Solving for matrices

Question

asked Nov 3, 2020 137k views

2 Answers

Avacariu · Answer 1 · 2020-11-06T06:28:17+0000

The correct answer is option d: (1, 5, 2).

To solve the system of equations using the augmented matrix and row operations, we can set up the augmented matrix for the system:

$\left[\begin{array}{ccc}3&-4&5\\5&2&-2\\5&-4&4\end{array}\right] \left[\begin{array}{ccc}-27\\11\\-7\end{array}\right]$

Now, let's perform row operations to get the reduced row-echelon form (RREF):

$R_2 =R_2 - (5)/(3) R_1\\R_3 =R_3- (5)/(3) R_1$

$R_3 =R_3 +R_2\\R_1 =R_1 -R_2\\R_1 =R_1 +R_3\\R_2 =R_2 +R_3$

After performing these row operations, the augmented matrix becomes:

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}10\\5\\2\end{array}\right]$

Now, we can read off the solution directly from the RREF matrix:

x=10

y=5

z=2

So, the correct answer is option d: (1, 5, 2).

Alexloehr · Answer 2 · 2020-11-06T21:19:42+0000

Answer:

D

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=(90)/(45)=2.

Substitute z=2 into the second equation:

$-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=(130)/(26)=5.$

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)