# Thread: Two matrix determinant questions

1. ## Two matrix determinant questions

I have two problems I've been assigned to assist in studying for an exam, but I can't figure out how to proceed. The first one is:

Prove the following:

|1 a bc| --|1 a a^2|----------------- |1 a a^2|
|1 b ac| = |1 b b^2| = (c-a)(b-a)(c-b)|0 1 b+a |=(c-a)(b-a)(c-b)
|1 c ab| --|1 c c^2|----------------- |0 0 1 |

Using only these rules:
1: If each element of a one row or column of a determinant is multiplied by a number k, the value of the determinant is multiplied by k.
2: The value of a determinant is zero if
a: all elements of one row or colum are zero, or if
b: two rows or columns are identical, or if
c: two rows or colums are proportional
3: If two rows or colums of a determinant are interchanged, the value of the determinant changes sign.
4: The value of a determinant is unchanged if
a: rows are written as columns and columns as rows, or if
b: we add to each element of one row or column, k times the corresponding element of another row or column, where k is any number.

My second question is more general.

When trying to determine the determinant of a matrix containing polynomials by hand, is there any way to do it other than brute forcing it with tons of polynomial multiplication?

2. Originally Posted by Fwahm
Prove the following:

|1 a bc| --|1 a a^2|----------------- |1 a a^2|
|1 b ac| = |1 b b^2| = (c-a)(b-a)(c-b)|0 1 b+a |=(c-a)(b-a)(c-b)
|1 c ab| --|1 c c^2|----------------- |0 0 1 |

Using only these rules:
1: If each element of a one row or column of a determinant is multiplied by a number k, the value of the determinant is multiplied by k.
2: The value of a determinant is zero if
a: all elements of one row or column are zero, or if
b: two rows or columns are identical, or if
c: two rows or columns are proportional
3: If two rows or columns of a determinant are interchanged, the value of the determinant changes sign.
4: The value of a determinant is unchanged if
a: rows are written as columns and columns as rows, or if
b: we add to each element of one row or column, k times the corresponding element of another row or column, where k is any number.
The set of displayed equations looks a mess on my monitor. I'm guessing that part of it should say: Prove the following:
$\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatr ix} = (c-a)(b-a)(c-b)\begin{vmatrix}1&a&a^2\\0&1&b+a\\0&0&1\end{vmatr ix} = (c-a)(b-a)(c-b)$.

To see why that is true, use Rule 4b to subtract the top row from the second row, getting $\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatr ix} = \begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\1&c&c^2\end{vmatrix}$. Now use Rule 1 to take a factor (b–a) out of the second row: $\begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\1&c&c^2\end{vmatrix} = (b-a)\begin{vmatrix}1&a&a^2\\0&1&b+a\\1&c&c^2\end{vma trix}$. Next two steps are to use the same two rules to subtract the top row from the bottom row and then take a factor (c–a) out of the bottom row. You then get $(b-a)(c-a)\begin{vmatrix}1&a&a^2\\0&1&b+a\\0&1&c+a\end{vma trix}$. Now subtract the second row from the bottom row and take out a factor (c-b), and you'll get the first part of the question. I'll leave you to do the rest.