Linear Algebra Questions and Answers – System of Equations and their Consistencies

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This set of Linear Algebra Interview Questions and Answers for freshers 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
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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.

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

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.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn