Linear Algebra Questions and Answers – System of Equation using Gauss Elimination Method

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This set of Linear Algebra Questions and Answers for Experienced people 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
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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
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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
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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.

Sanfoundry Global Education & Learning Series – Linear Algebra.

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