Matrices – CA Foundation Maths Study Material

This Matrices – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

Matrices – CA Foundation Maths Study Material

Previous Year Exam Questions (Memory Based)

Question 1.
The value of k for which the points (k, 1)., (5, 5) and (10, 7) may be collinear if: [1 Mark, May 2018]
(a) k = -5
(b) k = 7
(c) k = 9
(d) k = 1
Answer:
Formula
Points (x₁; y₁) ; (x₂; y₂) and (x₃; y₃) are collinear; if
\(\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\) = 0

⇒ \(\left|\begin{array}{ccc}
K & 1 & 1 \\
5 & 5 & 1 \\
10 & 7 & 1
\end{array}\right|\) = 0
or; K(5 – 7) – 1(5 – 10) + 1(35 – 50) = 0
or; -2K + 5 – 15 = 0 or K = -5
(a) is correct.

Question 2.
If |A| = 0, then A is: [1 Mark, May 2018]
(a) 0
(b) Uro matrix
(c) Singular matrix
(d) Non-Singular matrix
Answer:
(c)
If |A| = 0, Then,
Matrix – A is called Singular Matrix.

Question 3.
If A = \(\left[\begin{array}{cc}
2 i & i \\
i & -i
\end{array}\right]\) if i² = -1 then |A| = ?
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(b)
|A| = \(\left|\begin{array}{cc}
2 i & i \\
i & -i
\end{array}\right|\) = -2i² – i²
= -3i² = -3(-1) = 3

Question 4.
If A and B are matrices then which from the following is true ? [1 Mark, May 2018]
(«) A + B ≠B + A
(b) (A’)² ≠ A
(c) AB ≠ BA
(d) all are true
Answer:
(c)

Question 5.
Transpose of a rectangular matrix is a: [1 Mark, May 2018]
(a) Rectangular matrix
(b) Diagonal matrix
(c) Square matrix
(d) Scalar matrix
Answer:
(a)

Question 6.
If A = \(\left[\begin{array}{cc}
-5 & 2 \\
1 & -3
\end{array}\right]\), then adj. A is [1 Mark, Nov. 2018]
(a) \(\left[\begin{array}{ll}
-3 & -2 \\
-1 & -5
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
3 & -2 \\
-1 & 5
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
5 & 1 \\
2 & 3
\end{array}\right]\)
(d) \(\left[\begin{array}{ll}
3 & 2 \\
1 & 5
\end{array}\right]\)
Answer:
(a)

Question 7.
If A = \(\left[\begin{array}{ll}
5 & x \\
y & 0
\end{array}\right]\) and A = A^T, then [1 Mark, Nov. 2008]
(a) x = 0, y = 5
(b) x = y
(c) x + y =5
(d) None
Answer:
(b)
A = A^T
∴ \(\left[\begin{array}{ll}
5 & x \\
y & 0
\end{array}\right]=\left[\begin{array}{ll}
5 & y \\
x & 0
\end{array}\right]\)
⇒ x = y

Question 8.
Let A^T be the transpose of matrix A having order m × n, then (A^T. A) is a matrix of order: [1 Mark, Nov. 2018]
(a) m × m
(b) n × n
(c) m × n
(d) n × m
Answer:
(a)
Order of A^T is m × n
Order of A = n × m
Order of (A^T_{m × n} . A_{n × m})
= m × m

Question 9.
If \(\left(\begin{array}{cc}
x+y & 1 \\
1 & x-y
\end{array}\right)+\left(\begin{array}{cc}
2 & 3 \\
2 & -4
\end{array}\right)=\left(\begin{array}{cc}
12 & 4 \\
3 & 0
\end{array}\right)\) then [1 Mark, June 2019]
(a) x = 7, y = -3
(b) x = -7, y = -3
(c) x = -7, y = 3
(d) x = 7, y = 3
Answer:
(d)
TRICKS:
Go by Choices
(d) satisfies it

Question 10.
[1 2 3]\(\left[\begin{array}{l}
\log _{10} 2 \\
\log _{10} 3 \\
\log _{10} 4
\end{array}\right]\) = . [1 Mark, June 2019]
(a) log₁₀(1521)
(b) log₁₀(1152)
(c) log₁₀(5211)
(d) log₁₀(2151)
Answer:
(b)
[1 2 3]\(\left[\begin{array}{l}
\log _{10} 2 \\
\log _{10} 3 \\
\log _{10} 4
\end{array}\right]\)
1.log₁₀ 2 + 2log₁₀3 + 3log₁₀4
= log₁₀2 + log₁₀3² +log₁₀ 4³
= l0g₁₀ (2.9.64) = log₁₀1152