# Signals & Systems Questions and Answers – Basics of Linear Algebra

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.
advertisement
advertisement

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.
advertisement

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 = $$\begin{bmatrix} a & -e & -d\\ e & b & -f\\ d & f & c\\ \end{bmatrix}$$.
advertisement

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.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
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.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.