Matrices MCQ Quiz - Objective Question with Answer for Matrices - Download Free PDF

Concept:

A × A-1 = I, where I is an identity matrix

|A| = \(\rm 1\over {|A^{-1}|}\)

Calculation:

Given: \(\rm A=\begin{bmatrix} x & 2 \\\ 4 & 3 \end{bmatrix}\) and \(\rm A ^{-1}=\begin{bmatrix} {1\over8} & {-1\over 12} \\\ {-1\over 6}& {4\over 9} \end{bmatrix}\)

|A^-1| = \(\rm {4\over 72} - {1\over 72} = {3\over 72} = {1\over 24}\)

|A| = \(\rm {1 \over {|A^{-1}|}}\) = 24

⇒ 3x - 8 = 24

∴ x = \(\rm 32\over 3\)

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 \[ \text{det}(A) = a \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} - (-1) \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix} + (-3) \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} \] \[ \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2 \cdot 2 - 3 \cdot 1) = 4 - 3 = 1 \] \[ \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix} = (3 \cdot 2 - 3 \cdot 2) = 6 - 6 = 0 \] \[ \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = (3 \cdot 1 - 2 \cdot 2) = 3 - 4 = -1 \] \] \[ \text{det}(A) = a \cdot 1 - (-1) \cdot 0 + (-3) \cdot (-1) = a + 0 + 3 = a + 3 \] \] \[ a + 3 = 0 \implies a = -3 \] ```

Calculation

Given

AA^T = I = A^TA

\(\frac{1}{2}\)A \(\left[(A+A^T)^2+(A-A^T)^2\right]\) 

⇒ \(\frac{1}{2}\mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2\mathrm{AA^T}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2\mathrm{AA^T}\right]\)

⇒ A[A² + (A^T)²] = A³ + A^T

Hence option 4 is correct

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.

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: \(A = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&3\\ 3&4 \end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&1 \end{array}} \right)\)

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.

Concept:

Symmetric Matrix:

• Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A itself
• A^T = A or A’ = A

Where, A^T or A’ denotes the transpose of matrix

• A square matrix A is said to be symmetric if a_ij = a_ji for all i and j

Where a_ij and a_ji is an element present in matrix.

 

Calculation:

Given:

A is a symmetric matrix,

⇒ A^T = A or a_ij = a_ji

A = \(\left[ \begin{matrix} 2 & x-3 & x-2 \\ 3 & -2 & -1 \\ 4 & -1 & -5 \\ \end{matrix} \right]\)

So, by property of symmetric matrices

⇒ a₁₂ = a₂₁

⇒ x – 3 = 3

∴ x = 6

Concept:

• 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 A^T is 3 × 4 and order of C^T is 3 × 7

Now,

A^TB = {3 × 4} {4 × 5} = 3 × 5

⇒ Order of A^TB is 3 × 5

Hence order of (A^TB) ^T is 5 × 3

Now order of (A^TB) ^T C ^T = {5 × 3} {3 × 7} = 5 × 7

∴ Order of (A^TB) ^T C ^T is 5 × 7

Concept:

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 = A^T
• a_ij = a_ji

 

CALCULATION:

Given - \({\rm{A}} = \left( {\begin{array}{*{20}{c}} 4&{{\rm{x}} + 2}\\ {2{\rm{x}} - 3}&{{\rm{x}} + 1} \end{array}} \right)\)

A real square matrix A = (a_ij) is said to be symmetric, if A = A^T

Where A^T = transpose of matrix A

\({{\rm{A}}^{\rm{T}}} = \left( {\begin{array}{*{20}{c}} 4&{2{\rm{x}} - 3}\\ {{\rm{x}} + 2}&{{\rm{x}} + 1} \end{array}} \right)\)

∴ A = A^T

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} 4&{{\rm{x}} + 2}\\ {2{\rm{x}} - 3}&{{\rm{x}} + 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4&{2{\rm{x}} - 3}\\ {{\rm{x}} + 2}&{{\rm{x}} + 1} \end{array}} \right]\)

Compare A₂₁ element.

⇒ x + 2 =2x - 3 

⇒ x = 5

Concept:

For an invertible matrix A:

• A^-1 = \(\rm \frac{adj(A)}{|A|}\).
• |A-1| = |A|^-1 = \(\rm \frac{1}{|A|}\).

 

Calculation:

\(\rm A^{-1}=\begin{bmatrix} 1& 2& 3\\ 2& 4& 3\\ 3& 1& 6\end{bmatrix}=\frac{adj(A)}{k}\)         -----(1)

From the definition of the inverse of a matrix, 

A-1 = \(\rm \frac{adj(A)}{|A|}\)              -----(2)

Comparing equation (1) & (2), we get

k = |A|  

Using the properties of the determinant of inverse of a matrix, we have:

k = |A| = \(\rm \frac{1}{|A^{-1}|}\)        ----(3)

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 = \(\rm - \frac1{15}\).

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|

Concept:

A × A-1 = I, where I is an identity matrix

|A| = \(\rm 1\over {|A^{-1}|}\)

Calculation:

Given: \(\rm A=\begin{bmatrix} x & 2 \\\ 4 & 3 \end{bmatrix}\) and \(\rm A ^{-1}=\begin{bmatrix} {1\over8} & {-1\over 12} \\\ {-1\over 6}& {4\over 9} \end{bmatrix}\)

|A^-1| = \(\rm {4\over 72} - {1\over 72} = {3\over 72} = {1\over 24}\)

|A| = \(\rm {1 \over {|A^{-1}|}}\) = 24

⇒ 3x - 8 = 24

∴ x = \(\rm 32\over 3\)

Concept:

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. 

\(\rm tr\left( A \right) = \mathop \sum \limits_{n = 1}^n {A_{nn}}\)

Calculation:

Given: \(A=\begin{bmatrix} -1 & 4 \\\ 5 & 8 \end{bmatrix}\)

Trace of matrix = sum of elements on the main diagonal

= -1 + 8

= 7

Concept:

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∶

\(A\; = \;\left[ {\begin{array}{*{20}{c}} n&9\\ 2&1 \end{array}} \right] , B\; = \;\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

We cannot find anything from this statement.

Combining statement 1 and 2∶

\(AB\; = \;\left[ {\begin{array}{*{20}{c}} (n\times 1+9\times0)&(n\times0+9\times1)\\ (2\times1+1\times0)&(2\times0+1\times1) \end{array}} \right]\)

\(AB = \left[ {\begin{array}{*{20}{c}} n&9\\ 2&1 \end{array}} \right]\)

Also, \(A = \left[ {\begin{array}{*{20}{c}} n&9\\ 2&1 \end{array}} \right]\)

∴ We cannot find the value of n from both statements together.

Calculation:

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.

Concept:

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.

Concept:

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) = \(\rm \frac{1}{k}\;{A^{ - 1}}\) 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