This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “System of Equations and their Consistencies”.
1. Test for consistency and solve to find the value of x.
5x + 3y + 7z = 4 3x + 26y + 2z = 9 7x + 2y + 10z = 5
a) Consistent, x=1
b) Consistent, x=-1
c) Inconsistent system, solution does not exist
d) Consistent, infinite number of solutions possible
View Answer
Answer: d
Explanation: In this Question we have,
\(\begin{bmatrix}5&3&7\\3&26&2\\7&2&10\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\9\\5\end{bmatrix}\)
By 7R1 and 5R3
\(\begin{bmatrix}35&21&49\\3&26&2\\35&10&50\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\9\\25\end{bmatrix}\)
By R3-R1 and 5R2
\(\begin{bmatrix}35&21&49\\15&130&10\\0&-11&1\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\45\\-3\end{bmatrix}\)
By \(R_2 – \frac{3}{7} R_1\)
\(\begin{bmatrix}35&21&49\\0&121&-11\\0&-11&1\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\33\\-3\end{bmatrix}\)
By \(R_3 + \frac{1}{11}R_2\)
\(\begin{bmatrix}35&21&49\\0&121&-11\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\33\\0\end{bmatrix}\)
By \(\frac{1}{7} R_1 and \frac{1}{11} R_2\)
\(\begin{bmatrix}5&3&7\\0&11&-1\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\3\\0\end{bmatrix}\)
The rank of both the coefficient matrix and the augmented matrix is equal (i.e. 2)
Thus the equations are consistent.
However the rank of the matrix 2 is less than the total number of unknowns 3.
Hence the given set of equations has infinite number of solutions.
Explanation: In this Question we have,
\(\begin{bmatrix}5&3&7\\3&26&2\\7&2&10\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\9\\5\end{bmatrix}\)
By 7R1 and 5R3
\(\begin{bmatrix}35&21&49\\3&26&2\\35&10&50\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\9\\25\end{bmatrix}\)
By R3-R1 and 5R2
\(\begin{bmatrix}35&21&49\\15&130&10\\0&-11&1\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\45\\-3\end{bmatrix}\)
By \(R_2 – \frac{3}{7} R_1\)
\(\begin{bmatrix}35&21&49\\0&121&-11\\0&-11&1\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\33\\-3\end{bmatrix}\)
By \(R_3 + \frac{1}{11}R_2\)
\(\begin{bmatrix}35&21&49\\0&121&-11\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\33\\0\end{bmatrix}\)
By \(\frac{1}{7} R_1 and \frac{1}{11} R_2\)
\(\begin{bmatrix}5&3&7\\0&11&-1\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\3\\0\end{bmatrix}\)
The rank of both the coefficient matrix and the augmented matrix is equal (i.e. 2)
Thus the equations are consistent.
However the rank of the matrix 2 is less than the total number of unknowns 3.
Hence the given set of equations has infinite number of solutions.
2. Test for consistency and solve the system of equations if possible to get the value of z.
2x - 3y+ 7z = 5 3x + y - 3z = 13 2x + 19y - 47z = 32
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a) Consistent, z = -1
b) Consistent, z = 0
c) Inconsistent
d) Consistent, z = 5
View Answer
Answer: c
Explanation: In this Question we have,
\(\begin{bmatrix}2&-3&7\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}5\\13\\32\end{bmatrix}\)
By R1-R2
\(\begin{bmatrix}-1&-4&10\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-8\\13\\32\end{bmatrix}\)
By -R1
\(\begin{bmatrix}1&4&-10\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\13\\32\end{bmatrix}\)
By R2-3R1 and R3-2R1
\(\begin{bmatrix}1&4&-10\\0&-11&27\\0&11&-27\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-11\\16\end{bmatrix}\)
By R3+R2
\(\begin{bmatrix}1&4&-10\\0&-11&27\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-11\\5\end{bmatrix}\)
Here we see that the rank of the co-efficient matrix is 2, while the rank of the augmented matrix is 3.
Since the two ranks are not equal, the given system of equations is inconsistent.
Explanation: In this Question we have,
\(\begin{bmatrix}2&-3&7\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}5\\13\\32\end{bmatrix}\)
By R1-R2
\(\begin{bmatrix}-1&-4&10\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-8\\13\\32\end{bmatrix}\)
By -R1
\(\begin{bmatrix}1&4&-10\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\13\\32\end{bmatrix}\)
By R2-3R1 and R3-2R1
\(\begin{bmatrix}1&4&-10\\0&-11&27\\0&11&-27\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-11\\16\end{bmatrix}\)
By R3+R2
\(\begin{bmatrix}1&4&-10\\0&-11&27\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-11\\5\end{bmatrix}\)
Here we see that the rank of the co-efficient matrix is 2, while the rank of the augmented matrix is 3.
Since the two ranks are not equal, the given system of equations is inconsistent.
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