250+ TOP MCQs on System of Equation using Gauss Elimination Method and Answers [2024]

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

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
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&-3end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}1\2\0\3end{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&-12end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}1\0\-2\0end{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&-22end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}1\0\-2\0end{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
Answer: c
Explanation: Converting the equations into Matrix Form:
(begin{bmatrix}1&2&3\2&3&4\3&4&5end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\5\6end{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&-4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\-3\-6end{bmatrix})
R₃=R₃-5R₂
(begin{bmatrix}1&2&3\0&-1&-2\0&0&6end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}4\-3\9end{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

a) z=(frac{53}{41})
b) z=(frac{-143}{287})
c) z=(frac{79}{287})
d) z=(frac{-53}{41})
Answer: a
Explanation: Converting the equations into Matrix Form:
(begin{bmatrix}2&3&4\3&8&1\12&4&3end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}5\2\-1end{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&3end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-3\2\-1end{bmatrix})

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

R₃=R₃-8R₂
(begin{bmatrix}1&5&-3\0&-7&10\0&0&-41end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}-3\11\-53end{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.

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
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&3end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-1\4\1\3end{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&3end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-4\4\1\3end{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&-13end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-4\12\13\19end{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&117end{bmatrix}
begin{bmatrix}x\y\z\wend{bmatrix} = begin{bmatrix}-4\12\-83\-137end{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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