Problem17:Find an upper triangular matrix A, such that, |
> ;
3 [1 30]
A = [ ]
[0 -8]
|
|
SOLUTION:
| Just multiply a general 2 by 2 upper triangular matrix to see the pattern. |
> A := matrix(2,2,[a,b,0,c]);
[a b]
A := [ ]
[0 c]
> A2 := evalm(A^2);
[ 2 ]
[a a b + b c]
A2 := [ ]
[ 2 ]
[0 c ]
> A3 := evalm(A^3);
[ 3 2 ]
[a a b + (a b + b c) c]
A3 := [ ]
[ 3 ]
[0 c ]
| Hence, we need to find a,b and c so that A3 coincides with the matrix A^3 given in the problem. The three equations are, |
> Eq1 := a^3 = 1; Eq2 := c^3 = -8; Eq3 := b*(a^2+a*c+c^2) = 30;
3
Eq1 := a = 1
3
Eq2 := c = -8
2 2
Eq3 := b (a + c a + c ) = 30
| and the solution is straight forward, |
> a := 1: c := -2: Eq3;
3 b = 30
| and b = 10. Let's check, |
> matrix(2,2,[1,10,0,-2])^3;
[1 10]3
[ ]
[0 -2]
| evaluates to: |
> evalm(%);
[1 30]
[ ]
[0 -8]
| cool! |