Signals & Systems Questions and Answers – Basics of Linear Algebra

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This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Basics of Linear Algebra”.

1. Find the values of x, y, z and w from the below condition.
\(5\begin{bmatrix}
x & z \\
y & w \\
\end{bmatrix} = \begin{bmatrix}
2 & 10 \\
3 & 2x+y \\
\end{bmatrix} + \begin{bmatrix}
z & 5 \\
7 & w \\
\end{bmatrix} \).
a) x=1, y=3, z=4, w=0
b) x=2, y=3, z=8, w=1
c) x=1, y=2, z=3, w=1
d) x=1, y=2, z=4, w=1
View Answer

Answer: d
Explanation: 5z=10+5 => 5z=15 => z=3
5x=2+z => 5x=5 => x=1
5y=3+7 => 5y=10 => y=2
5w=2+2+w => 4w=4 => w=1.
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2. The matrix A is represented as \(\begin{bmatrix}
1 & 4 \\
-2 & 9 \\
-3 & -8 \\
\end{bmatrix}\). The transpose of the matrix of this matrix is represented as?
a) \(\begin{bmatrix}
1 & 4 \\
-2 & 9 \\
\end{bmatrix}\)
b) \(\begin{bmatrix}
1 & 4 \\
-2 & 9 \\
-3 & 8 \\
\end{bmatrix}\)
c) \(\begin{bmatrix}
1 & -2 & -3\\
4 & 9 & 8\\
\end{bmatrix}\)
d) \(\begin{bmatrix}
-1 & 2 & 3\\
-4 & -9 & 8\\
\end{bmatrix}\)
View Answer

Answer: c
Explanation: Given matrix is a 3×2 matrix and the transpose of the matrix is 3×2 matrix.
The values of matrix are not changed but, the elements are interchanged, as row elements of a given matrix to the column elements of the transpose matrix and vice versa but the polarities of the elements remains same.

3. Find the inverse of the matrix \(A = \begin{bmatrix}
8 & 5 & 2\\
4 & 6 & 3\\
7 & 4 & 2\\
\end{bmatrix}\).
a) \(\frac{1}{13}*\begin{bmatrix}
90 & 65 & 80\\
65 & 61 & 54\\
80 & 58 & 69\\
\end{bmatrix}\)
b) \(\frac{1}{14}*\begin{bmatrix}
93 & 68 & 80\\
68 & 61 & 58\\
80 & 58 & 69\\
\end{bmatrix}\)
c) \(\frac{1}{13}*\begin{bmatrix}
94 & 67 & 80\\
67 & 60 & 56\\
80 & 58 & 69\\
\end{bmatrix}\)
d) \(\frac{1}{13}*\begin{bmatrix}
93 & 68 & 80\\
68 & 61 & 58\\
80 & 58 & 69\\
\end{bmatrix}\)
View Answer

Answer: d
Explanation: The inverse of matrix A = \(\frac{adjA}{|A|}\),
adjA=AA-1,
adjA = \(\frac{1}{13}*\begin{bmatrix}
93 & 68 & 80\\
68 & 61 & 58\\
80 & 58 & 69\\
\end{bmatrix}\), |A|=13.
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4. Given the equations are 4x+2y+z=8, x+ y+ z=3, 3x+y+3z=9. Find the values of x, y and z.
a) 5/3, 0, 2/3
b) 1, 2, 3
c) 4/3, 1/3, 5/3
d) 2, 3, 4
View Answer

Answer: a
Explanation: The matrix from the equations is represented as M=\(\begin{bmatrix}
4 & 2 & 1\\
1 & 1 & 1\\
3 & 1 & 3\\
\end{bmatrix}\)
The another matrix is X = \(\begin{bmatrix}
8\\
3\\
9\\
\end{bmatrix}\)
Then |M| = 6
For x=\(\begin{bmatrix}
8 & 2 & 1\\
3 & 1 & 1\\
9 & 1 & 3\\
\end{bmatrix}\) = 5/3
Similarly, y=0, z=-2/3.

5. Find the adjacent A as A=\(\begin{bmatrix}
1 & 7 & -3\\
5 & 4 & -2\\
6 & 8 & -6\\
\end{bmatrix}\).
a) \(\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 3\\
2 & 3 & 4\\
\end{bmatrix}\)
b) \(\begin{bmatrix}
31 & 39 & 80\\
39 & 45 & 74\\
80 & 74 & 136\\
\end{bmatrix}\)
c) \(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix}\)
d) \(\begin{bmatrix}
35 & 34 & 32\\
56 & 67 & 48\\
98 & 74 & 52\\
\end{bmatrix}\)
View Answer

Answer: b
Explanation: The adjacency of A is given by AAT
AT = \(\begin{bmatrix}
1 & 5 & 6\\
7 & 4 & 8\\
-3 & -2 & -6\\
\end{bmatrix}\),
AAT = \(\begin{bmatrix}
1 & 7 & -3\\
5 & 4 & -2\\
6 & 8 & -6\\
\end{bmatrix} × \begin{bmatrix}
1 & 5 & 6\\
7 & 4 & 8\\
-3 & -2 & -6\\
\end{bmatrix}\)
adjA=\(\begin{bmatrix}
31 & 39 & 80\\
39 & 45 & 74\\
80 & 74 & 136\\
\end{bmatrix}\).
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6. Find the rank of the matrix A=\(\begin{bmatrix}
1 & 3 & 5\\
4 & 6 & 7\\
1 & 2 & 2\\
\end{bmatrix}\).
a) 3
b) 2
c) 1
d) 0
View Answer

Answer: a
Explanation: To find out the rank of the matrix first find the |A|
If the value of the|A| = 0 then the matrix is said to be reduced
But, as the determinant of A has some finite value then, the rank of the matrix is 3.

7. The rank of the matrix (m × n) where m<n cannot be more than?
a) m
b) n
c) m*n
d) m-n
View Answer

Answer: a
Explanation: let us consider a 2×3 matrix \(\begin{bmatrix}
1 & 1 & 1\\
4 & 5 & 6\\
\end{bmatrix}\)
Where R1≠R2 rank is 2
Another 2×3 matrix \(\begin{bmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
\end{bmatrix}\)
Here, R1=R2 rank is 1
And the rank of these two matrices is 1, 2
So rank is cannot be more than m.
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8. Given A=\(\begin{bmatrix}
2 & -0.1 \\
0 & 3 \\
\end{bmatrix} A^{-1} = \begin{bmatrix}
1/2 & a \\
0 & b \\
\end{bmatrix}\) then find a + b.
a) \(\frac{6}{20}\)
b) \(\frac{7}{20}\)
c) \(\frac{8}{20}\)
d) \(\frac{5}{20}\)
View Answer

Answer: b
Explanation: AA-1 = I = \(\begin{bmatrix}
1 & 2-0.1b \\
0 & 3b \\
\end{bmatrix}=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}\)
Therefore, a = \(\frac{1}{60}\) and b = \(\frac{1}{3}\) and a + b = \(\frac{7}{20}\).

9. If a square matrix B is skew symmetric then.
a) BT = -B
b) BT = B
c) B-1 = B
d) B-1 = BT
View Answer

Answer: a
Explanation: The transpose of a skew symmetric matrix should be equal to the negative of the matrix
Example: let us consider a matrix B = \(\begin{bmatrix}
a & e & d\\
-e & b & f\\
-d & -f & c\\
\end{bmatrix}\), BT = \(\begin{bmatrix}
a & -e & -d\\
e & b & -f\\
d & f & c\\
\end{bmatrix}\).
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10. For the following set of simultaneous equations 1.5x-0.5y=2, 4x+2y+3z=9, 7x+y+5=10.
a) The solution is unique
b) Infinitely many solutions exist
c) The equations are incompatible
d) Finite number of multiple solutions exist
View Answer

Answer: a
Explanation: The equations can be written as \(\begin{bmatrix}
1.5 & -0.5 & 0\\
4 & 2 & 3\\
7 & 1 & 5\\
\end{bmatrix}\)
It can also be written as A = \(\begin{bmatrix}
3 & -2 & 0\\
4 & 2 & 3\\
7 & 1 & 5\\
\end{bmatrix}\), |A|=19
Hence, it has a unique solution.

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Manish Bhojasia - Founder & CTO at Sanfoundry
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 | Youtube | Instagram | Facebook | Twitter