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Consider the accompanying matrix as the augmented matrix of a linear system. State in words the next two elementary row operations that should be performed in the process of solving the system.
(1 -4 4 0 -2
0 2 -6 0 5
0 0 1 2 - 4
0 0 4 5 -1]
(Type an integer or a simplified fraction)
A. Replace row 4 by its sum with - 4 times row 3.
(Type an integer or a simplified fraction)
B. Replace row 2 by its sum with times row 4.
(Type an integer or a simplified fraction.)
C. Interchange row 3 and row 2.

Respuesta :

Answer:

The correct option is (A)

A. Replace row 4 by its sum with - 4 times row 3.

[tex]\left[\begin{array}{ccccc}1&-4&4&0&-2\\0&2&-6&0&5\\0&0&1&2&-4\\0&0&0&-3&15\end{array}\right][/tex]

w = 8

x = 17/2

y = 6

z = -5

Step-by-step explanation:

The given matrix is

[tex]\left[\begin{array}{ccccc}1&-4&4&0&-2\\0&2&-6&0&5\\0&0&1&2&-4\\0&0&4&5&-1\end{array}\right][/tex]

To solve this matrix we need to create a zero at the 4th row and 3rd column which is 4 at the moment.

Multiply 3rd row by -4 and add it to the 4th row.

Mathematically,

[tex]R_4 = R_4 - 4R_3[/tex]

So the correct option is (A)

A. Replace row 4 by its sum with - 4 times row 3.

So the matrix becomes,

[tex]\left[\begin{array}{ccccc}1&-4&4&0&-2\\0&2&-6&0&5\\0&0&1&2&-4\\0&0&0&-3&15\end{array}\right][/tex]

Now the matrix may be solved by back substitution method.

Bonus:

The solution is given by

Eq. 1

-3z = 15

z = -15/3

z = -5

Eq. 2

y + 2z = -4

y + 2(-5) = -4

y - 10 = -4

y = -4 + 10

y = 6

Eq. 3

2x - 6y + 0z = 5

2x - 6(6) = 5

2x - 12 = 5

2x = 12 + 5

2x = 17

x = 17/2

Eq. 4

w - 4x + 4y + 0z = -2

w - 4(17/2) + 4(6) = -2

w - 34 + 24 = -2

w - 10 = -2

w = -2 + 10

w = 8

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