Have trouble with this determinant....

[ash_wolf]

Hey there....

How does one solve this determinant?

 

10 -10 0 0

-10 17 -2 -5

0 -2 7 -1

0 -5 -1 26

 

Since the lines arent coming straight....

Row #1: 10, -10, 0, 0

Row #2: -10, 17, -2, -5

Row #3: 0, -2, 7, -1

Row #4: 0, -5, -1, 26

 

How do i solve this?

[the tree]

If you want your notation to render clearly on this forum, then you'd do well to read the LaTeX Tutorial

For instance, that matrix of yours, easier to read when it looks like this: [math]\begin{bmatrix}10&-10&0&0\\

-10&17&-2&-5\\0&-2&7&-1\\0&-5&-1&26\end{bmatrix}[/math]

Now unfortunately my maths course skips fun stuff like matrices, so I can't really help you much here. But I always find mathworld useful for comprehensive explanations.

[Dave]

I would use expansion by minors:

 

[math]\left| \begin{array}{cccc}

10 & -10 & 0 & 0 \\

-10 & 17 & -2 & -5 \\

0 & -2 & 7 & -1 \\

0 & 5 & -1 & 26 \end{array} \right| = 10 \left| \begin{array}{ccc} 17 & -2 & -5 \\

-2 & 7 & -1 \\

5 & -1 & 26 \end{array} \right| + 10 \left| \begin{array}{ccc} -10 & -2 & -5 \\

0 & 7 & -1 \\

0& -1 & 26 \end{array} \right|[/math]

 

Notice the change in sign on the second determinant.

[Pre4edgc]

Expansion by diagonals work well too.

10 -10 0 0

-10 17 -2 -5

0 -2 7 -1

0 -5 -1 26

(Sorry, don't know how to write the math like you did.)

Expand it like this. Take the first three rows, copy them, and put them at the end like this:

10 -10 0 0 10 -10 0

-10 17 -2 -5 -10 17 -2

0 -2 7 -1 0 -2 7

0 -5 -1 26 0 -5 -1

Then take the downward diagonals, first to last, and then the upward, and multiply the numbers in those diagonals. Kind of a big number, but it still works. In this case, a downward would be 10, 17, 7, and 26. Multiply them, and keep record of the number. After you do that for all of them, add the products from the downward diagonals, and subtract the products of the upward diagonals. Your determinant is that. Not very mathy, and I had to look up some of it in my math textbook, but it should work.