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

$$-x^3+ax^2+bx+c$$
 Recognitions: Gold Member Science Advisor Staff Emeritus What part of "Show us the work!!!! How did you get that wrong answer?" did you not understand?

 Quote by HallsofIvy What part of "Show us the work!!!! How did you get that wrong answer?" did you not understand?
I am sorry.

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

 Quote by Bob I am sorry. $$(a-x)(x*x-1)-1(-bx-c)+0\\=ax^2-a-x^3+x+bx+c\\=-x^3+ax^2-(a-b)x-a+c$$
Recheck your first term, (a-x)(x*x-1) isn't quite right.

Recognitions:
Gold Member
 Quote by Bob I am sorry. $$(a-x)(x*x-1)-1(-bx-c)+0\\=ax^2-a-x^3+x+bx+c\\=-x^3+ax^2-(a-b)x-a+c$$
$$(a-x)\left|\begin{array}{cc}-x & 0 \\1 & -x\end{array}\right|- (1)\left|\begin{array}{cc}b & c \\ 1 & -x\end{array}\right|$$