# Another check:

• May 11th 2008, 03:28 AM
deragon999
Another check:
Been given the matrix A=[k,k,0;k^2,2,k;0,k,k] ; indicates an new row

Need to find: a) the determinant of A using cofactor expansion, and
b) the values of k for which A is invertible...

a)

detA=kdet[2,k;k,k]-kdet[k^2,k;0,k]-0det[k^2,2;0,k]
detA=k(2k-k^2)-k(K^3-0)-0
detA=-k^4-k^3+2k^2

b)

0=-k^4-k^3+2k^2

k= -2,1,0(repeated)
Therefore A is invertable where DetA doesn't = zero, which is for values of k>1, 0<k<1, -2<k<0, and k<-2
• May 11th 2008, 03:35 AM
Moo
Hello,

Quote:

Originally Posted by deragon999
Been given the matrix A=[k,k,0;k^2,2,k;0,k,k] ; indicates an new row

Need to find: a) the determinant of A using cofactor expansion, and
b) the values of k for which A is invertible...

a)

detA=kdet[2,k;k,k]-kdet[k^2,k;0,k]-0det[k^2,2;0,k]
detA=k(2k-k^2)-k(K^3-0)-0
detA=-k^4-k^3+2k^2

b)

0=-k^4-k^3+2k^2

k= -2,1,0(repeated)
Therefore A is invertable where DetA doesn't = zero, which is for values of k>1, 0<k<1, -2<k<0, and k<-2

It's OK ! (Clapping)
• May 12th 2008, 08:05 PM
Dr Zoidburg
fyi,
-k4 – k³ + 2k²

= -k²( k² + k – 2)
= -k²(k + 2)(k - 1)

which is easier to solve in (b) than leaving it as -k4 – k³ + 2k².

Good luck with the assignment. due tomorrow is it not?
• May 12th 2008, 11:44 PM
Moo
Quote:

Originally Posted by Dr Zoidburg
fyi,
-k4 – k³ + 2k²

= -k²( k² + k – 2)
= -k²(k + 2)(k - 1)

which is easier to solve in (b) than leaving it as -k4 – k³ + 2k².

Good luck with the assignment. due tomorrow is it not?

I think he did it, but didn't show it, right ?
• May 13th 2008, 12:31 AM
Dr Zoidburg
most likely. But I figured it looks better assignment-wise to show the factorisation.