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I Guess It’s Time
Aug 11
No Solution
Jul 17
There is no solution. To find the solution you could use normal algebra but that would take too long and very long if there is no solution.
So use linear algebra instead.
rewrite the equation as to matricies.
|1 4 -2| | 6|
A= |2 3 2 | B= | 7|
|1 -1 4| | 1|
Multiplying the inverse of matrix A to matrix B will yield a 3×1 matrix | x|
| y|
| z|
to find the inverse of A
find the determinate of A
find the transpose of A
find the resulting matrix of cofactors
multiply by matrix |1 -1 1 |
|-1 1 -1 |
|1 -1 1 |
now multiply the resulting matrix by (1 divided by the determinate)
To find the deteminate of |x1 x2 x3|
|y1 y2 y3|
|z1 z2 z3|
x1[(y2)(z3) - (y3)(z2)] - x2[(y1)(z3) - (y3)(z1)] + x3[(y1)(z2) - (y2)(z1)]
the determinate of matrix a becomes
1[12-(-2)] – 4[8 - 2] + (-2)[ -2 - 3]
14 -24 + 10
0
but we need to multiply the final matrix by 1/0
can’t divide by 0
no solution
thanks to Nick
