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.

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=\(\begin{bmatrix}
31 & 39 & 80\\
39 & 45 & 74\\
80 & 74 & 136\\
\end{bmatrix}\).

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.

Sanfoundry Global Education & Learning Series – Signals & Systems.

To practice all areas of Signals & Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.

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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 lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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