Linear Algebra Questions and Answers - System of Equation using Gauss Elimina...

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “System of Equation using Gauss Elimination Method”.

1. Solve the following equations using Gauss Elimination Method and find the value of x and z.

x + y + 2z + 3w = 1
2x + 3y - 2z + 4w = 2
2x + 3y + z - w = 0
3x - 2y + z - 3w = 3

a) x=1.1 and z=-0.2
b) x=0.3 and z=-0.2
c) x=1.1 and z=0.3
d) x=0.3 and z=-0.6
View Answer

Answer: a
Explanation: Converting the equations into Matrix Form:
\(\begin{bmatrix}1&1&2&3\\2&3&-2&4\\2&3&1&-1\\3&-2&1&-3\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}1\\2\\0\\3\end{bmatrix}\)
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R₂=R₂-R₁ and R₃=R₃-2R₁ and R₄=R₄-3R₁
\(\begin{bmatrix}1&1&2&3\\0&1&-6&-2\\0&1&-3&-7\\0&-5&-5&-12\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}1\\0\\-2\\0\end{bmatrix}\)
R₃=R₃-R₂ and R₄=R₄+5R₂
\(\begin{bmatrix}1&1&2&3\\0&1&-6&-2\\0&0&3&-5\\0&0&-35&-22\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}1\\0\\-2\\0\end{bmatrix}\)

From this, we get
x+y+2z+3w=1 – (1)
y-6z-2w=0 – (2)
3z-5w=-2 – (3)
-35z-22w=0 – (4)
From equations (3) and (4) we get
z=-0.2 and w=0.3
Substituting these values in equation (2)
y=6(-0.2)+2(0.3)
y=-0.6
Substituting these values in equation (3)
x+(-0.6)+2(-0.2)+3(0.3)=1
x=1.1
Thus, the solution of the following equation is
x=1.1
y=-0.6
z=-0.2
w=0.3.

2. Solve the following equations using Gauss Elimination Method.

x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6

a) x=0.5 y=0 and z=1
b) x=-0.5 y=0 and z=-1.5
c) x=-0.5 y=0 and z=1.5
d) x=0.5 y=0 and z=1.5
View Answer

Answer: c
Explanation: Converting the equations into Matrix Form:
\(\begin{bmatrix}1&2&3\\2&3&4\\3&4&5\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\5\\6\end{bmatrix}\)
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R₂=R₂-2R₁ and R₃=R₃-3R₁
\(\begin{bmatrix}1&2&3\\0&-1&-2\\0&-5&-4\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\-3\\-6\end{bmatrix}\)
R₃=R₃-5R₂
\(\begin{bmatrix}1&2&3\\0&-1&-2\\0&0&6\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\-3\\9\end{bmatrix}\)
From the above Matrix, we get –
x+2y+3z=4 -(2)
-y-2z=-3 -(1)
6z=9
Thus, z=1.5
Substituting in (1)
y=3-2(1.5)
y=0
Substituting in (2)
x=4-2(0)-3(1.5)
x=-0.5
Thus, x=-0.5 y=0 and z=1.5.

3. Find the value of z that satisfies all the given equations using Gauss Elimination Method.

12x + 4y + 3z = -1
3x + 8y + z = 2
2x + 3y + 4z = 5

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a) z=\(\frac{53}{41}\)
b) z=\(\frac{-143}{287}\)
c) z=\(\frac{79}{287}\)
d) z=\(\frac{-53}{41}\)
View Answer

Answer: a
Explanation: Converting the equations into Matrix Form:

\(\begin{bmatrix}2&3&4\\3&8&1\\12&4&3\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}5\\2\\-1\end{bmatrix}\)
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R₂=R₂-2R₁
\(\begin{bmatrix}1&5&-3\\3&8&1\\12&4&3\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-3\\2\\-1\end{bmatrix}\)

R₂=R₂-3R₁ and R₃=R₃-12R₁
\(\begin{bmatrix}1&5&-3\\0&-7&10\\0&-56&39\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-3\\11\\35\end{bmatrix}\)

R₃=R₃-8R₂
\(\begin{bmatrix}1&5&-3\\0&-7&10\\0&0&-41\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-3\\11\\-53\end{bmatrix}\)
From the above Matrix, we get-
x+5y-3z=-3-(2)
-7y+10z=11-(1)
-41z=-53
Thus, z=\(\frac{53}{41}\)
Substituting in (1)
y=\(\frac{79}{287}\)
Substituting in (2)
x=\(\frac{-143}{287}\).

4. Find the value of x and y that satisfy the given sets of equation using Gauss Elimination Method.

2x - 3y + 2z - 2w = 4
3x + 2y - 3z + 5w = 1
4x + 5y + 11z + 3w = 3
5x + 3y + 2z + 7w = - 1

a) x=0.85 and y=-1.2
b) x=0.85 and y=0.38
c) x=0.27 and y=0.38
d) x=0.27 and y=-1.2
View Answer

Answer: d
Explanation: Converting the equations into Matrix Form:

\(\begin{bmatrix}5&3&2&7\\2&-3&2&-2\\3&2&3&5\\4&5&11&3\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}-1\\4\\1\\3\end{bmatrix}\)
We now need to convert this matrix into Upper Triangular Matrix by Elementary Row Transformation only.
R₁=R₁-R₄
\(\begin{bmatrix}1&-2&-9&4\\2&-3&2&-2\\3&2&3&5\\4&5&11&3\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}-4\\4\\1\\3\end{bmatrix}\)

R₂=R₂-2R₁ and R₃=R₃-3R₁ and R₄=R₄-4R₁
\(\begin{bmatrix}1&-2&-9&4\\0&1&20&-10\\0&8&30&-7\\0&13&47&-13\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}-4\\12\\13\\19\end{bmatrix}\)
R₃=R₃-8R₂ and R₄=R₄-13R₂
\(\begin{bmatrix}1&-2&-9&4\\0&1&20&-10\\0&0&-130&73\\0&13&-213&117\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}-4\\12\\-83\\-137\end{bmatrix}\)
From this, we get
x-2y-9z+4w=-4 – (1)
y+20z-10w=12 – (2)
-130z+73w=-83 – (3)
-213z-117w=-137 – (4)
From equations (3) and (4) we get
z=0.85 and w=0.38
Substituting these values in equation (2)
y=12-20(0.85)+10(0.38)
y=-1.2
Substituting these values in equation (3)
x+(-2×(-1.2))-9(0.85)+4(0.38)=-4
x=0.27
Thus, the solution of the following equation is
x=0.27
y=-1.2
z=0.85
w=0.38.

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