# Determinant

• Nov 22nd 2007, 08:01 PM
jhonwashington
Determinant
Hello.

I'm asked to find the value of the following determinant using cramer's rule.

0 , 1 , 2
4, -5 , 8
8, 0 , -9

so I repeat the first two columns

0 1 2 0 1
4 -5 8 4 -5
8 0 -9 8 0

and then I multiply diagonally and add and then subtract and I get

0+64+0- -80+0-36=
64 +80-36 = 108

so My answer is 108, but all calculators tell me that my answer should be 180, why? I suspect that I should be adding 36 in that case I would get 180, but I don't see how I could be adding 36 if 36 has a negative sign. please help me.
• Nov 22nd 2007, 08:33 PM
earboth
Quote:

Originally Posted by jhonwashington
Hello.

I'm asked to find the value of the following determinant using cramer's rule.

0 , 1 , 2
4, -5 , 8
8, 0 , -9

so I repeat the first two columns

0 1 2 0 1
4 -5 8 4 -5
8 0 -9 8 0

and then I multiply diagonally and add and then subtract and I get

0+64+0- (-80+0-36)=
64 +80-36 = 108

so My answer is 108, but all calculators tell me that my answer should be 180, why? I suspect that I should be adding 36 in that case I would get 180, but I don't see how I could be adding 36 if 36 has a negative sign. please help me.

Hello,

you've forgotten to use brackets. I've added them in the quoted text.
• Nov 22nd 2007, 08:47 PM
Soroban
Hello, jhonwashington!

You made a teensy error . . .

Quote:

Evaluate the determinant: .$\displaystyle \begin{vmatrix}\: 0 & 1 & 2\: \\ \:4 & \text{-}5 & 8\: \\ \:8 & 0 & \text{-}9\:\end{vmatrix}$

So I repeat the first two columns: .$\displaystyle \begin{array}{ccccc}0 & 1 & 2 & 0 & 1 \\ 4 & \text{-}5 & 8 & 4 & \text{-}5 \\ 8 & 0 & \text{-}9 & 8 & 0 \end{array}$

and then I multiply diagonally and add and then subtract and I get:

. . $\displaystyle 0 + 64 + 0 - 80 + 0 \underbrace{-36}_{{\color{blue}\text{Here!}}}$

Multiply diagonally: .$\displaystyle \underbrace{(0)(\text{-}5)(\text{-}9)}_{+0} + \underbrace{(1)(8)(8)}_{+64} + \underbrace{(2)(4)(0)}_{+0} \underbrace{- (2)(\text{-}5)(8)}_{+80} - \underbrace{(0)(8)(0)}_0 \underbrace{- (1)(4)(\text{-}9)}_{{\color{red}+36}}$

And we get: .$\displaystyle 0 + 64 + 0 + 80 - 0 + 36 \;=\;\boxed{180}$

• Nov 23rd 2007, 05:15 AM
topsquark
Quote:

Originally Posted by jhonwashington
I'm asked to find the value of the following determinant using cramer's rule.

Am I missing something? I thought Cramer's rule was used to find the solution to a system of equations, not to find determinants?

-Dan