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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.

Respuesta :

Answer

(a)

[tex]A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}[/tex]

(b)

[tex]A^{-1} b = \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}[/tex]

Step-by-step explanation:

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

[tex]Ax = b[/tex]

So, if you know the inverse of   [tex]A[/tex]   then

[tex]x = A^{-1} b[/tex]

For this case

[tex]A = \begin{pmatrix} 15 && 5 && -1 \\ -5 && 18 && -6 \\ -4 && -1 && 12 \end{pmatrix}[/tex]

Now for this case the inverse of A would be

[tex]A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}[/tex]

Then when you multiply with the vector solution

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

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

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