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## #1 2006-03-13 17:02:34

4littlepiggiesmom
Member
Registered: 2006-01-09
Posts: 42

### 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

Answer options
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

Answer options
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?

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## #2 2006-03-13 17:17:58

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### 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

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## #3 2006-03-13 21:36:00

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### 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 (2006-03-13 21:37:27)

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## #4 2006-03-14 00:08:59

4littlepiggiesmom
Member
Registered: 2006-01-09
Posts: 42

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

Thanks so much guys/gals!

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## #5 2006-03-15 15:03:06

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

My pleasure

X'(y-Xβ)=0

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