What is the polynomial obtained by evaluating the determinant of a matrix?

[Bob]

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

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


Please help me! thanks.

 

[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.

 

[HallsofIvy]

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!)

 

[Data]

clever problem

 

[Bob]

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

 

[Data]

not quite

 

[interested_learner]

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

 

[Bob]

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

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

 

[HallsofIvy]

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

 

[Bob]

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

I am sorry.


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

 

[d_leet]

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.

 

[HallsofIvy]

I am sorry.


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

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!