Question

How do you solve the system `3a+2b=27, 6a-7b+c=5, -2a+10b+5c=-29` using matrices?

Answer 1

`a=1323/199, b=702/199, c= -2029/199`

Explanation:

3a + 2b = 27
6a - 7b + c = 5
-2a + 10b +5c = -29

First, convert the left side of the equations into a matrix called 'A'. Inside the matrix, put in the coeffecients of each of the variables in the equations above.

A= `[(3,2,0),(6,-7, 1),(-2, 10, 5)]`

Now, do the same for right side of the equation and convert the numbers into a matrix called 'b'.

b= `[(27),(5),(-29)]`

Use the formula `Ax=b`, where `x` is the solution for the system of equations.

Rearrange for `x` `->` `x = A^-1 * b`

`A^-1` refers to the inverse of the `A` matrix. It can easily be found using a graphical calculator.

If you need to do it by hand, here are some videos from Khan Academy on how to do it. I did mine using Symbolab.

`x= [(3,2,0),(6,-7, 1),(-2, 10, 5)]^-1 * [(27),(5),(-29)]`

`x = [(1323/199),(702/199),(-2029/199)]`

Therefore,

`a=1323/199, b=702/199, c= -2029/199`

Answer 2

The solution is `((a),(b),(c))` `=` `((1323/199),(702/199),(-2029/199))`

Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

`((3,2,0,:,27),(6,-7,1,:,5),(-2,10,5,:,-29))`

`R3larr3R3+R2`, `=>`, `((3,2,0,:,27),(6,-7,1,:,5),(0,23,16,:,-82))`

`R2larrR2-2R1`, `=>`, `((3,2,0,:,27),(0,-11,1,:,-49),(0,23,16,:,-82))`

`R2larr(R2)/(-11)`, `=>`, `((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,23,16,:,-82))`

`R3larr(R3)/(23)`, `=>`, `((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,1,16/23,:,-82/23))`

`R3larrR3-R2`, `=>`, `((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,0,199/253,:,-2029/253))`

`R3larr(R3)*(253/199)`, `=>`, `((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,0,1,:,-2029/199))`

`R2larrR2+(1/11)R3`, `=>`, `((3,2,0,:,27),(0,1,0,:,702/99),(0,0,1,:,-2029/199))`

`RlarrR1-(2)R2`, `=>`, `((3,0,0,:,3969/199),(0,1,0,:,702/99),(0,0,1,:,-2029/199))`

`Rlarr(R1)/(3)`, `=>`, `((1,0,0,:,1323/199),(0,1,0,:,702/199),(0,0,1,:,-2029/199))`