Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## #1 2006-03-14 16:02:34

4littlepiggiesmom
Member

Offline

### Computing 2 deteminant problem.....

I can't get the lines to work right but I think you'll know what they are....

1.For what value of X is the folloing determinant NOT equal to 0?
x+1        -2           -1
0           x-2          -2
0            0            x+3

a. 1
b 2
c. -3
d. -1
e. none of these       I came up with d

2. Compute the following determinant:
x-1     -2         -1
-1      x-2       -2
3         3        x+3

a x^3 +3
b. x^3 + 6 x^2 - 9x + 33
c. X63 +14x +3
d. x^3 + 3
e. none of these                      Yes a and D are the same but I came up with c?

## #2 2006-03-14 16:17:58

George,Y
Super Member

Offline

### Re: Computing 2 deteminant problem.....

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=matrices&s2=determinant&s3=basic

X'(y-Xβ)=0

## #3 2006-03-14 20:36:00

gnitsuk
Full Member

Offline

### Re: Computing 2 deteminant problem.....

1. Let's evaluate the determinant:

Remember DET|a1 b1 c1|
|a2 b2 c2|
|a3 b3 c3|    =   a1(b2c3 - c2b3) - b1(a2c3 - c2a3) + c1(a2b3 - b2a3)

Using this we can calucluate our determinant as:

(x+1)(x-2)(x+3) -0 + 0 = (x+1)(x-2)(x+3) so we straight away see that the only value of x listed for which this is NOT zero is 1. Therefore the answer is a. The link provided by George confirms this.

2. Lets just expand the determinant again using the same formula.

(x-1)[(x-2)(x+3) + 6] + 2(-x - 3 + 6) - 1(-3 -3x + 6) =

x^3 + x^2 - 6x + 6x - x^2 - x + 6 - 6 - 2x - 6 + 12 + 3 + 3x - 6 =

x^3 + 3

So the answer is a or d

Mitch.

Last edited by gnitsuk (2006-03-14 20:37:27)

## #4 2006-03-14 23:08:59

4littlepiggiesmom
Member

Offline

### Re: Computing 2 deteminant problem.....

Thanks so much guys/gals!

George,Y
Super Member

Offline

My pleasure

X'(y-Xβ)=0