Question

Suppose that the augmented matrix for a linear system has been reduced by row operations to the given row echelon form. Solve the system. (a) [1 -3 4 7, 0 1 2 2, 0 0 1 5] (b) [1 0 8 -5 6, 0 1 4 -9 3, 0 0 1 1 2] (c) [1 7 -2 0 -8 -3, 0 0 1 1 6 5, 0 0 0 1 3 9, 0 0 0 0 0 0] (d) [1 -3 7 1, 0 1 4 0, 0 0 0 1]

Answer 1

Answer:

a) x = -37 , y = -8, z = 5.

b) w = 13z-10, x = 13z-5, y= 2-z, z = free parameter.

c) v = 2z-7w-11, w = free parameter, x = -3z-4, y = 9-3z, z = free parameter

d) No solution

Step-by-step explanation:

a) $\left[\begin{array}{cccc}1&-3&4&7\\0&1&2&2\\0&0&1&5\end{array}\right]$

The following set of equations are:

z = 5

y+2z=2

x-3y+4z=7

Solving above equations:

y = -8, x = 7-4(5)+3(-8) = -37

b) $\left[\begin{array}{ccccc}1&0&8&-5&6\\0&1&4&-9&3\\0&0&1&1&2\end{array}\right]$

The following set of equations are:

y+z = 2

x+4y-9z=3

w+8y-5z=6

Solving above equations:

z = free parameter - No. of columns > No. of rows; hence, multiple solutions.

y = 2-z,  x = 3+9z-4(2-z) = -5+13z,  w = 6+5z-8(2-z) = 13z-10

c) $\left[\begin{array}{cccccc}1&7&-2&0&-8&-3\\0&0&1&1&6&5\\0&0&0&1&3&9\end{array}\right]$

The following set of equations are:

y + 3z = 9

x + y + 6z = 5

v + 7w -2x -8z = -3

Solving above equations:

z & w = free parameter - No. of columns > No. of rows; hence, multiple solutions.

y = 9-3z,  x = -3z-4,  v = -11+2z-7w

d) $\left[\begin{array}{cccc}1&-3&7&1\\0&1&4&0\\0&0&0&1\end{array}\right]$

The following set of equations are:

0 = 1

y + 4z = 0

x-3y+7z = 1

Solving above equations:

No Solution as the first equation is inconsistent