Question

Solve these three equations using matrix inversion method:
5x-6y-7z=7
6x-4y+10z=-34
2x+4y-3z=29​

Answer 1

Given:

$5x - 6y - 7z = 7 \\ 6x - 4y + 10z = - 34 \\ 2x + 4y - 3z = 29$

To find: Solve these three equations using matrix inversion method.

Solution:

Step 1:Write these equations in matrix form AX=B

as shown

$\left[\begin{array}{ccc}5&-6&-7\\6&-4&10\\2&4&-3\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}7\\-34\\29\end{array}\right]$

Step 2: Solution is $X=A^{-1}B$

Thus, Find A inverse first.

Step 3: Find determinant.

$|A|=\left|\begin{array}{ccc}5&-6&-7\\6&-4&10\\2&4&-3\end{array}\right|\\\\|A|=5(12-40)-(-6)(-18-20)-7(24+8)\\\\|A|=5(-28)+6(-38)-7(32)\\\\|A|=-140-228-224\\\\|A|=-592$

Find A inverse

$A^{-1}=\frac{-1}{592}\left[\begin{array}{ccc}-28&-46&-88\\38&-1&-92\\32&-32&16\end{array}\right]$

or

$A^{-1}=\frac{1}{592}\left[\begin{array}{ccc}28&46&88\\-38&1&92\\-32&32&-16\end{array}\right]$

Step 4: Multiply A inverse with B

$A^{-1}B=\frac{1}{592}\left[\begin{array}{ccc}28&46&88\\-38&1&92\\-32&32&-16\end{array}\right]\left[\begin{array}{ccc}7\\-34\\29\end{array}\right]$

$A^{-1}B=\frac{1}{592}\left[\begin{array}{ccc}1184\\2368\\-1776\end{array}\right]$

$X=\left[\begin{array}{ccc}\frac{1184}{592}\\\\\frac{2368}{592}\\\\\frac{-1776}{592}\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}2\\4\\-3\end{array}\right]$

Final answer:

$x= 2 \\ y = 4 \\ z = - 3$

Hope it helps you.

To learn more:

1)solve

5X-Y+4Z=5,2x+3y+5z=2,7x-2y+6z=5 by cramer's rule differentiate tha following with respect x

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2) if A=(3 0/5 1) and B=(-4 2/1 0) find A^2-2ab+b^2

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