250+ TOP MCQs on Basics of Linear Algebra and Answers

Answer: D
Clarification: 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.

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
Answer: A
Clarification: 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})
Answer: b
Clarification: The adjacency of A is given by AA^T
A^T = (begin{bmatrix}
1 & 5 & 6\
7 & 4 & 8\
-3 & -2 & -6\
end{bmatrix}),
AA^T = (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
Answer: A
Clarification: 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 mA. m
B. n
C. m*n
D. m-n
Answer: A
Clarification: 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.

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})
Answer: B
Clarification: 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. B^T = -B
B. B^T = B
C. B^-1 = B
D. B^-1 = B^T
Answer: A
Clarification: 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}), B^T = (begin{bmatrix}
a & -e & -d\
e & b & -f\
d & f & c\
end{bmatrix}).

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
Answer: A
Clarification: 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.