Question

g Given the system of equations: 15 c1 – 5 c2 – c3 = 2400 - 5 c1 + 18 c2 – 6 c3 = 3500 - 4 c1 - c2 + 12 c3 = 4000 (a) Calculate the inverse of the matrix? (b) Use the inverse to solve the problem. (c) Determine how much the load in equations 3 needs to be changed by to achieve a 15 % increase in c1.

Answer 1

(a)

`A^(-1) = (1)/(3493) \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}`

(b)

`A^(-1) b = \begin{pmatrix} (500)/(7) \\\\ (2400)/(7) \\\\ (2700)/(7) \end{pmatrix}`

Explanation:

Remember that when you want to solve a problem like this, you express the equation as following

`Ax = b`

So, if you know the inverse of
`A` then

`x = A^(-1) b`

For this case

`A = \begin{pmatrix} 15 && 5 && -1 \\ -5 && 18 && -6 \\ -4 && -1 && 12 \end{pmatrix}`

Now for this case the inverse of A would be

`A^(-1) = (1)/(3493) \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}`

Then when you multiply with the vector solution

`A^(-1) b = (1)/(3493) \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix} * \begin{pmatrix} 2400 \\ 3500 \\ 4000 \end{pmatrix} = (1)/(3493) \begin{pmatrix} 249500 \\ 1197600 \\ 1347300 \end{pmatrix}\\\\\\= \begin{pmatrix} (500)/(7) \\\\ (2400)/(7) \\\\ (2700)/(7) \end{pmatrix}`

So from that information you can conclude that the solution to the system of equations is x = 500/7 y = 2400/7 and z = 2700/7