# Thread: Find Determinant by Upper Trianglular matrix

1. ## Find Determinant by Upper Trianglular matrix

So the matrix is this:
[ 1 1 1 1 ]
[ a b c d ]
[ a^2 b^2 c^2 d^2 ]
[ a^3 b^3 c^3 d^3 ]

The method I was using was to multiple row 2, 3, 4 by 1/a, 1/a^2, 1/a^3, then subtract from row 1. Then multiply row 3 and 4 such that I can subtract from row 2, etc.

The problem with this method is that I'm getting a lot of computation and i'm starting to wonder if i'm do this the right way. Is there a better way of doing this?

2. Originally Posted by hashshashin715
So the matrix is this:
[ 1 1 1 1 ]
[ a b c d ]
[ a^2 b^2 c^2 d^2 ]
[ a^3 b^3 c^3 d^3 ]

The method I was using was to multiple row 2, 3, 4 by 1/a, 1/a^2, 1/a^3, then subtract from row 1. Then multiply row 3 and 4 such that I can subtract from row 2, etc.

The problem with this method is that I'm getting a lot of computation and i'm starting to wonder if i'm do this the right way. Is there a better way of doing this?
When I was in school 55 years ago they taught us to do it this way. You can see that the determinant must be a polynomial of degree 6 in the variables a,b,c,d. If a=b then the first and second columns of the matrix are equal and so the determinant is 0. Therefore (by the factor theorem) $a-b$ is a factor of the determinant. For the same reason, $a-c$, $a-d$, $b-c$, $b-d$ and $c-d$ are also factors. Thus the determinant is a multiple of $(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$. But since that is already a polynomial of degree 6, there cannot be any further factors apart from a numerical constant.

Thus $\begin{vmatrix}1&1&1&1\\ a&b&c&d\\ a^2&b^2&c^2&d^2\\ a^3&b^3&c^3&d^3\end{vmatrix} = k(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$. To find the numerical constant k, compare the coefficients of $bc^2d^3$ on both sides of that equation, and you find that $k=-1$. So $\begin{vmatrix}1&1&1&1\\ a&b&c&d\\ a^2&b^2&c^2&d^2\\ a^3&b^3&c^3&d^3\end{vmatrix} = -(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$.

3. Thanks for this method, but the problem states that I have to find the determinant by transforming the matrix to an upper triangular matrix. Sorry for not stating that in OP.

4. First reduce the first row by subtracting a times the first row from the second, $a^2$ times the first row from the third row, and $a^3$ times the first row from the fourth to get
$\begin{bmatrix}1 & 1 & 1 & 1 \\0 & b-a & c-a & d-a\\ 0 & b^2- a^2 & c^2- a^2 & d^2- a^2 \\ 0 & b^3- a^3 & c^3- a^3 & d^3- a^3\end{bmatrix}$

Now [tex]b^2- a^2= (b- a)(b+ a) so we need to multiply b+ a times the second row and subract that from the third row.
[tex](c^2- a^2)- (c- a)(b+ a)= (c- a)(c+ a)- (c- a)(b+ a)= (c- a)(c- b)[tex] and $(d^2- a^2)- (d- a)(b+ a)= (d-a)(d- b)$

[tex]b^3- a^3= (b- a)(b^2+ ab+ a) so need to multiply $b^2+ ab+ a^2$ times the second row and subtract that from the fourth row.
$(c^3- a^3)- (c- a)(b^2+ ab+ a^2)= (c-a)(c^2+ ac+ a^2)- (c-a)(b^2+ ab+ a^2)= (c- a)(c^2- b^2+ ac-ab)= (c-a)(c-b)(a+ b+ c)$

Similarly, $(d^3- a^3)- (d-a)(b^2+ ab+ a^2)= (d-a)(d-b)(a+ b+ d)$
Clearing the second column, we have
$\begin{bmatrix}1 & 1 & 1 &1 \\ 0 & b-a & c-a & d-a \\ 0 & 0 & (c-a)(c-b) & (d-a)(d-b) \\ 0 & 0 & (c-a)(c-b)(a+ b+ c) & (d-a)(d-b)(a+ b+ d)\end{bmatrix}$.

Now you try the third column. (Multiply the third row by (a+ b+ c) and subtract from the fourth row.)