Question

Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.

3x + 5y - 2w = -13
2x + 7z - w = -1
4y + 3z + 3w = 1
-x + 2y + 4z = -5

A. {(-1, -1 , 0, )}

B. {(1, -2, 0, 3)}

C. {( , -2, 0, )}

D. {( , - , 0, )}

Answer 1

Answer with explanation:

The given system of equation are

3x + 5y - 2w = -13

2x + 7z - w = -1

4y + 3z + 3w = 1

-x + 2y + 4z = -5

Writing the system of equation in terms of Augmented Matrix

`\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\0&4&3&3&1\\-1&2&4&0&-5\end{array}\right]\\\\R_(3) \leftrightarrow R_(4)\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\-1&2&4&0&-5\\0&4&3&3&1\end{array}\right]\\\\R_(3) \rightarrow 2R_(3)+R_(2)\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\0&4&15&-1&-11\\0&4&3&3&1\end{array}\right]\\\\R_(4)\rightarrow R_(4)-R_(3)\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\2&0&7&-1&-1\\0&4&15&-1&-11\\0&0&-12&4&12\end{array}\right]`

`R_(2)\rightarrow 3R_(2)-2R_(1)\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\0&-10&21&1&23\\0&4&15&-1&-11\\0&0&-12&4&12\end{array}\right]\\\\R_(3)\rightarrow 5R_(3)+2R_(2)\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\0&-10&21&1&23\\0&0&117&-3&-9\\0&0&-12&4&12\end{array}\right]\\\\ R_(4)\rightarrow 3R_(4)+4R_(3)\\\\\left[\begin{array}{ccccc}3&5&0&-2&-13\\0&-10&21&1&23\\0&0&117&-3&-9\\0&0&432&0&0\end{array}\right]`

→432 z=0

z=0

⇒117 z-3 w=-9

-3 w=-9

Dividing both sides by -3

w=3

⇒-10 y+21z+w=23

-10 y+0+3=23

-10 y=23-3

-10 y= 20

y=-2

⇒3 x+5 y-2w=-13

3 x+5 ×(-2)-2 ×3= -13

3 x-10-6= -13

3 x=16-13

3 x=3

x=1

Option B. {(1, -2, 0, 3)}