Using matrices to solve simulateneous equations?

Question

If I have the question:

4x1 + 3x2 = 18
3x1 - x2 = 7

How would I solve it using matrices?

Note: the 1 and 2 aren't 'to the power', its to differentiate between two different variables.

Answer 1

Matrix A
[4___3]
[3__-1]

Matrix A^(-1)
(-1/13)[-1__- 3]
_____[- 3 __4]

[x1] = A^(-1) [18]
[x2]_______ [7]

[x1] = [3]
[x2]__[2]

x1 = 3
x2 = 2

Difficult to type but hope you can follow.

Answer 2

you're explicitly being asked to remedy those utilising a matrix, not substitution or addition/removing. So each physique doing it those different techniques is misguided. The matrix for the 1st is [ 4 5 | -7 ] [ 3 -6 | 24 ] Divide the 2nd row by ability of three to get [ 4 5 | -7 ] [ a million -2 | 8 ] Multiply the final row by ability of -4 and upload it to the 1st row to get [ 4-4 5+8 | -7-32] [ a million -2 | 8 ] [ 0 13 | -39] [ a million -2 | 8 ] Divide the 1st row by ability of 13 to get [ 0 a million | -3] [ a million -2 | 8 ] So y= -3. Now you basically have one extra matrix operation to make to remedy for x.

Answer 3

1. Express the system in matrix form:

4 3 x1 18

3 -1 x2 7


2 a) Find the determinant of the matrix of coefficients:

det(M) = (4 * -1) - (3 * 3) = -13


2 b) Find the inverse of the matrix M:


3. Rewrite the system:

(x1, x2) = (M^-1) * (18, 7) = (3,2)

Answer 4

I4 3IIx1I=I18I
I3 -1Ix2I=I 7I .suposed2 b matrices..
(using the rules M1xM2=M3 and gives u ur orig simul equns..)
Mult row 2 by 3...making M4
I4 3IIx1I=I18I
I9-3II3X2I=I21I....M4
add rows..making new 2x1 M5
>I13 0IIx1I=I39I
......... Ix2I
>13X1=39..X1=3..and X2=2 by subst in orig equation..
this is hopeless format for mathematical representational symbols!