Matrices MCQ Quiz - Objective Question with Answer for Matrices - Download Free PDF
Last updated on Apr 22, 2025
Latest Matrices MCQ Objective Questions
Matrices Question 1:
If
Answer (Detailed Solution Below)
Matrices Question 1 Detailed Solution
Concept:
A × A-1 = I, where I is an identity matrix
|A| =
Calculation:
Given:
|A-1| =
|A| =
⇒ 3x - 8 = 24
∴ x =
Matrices Question 2:
Find the values of "a", if the given matrix is singular
Answer (Detailed Solution Below)
Matrices Question 2 Detailed Solution
To determine the values of \( a \) that make the given matrix singular, we need to find the determinant of the matrix and set it to zero. A matrix is singular if and only if its determinant is zero. Given the matrix: \[ A = \begin{bmatrix} a & -1 & -3 \\ 3 & 2 & 3 \\ 2 & 1 & 2 \end{bmatrix} \] The determinant of matrix \( A \) can be computed using the formula for the determinant of a 3x3 matrix: \[ \text{det}(A) = a \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} - (-1) \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix} + (-3) \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2 \cdot 2 - 3 \cdot 1) = 4 - 3 = 1 \] \[ \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix} = (3 \cdot 2 - 3 \cdot 2) = 6 - 6 = 0 \] \[ \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = (3 \cdot 1 - 2 \cdot 2) = 3 - 4 = -1 \] Substituting these values back into the determinant formula: \[ \text{det}(A) = a \cdot 1 - (-1) \cdot 0 + (-3) \cdot (-1) = a + 0 + 3 = a + 3 \] For the matrix to be singular, the determinant must be zero: \[ a + 3 = 0 \] Solving for \( a \): \[ a = -3 \] Thus, the value of \( a \) that makes the matrix singular is \( -3 \). Therefore, the correct option is option 1. Here is the LaTeX code for the determinant calculation: ```latex \[ \text{det}(A) = a \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} - (-1) \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix} + (-3) \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} \] \[ \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2 \cdot 2 - 3 \cdot 1) = 4 - 3 = 1 \] \[ \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix} = (3 \cdot 2 - 3 \cdot 2) = 6 - 6 = 0 \] \[ \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = (3 \cdot 1 - 2 \cdot 2) = 3 - 4 = -1 \] \] \[ \text{det}(A) = a \cdot 1 - (-1) \cdot 0 + (-3) \cdot (-1) = a + 0 + 3 = a + 3 \] \] \[ a + 3 = 0 \implies a = -3 \] ```
Matrices Question 3:
Let A be a square matrix such that AAT = I. Then
Answer (Detailed Solution Below)
Matrices Question 3 Detailed Solution
Calculation
Given
AAT = I = ATA
⇒
⇒ A[A2 + (AT)2] = A3 + AT
Hence option 4 is correct
Matrices Question 4:
If A and B are two non-zero n × n matrices such that A2 + B = A2 B, then
Answer (Detailed Solution Below)
Matrices Question 4 Detailed Solution
Concept:
For a matrix A, AI = IA = A, where I is the Identity matrix.
Calculation:
Given, A2 + B = A2 B ... (1)
⇒ A2 – A2 B + B = 0
⇒ A2 – A2 B + B – I = – I
⇒ A2 (I – B) – (I – B) = – I
⇒ (A2 – I) (I – B) = – I
⇒ (I – A2) (I – B) = I
⇒ (I – B) (I – A2) = I
⇒ I – B – A2 + BA2 = I
⇒ BA2 – B – A2 = 0
⇒ A2 + B = BA2 ...(2)
From (1) + (2),
⇒ A2 B = BA2
∴ A2B = BA2
The correct answer is Option 4.
Matrices Question 5:
If
Answer (Detailed Solution Below)
Matrices Question 5 Detailed Solution
Concept:
Let A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p. Similarly, the product BA will exist if and only if q = m.
Calculation:
Given:
So, order of the matrix A is 3 × 2 whereas the order of the matrix A is 2 × 2.
As we know that, if A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p
Here, n = 2 = p ⇒ The product AB exists.
As we know that, if A and B be any two matrices of order m × n and p × q respectively. Then the product BA exists if and only if q = m.
Here, q = 2 and m = 3 ⇒ m ≠ q ⇒ The product BA does not exists.
Top Matrices MCQ Objective Questions
If A =
Answer (Detailed Solution Below)
Matrices Question 6 Detailed Solution
Download Solution PDFConcept:
Symmetric Matrix:
- Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A itself
- AT = A or A’ = A
Where, AT or A’ denotes the transpose of matrix
- A square matrix A is said to be symmetric if aij = aji for all i and j
Where aij and aji is an element present in matrix.
Calculation:
Given:
A is a symmetric matrix,
⇒ AT = A or aij = aji
A =
So, by property of symmetric matrices
⇒ a12 = a21
⇒ x – 3 = 3
∴ x = 6
lf the order of A is 4 × 3, the order of B is 4 × 5 and the order of C is 7 × 3, then the order of (ATB)T C T is
Answer (Detailed Solution Below)
Matrices Question 7 Detailed Solution
Download Solution PDFConcept:
- To multiply an m × n matrix by an n × p matrix, the n must be the same, and the result is an m × p matrix.
- If A be a matrix of order m × n than the order of transpose matrix is n × m
Calculation:
Given:
Order of A is 4 × 3, the order of B is 4 × 5 and the order of C is 7 × 3
The transpose of the matrix obtained by interchanging the rows and columns of the original matrix.
So, order of AT is 3 × 4 and order of CT is 3 × 7
Now,
ATB = {3 × 4} {4 × 5} = 3 × 5
⇒ Order of ATB is 3 × 5
Hence order of (ATB) T is 5 × 3
Now order of (ATB) T C T = {5 × 3} {3 × 7} = 5 × 7
∴ Order of (ATB) T C T is 5 × 7
If
Answer (Detailed Solution Below)
Matrices Question 8 Detailed Solution
Download Solution PDFConcept:
Symmetric Matrix: If the transpose of a matrix is equal to itself, that matrix is said to be symmetric.
Or, A matrix A is symmetric if and only if swapping indices doesn't change its components
- A = AT
- aij = aji
CALCULATION:
Given -
A real square matrix A = (aij) is said to be symmetric, if A = AT
Where AT = transpose of matrix A
∴ A = AT
Compare A21 element.
⇒ x + 2 =2x - 3
⇒ x = 5
If
Answer (Detailed Solution Below)
Matrices Question 9 Detailed Solution
Download Solution PDFConcept:
For an invertible matrix A:
- A-1 =
. - |A-1| = |A|-1 =
.
Calculation:
From the definition of the inverse of a matrix,
A-1 =
Comparing equation (1) & (2), we get
k = |A|
Using the properties of the determinant of inverse of a matrix, we have:
k = |A| =
We know,
A.A-1 = I
⇒ |A.A-1| = |I| = 1
⇒ |A| |A-1| = 1
⇒ |A| = 1/ |A-1| ....(4)
Now,
|A-1| = 1(24 - 3) + 2(9 - 12) + 3(2 - 12) = 21 - 6 - 30 = - 15.
|A-1| = -15
Therefore, from equation (3)
k =
Mistake PointsNote that, we have A-1 matrix, not an A matrix. So to find the value of k, don't you have to use relation |A| = 1/|A-1|
If
Answer (Detailed Solution Below)
Matrices Question 10 Detailed Solution
Download Solution PDFConcept:
A × A-1 = I, where I is an identity matrix
|A| =
Calculation:
Given:
|A-1| =
|A| =
⇒ 3x - 8 = 24
∴ x =
If
Answer (Detailed Solution Below)
Matrices Question 11 Detailed Solution
Download Solution PDFConcept:
Trace of a matrix:
Trace of a matrix is the sum of elements on the main diagonal.
The trace is only defined for a square matrix (n × n).
Let A be n × n matrix.
Calculation:
Given:
Trace of matrix = sum of elements on the main diagonal
= -1 + 8
= 7
Consider the following question and decide which of the statements is sufficient to answer the question.
Find the value of n, if
Statements∶
1. AB = A
2.
Answer (Detailed Solution Below)
Matrices Question 12 Detailed Solution
Download Solution PDFConcept:
Multiplication of matrices:
- The number of columns of the 1st matrix must equal the number of rows of the 2nd matrix.
- The result will have the same number of rows as the 1st matrix, and the same number of columns as the 2nd matrix.
- To multiply an m × n matrix by an n × p matrix, the n must be the same, and the result is an m × p matrix.
Calculation:
From statement 1∶
AB = A
We cannot find anything from this statement.
From statement 2∶
We cannot find anything from this statement.
Combining statement 1 and 2∶
Also,
∴ We cannot find the value of n from both statements together.
Each entry is the number of all possible matrices of 3 × 3 with 0 or 1, respectively.
Answer (Detailed Solution Below)
Matrices Question 13 Detailed Solution
Download Solution PDFCalculation:
As we know that
Number of possible entries of 3 × 3 = 9
And, every entry has two choices = 0 and 1
Now,
Total number of choices = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
⇒ 29
⇒ 512
∴ The total number of choices is 512.
If A is an Involuntary matrix and I is a unit matrix of same order, then (I − A) (I + A) is
Answer (Detailed Solution Below)
Matrices Question 14 Detailed Solution
Download Solution PDFConcept:
Involuntary matrix:
- Matrix A is said to be Involuntary if A2 = I, where I is an Identity matrix of same order as of A.
- Involuntary matrix is a matrix that is equal to its own inverse. ⇔ A-1 = A
Calculation:
Given that A is involuntary matrix,
⇒ A2 = I
Now,
(I − A) (I + A) = I2 – IA + AI − A2
⇒ I – A + A – I (∵ A2 = I)
⇒ 0
∴ (I − A) (I + A) is zero matrix.If A2 - 2A - I = 0,then inverse of A is
Answer (Detailed Solution Below)
Matrices Question 15 Detailed Solution
Download Solution PDFConcept:
Properties of Matrices Inverse:
If A and B are the non-singular matrices, then the inverse matrix should have the following properties
- (AB) - 1 = B - 1 A - 1
- (A - 1) - 1 = A
- (AT) - 1 = (A - 1)T
- (KA - 1) =
for any K ≠ 0 - (An) - 1 = (A - 1)n
- AA - 1 = A - 1A = I
Calculation:
Given: A2 - 2A - I = 0
⇒ A.A - 2A = I
Post multiply by A-1, we get
⇒ AAA-1 - 2AA-1 = IA-1
⇒ AI - 2I = A-1 [∵ AA - 1 = A - 1A = I]
∴ A-1 = A - 2
the inverse of A is A - 2