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icang [17]
3 years ago
15

Solving for matrices

Mathematics
1 answer:
muminat3 years ago
4 0

Answer:

D

Step-by-step explanation:

The augmented matrix for the system of three equaitons is

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)

Multiply the first row by 5, the second row by -3 and add these two rows:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)

Subtract the third row from the second:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)

Divide the third row by 6:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)

Now multiply  the third equation by 26 and add it to the second row:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)

You get the system of three equations:

\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.

From the third equation

z=\dfrac{90}{45}=2.

Substitute z=2 into the second equation:

-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.

Now substitute z=2 and y=5 into the first equation:

3x-4\cdot 5-5\cdot 2=-27\\ \\3x-20-10=-27\\ \\3x-30=-27\\ \\3x=-27+30=3\\ \\x=1.

The solution is (1,5,2)

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