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

Matrix Multiplication and Properties:

• Matrix multiplication involves the dot product of rows and columns.
• A null matrix is a matrix in which all elements are zero.
• An identity matrix is a square matrix with 1's on the diagonal and 0's elsewhere.
• For matrices P and Q, PQ = QP does not generally hold unless P and Q commute.

Matrix Definitions:

• Null Matrix: A matrix where all elements are zero.
• Identity Matrix: A square matrix with 1's on the main diagonal and 0's elsewhere.

Calculation:

\(\rm P=\begin{bmatrix}0&c&-b\\\ -c&0&a\\\ b&-a&0\end{bmatrix}\ and\ \rm Q=\begin{bmatrix}a^2&ab&ac\\\ ab&b^2&bc\\\ ac&bc&c^2\end{bmatrix}\)

⇒ PQ = \(=\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}\ \)

⇒QP = \(=\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}\ \)

Then PQ = QP

∴ Option (c) is correct

Concept:

Matrix Multiplication:

• Matrix multiplication is defined only when the number of columns in the first matrix is equal to the number of rows in the second matrix.
• For two matrices \( A_{m \times n} \) and \( B_{n \times p} \), their product \( AB \) will result in a matrix \( C_{m \times p} \).
• In general, the product of \( X \) and \( Y \) is defined if and only if the number of columns in \( X \) equals the number of rows in \( Y \).

Calculation:

We have the following matrix operations:

\( A_{m \times n} B_{n \times p} = (AB)_{m \times p} \)

Now, consider the matrix multiplications:

\([Z_{3 \times 2} . Y_{2\times3}].X_{3 \times 3}] = [ZYX]_{3 \times 3}\)

\( Y_{2 \times 3} [X_{3 \times 3} Z_{3 \times 2}] = [YXZ]_{2 \times 2} \)

Finally, we observe that:

\( X_{3 \times 3} [Y_{2 \times 3} Z_{3 \times 2}] = X_{3 \times 3} [YZ]_{2 \times 2} \)

Conclusion:

\(\text{No. of columns in X} \neq \text{No. of rows in (YZ)}\)

Hence, X(YZ) is not defined.

∴ the Correct answer is Option 4

Calculation:

Given, \(\rm A=\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}\) and \(\rm B=\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\)

∴ A + B = \(\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}+\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\) = \(\begin{bmatrix}i&-i\\\ -i&i\end{bmatrix}\)

A – B = \(\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}-\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\) = \(\begin{bmatrix}i&i\\\ i&i\end{bmatrix}\)

∴ (A + B) (A – B) = \(\begin{bmatrix}i&-i\\\ -i&i\end{bmatrix}\begin{bmatrix}i&i\\\ i&i\end{bmatrix}\) = \(\begin{bmatrix}i^2-i^2&i^2-i^2\\\ -i^2+i^2&-i^2+i^2\end{bmatrix}\) = \(\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}\)

Now, A² = \(\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}\) = \(\begin{bmatrix}i^2&0\\\ 0&i^2\end{bmatrix}\) = \(\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\)

B2 = \(\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\) = \(\begin{bmatrix}i^2&0\\\ 0&i^2\end{bmatrix}\) = \(\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\)

⇒ A2 - B2 = \(\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}-\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\) = \(\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}\)

⇒ (A + B) (A – B) = A2 - B2 

∴ The value of (A + B) (A – B) is A2 - B2 

The correct answer is Option 1. 

Concept:

• We are given a matrix A such that A² = A. This means A is an idempotent matrix.
• We are asked to evaluate: A(I − 2A)³ + 2A³
• We will use the identity A² = A to simplify higher powers like A³.

Calculation:

Step 1: Use A² = A ⇒ A³ = A·A² = A·A = A

So, A³ = A

Now simplify the expression: A(I − 2A)³ + 2A³

Step 2: Expand (I − 2A)³ using binomial expansion:

(I − 2A)³ = I − 3(2A) + 3(2A)² − (2A)³

= I − 6A + 12A² − 8A³

Now substitute A² = A and A³ = A:

(I − 2A)³ = I − 6A + 12A − 8A = I − 2A

Step 3: Multiply A with the simplified expression

A(I − 2A)³ = A(I − 2A) = A − 2A²

Since A² = A ⇒ A − 2A = −A

Step 4: Add 2A³

A(I − 2A)³ + 2A³ = −A + 2A = A

∴ The correct answer is: (4) A

Calculation:

Given, \(\rm A=\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}\) and \(\rm B=\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\)

∴ A + B = \(\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}+\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\) = \(\begin{bmatrix}i&-i\\\ -i&i\end{bmatrix}\)

A – B = \(\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}-\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\) = \(\begin{bmatrix}i&i\\\ i&i\end{bmatrix}\)

∴ (A + B) (A – B) = \(\begin{bmatrix}i&-i\\\ -i&i\end{bmatrix}\begin{bmatrix}i&i\\\ i&i\end{bmatrix}\) = \(\begin{bmatrix}i^2-i^2&i^2-i^2\\\ -i^2+i^2&-i^2+i^2\end{bmatrix}\) = \(\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}\)

Now, A² = \(\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}\begin{bmatrix}i&0\\\ 0&i\end{bmatrix}\) = \(\begin{bmatrix}i^2&0\\\ 0&i^2\end{bmatrix}\) = \(\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\)

B2 = \(\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\begin{bmatrix}0&-i\\\ -i&0\end{bmatrix}\) = \(\begin{bmatrix}i^2&0\\\ 0&i^2\end{bmatrix}\) = \(\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\)

⇒ A2 - B2 = \(\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}-\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\) = \(\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}\)

⇒ (A + B) (A – B) = A2 - B2 

∴ The value of (A + B) (A – B) is A2 - B2 

The correct answer is Option 1. 

Calculation:

Given: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\)

\({{\rm{A}}^2} = {\rm{AA}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\)

\(\Rightarrow {{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {0 + 1}&{0 + 0}\\ {0 + 0}&{1 + 0} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

Now,

\(\Rightarrow {{\rm{A}}^4} = {{\rm{A}}^2}{{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

Hence Option 1^st is correct answer.

Concept:

Matrix Multiplication:

Multiplication is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix.

A m×n matrix multiplied by a n×p matrix results in a m×p matrix.

Matrices are multiplied by multiplying each element of a row of the first m×n matrix with the corresponding elements of all the columns of the second n×p matrix to obtain the first row of the product matrix with p columns, and so on for all the m rows of the first matrix.

Calculation:

\(\rm \begin{bmatrix} \rm 2x & 3 \end{bmatrix}\begin{bmatrix} \ \ 1 & 2 \\ -3 & 0 \end{bmatrix}\) = [2x - 9   4x + 0]

= [2x - 9   4x]

∴ \(\rm \begin{bmatrix} \rm 2x & 3 \end{bmatrix}\begin{bmatrix} \ \ 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} \rm x \\ 8 \end{bmatrix}=0\)

\(\rm \Rightarrow \begin{bmatrix}\rm 2x-9 & \rm 4x\end{bmatrix}\begin{bmatrix}\rm x \\ 8\end{bmatrix}\) = 0

⇒ [(2x - 9)x + 8×4x] = 0

⇒ [2x² - 9x + 32x] = 0

⇒ 2x² + 23x = 0

⇒ x(2x + 23) = 0

⇒ x = 0 or \(\rm -\dfrac{23}{2}\).

Concept:

The associative property of matrix is given by:

X (YZ) = (XY) Z      ----(1)

Given:

AB = B and BA = A      ----(2)

Calculation:

A2 + B2

⇒ AA + BB

⇒ A (BA) + B (AB)      [using (2)]

⇒ (AB) A + (BA) B      [using (1)]

⇒ BA + AB

⇒ A + B

Hence, A2 + B2 = A + B.

Concept:

Scalar Multiplication:

If any matrix is multiplied by a scalar k ∈ R, then each element of the matrix is multiplied by k i.e If A = [a_ij]_{m × n} then k × A = A = [k × a_ij]_{m × n}

Multiplication of Matrices:

If A and B are two matrices such that the no. of columns of A is equal to the no. of rows of B. If A = [a_ij] is a m × n matrix and B = [b_ij] be a n × p matrix, then the product AB is the resultant matrix of order m × p and is defined as:

\({\left( {AB} \right)_{ij}} = \;\mathop \sum \limits_{k = 1}^n {a_{ik}} \times {b_{kj}}\forall \;i = 1,\;2, \ldots ,m\;and\;j = 1,\;2,\; \ldots .,\;p\)

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 2&{ - \;2}\\ { - \;3}&4 \end{array}} \right]\)

Here, we have to find the value of the expression: (\(\rm -\) A2 + 6 A)

\(\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {10}&{ - \;12}\\ { - \;18}&{22} \end{array}} \right]\) and \(6 \cdot A = \left[ {\begin{array}{*{20}{c}} {12}&{ - \;12}\\ { - \;18}&{24} \end{array}} \right]\)

\(\Rightarrow - \;{A^2} + 6 \cdot A = \left[ {\begin{array}{*{20}{c}} { - \;10}&{12}\\ {18}&{ - \;22} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {12}&{ - \;12}\\ { - \;18}&{24} \end{array}} \right]\)

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&2 \end{array}} \right]\)

Hence, \(\rm (- A^2 + 6 \cdot A)= \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&2 \end{array}} \right]\)

Calculation:

We have to find the value of \(\left[ {{\rm{x\;\;y\;\;z}}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\)

\( \Rightarrow \left[ {{\rm{x\;\;y\;\;z}}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\)

= [ax + hy + gz    hx + by + fz    gx + fy + cz]

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:

Given: AB = C

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

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

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

So, 3x + y = 4   …. (1)

x + 2y = -2       …. (2)

Solving equation 1 and 2, we get

x = 2 and y = -2

Now,

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{\rm{x}} + {\rm{y}}}&{\rm{y}}\\ {\rm{x}}&{{\rm{x}} - {\rm{y}}} \end{array}} \right] \)

\(= {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {2 - 2}&{ - 2}\\ 2&{2 - \left( { - 2} \right)} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 2&4 \end{array}} \right]\)

\({{\rm{A}}^2} = {\rm{A}} \times {\rm{A}} \)

\(= {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 2&4 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 2&4 \end{array}} \right]\)

\( \Rightarrow {{\rm{A}}^2} = \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 8}\\ 8&{12} \end{array}} \right]\)

Calculation:

Given: 

\(\rm A = \left[ {\begin{array}{*{20}{c}} 1&\rm a\\ 0&1 \end{array}} \right]\)

\(\rm {A^2} = A.A = \left[ {\begin{array}{*{20}{c}} 1&\rm a\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1& \rm a\\ 0&1 \end{array}} \right]\)

=\( \left[ {\begin{array}{*{20}{c}} {1 + 0}&{\rm a + a}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&2\rm a\\ 0&1 \end{array}} \right]\)

\(\rm{A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}} 1&2\rm a\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&\rm a\\ 0&1 \end{array}} \right]\)

=\( \left[ {\begin{array}{*{20}{c}} 1&{\rm 2a + 1a}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&3\rm a\\ 0&1 \end{array}} \right]\)

Seeing the pattern here

Concept:

The product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix.

Calculations:

Given \(A = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]_{3\times 3}\) and \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]_{3\times1}\)

The product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix.

i.e AB is posible and its order is \(\rm 3\times 1\)

Now, To obtain the entry in row 1, column 1 of $\displaystyle AB,\text{}$ multiply the first row in $\displaystyle A$ by the first column in $\displaystyle B$, and add.

To obtain the entry in row 2, column 1 of $\displaystyle AB,\text{}$ multiply the second row in $\displaystyle A$ by the first column in $\displaystyle B$, and add.

To obtain the entry in row 3, column 1 of $\displaystyle AB,\text{}$ multiply the third row in $\displaystyle A$ by the first column in $\displaystyle B$, and add.

\(AB = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]_{3\times 3}​​\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]_{3\times1}\)

\(AB = \left[ {\begin{array}{*{20}{c}} {{\rm{ax}} + {\rm{hy}} + {\rm{gz}}}\\ {{\rm{hx}} + {\rm{by}} + {\rm{fz}}}\\ {{\rm{gx}} + {\rm{fy}} + {\rm{cz}}} \end{array}} \right]_{3\times1}\)

Concept:

Matrix Multiplication:

• Multiplication is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix.
• A m × n matrix multiplied by a n × p matrix results in a m × p matrix.
• Matrices are multiplied by multiplying each element of a row of the first m × n matrix with the corresponding elements of all the columns of the second n × p matrix to obtain the first row of the product matrix with p columns, and so on for all the m rows of the first matrix.
• For two matrices A and B, AB ≠ BA.

Calculation:

For two matrices A and B, we have:

(A + B)2 = (A + B)(A + B) = A2 + AB + BA + B2,

Since it is given that (A + B)2 = A2 + B2, we must have:

AB + BA = 0

⇒ \(\begin{bmatrix}1 &\ \ \ 2\\1&-1\end{bmatrix}\times\begin{bmatrix}\ \ \ \rm a &\rm b\\-1&1\end{bmatrix}+\begin{bmatrix}\ \ \ \rm a &\rm b\\-1&1\end{bmatrix}\times \begin{bmatrix}1 &\ \ \ 2\\1&-1\end{bmatrix}=\begin{bmatrix}0 &0\\0&0\end{bmatrix}\)

⇒ \(\begin{bmatrix}\rm a-2 &\rm b+2\\\rm a +1&\rm b-1\end{bmatrix}+\begin{bmatrix}\rm a+b &\rm 2a-b\\-1+1&-2-1\end{bmatrix}=\begin{bmatrix}0 &0\\0&0\end{bmatrix}\)

⇒ \(\begin{bmatrix}\rm 2a+b-2 &\rm 2a+2\\\rm a +1&\rm b-4\end{bmatrix}=\begin{bmatrix}0 &0\\0&0\end{bmatrix}\)

Comparing the elements of equal matrices, we get:

a + 1 = 0 and b - 4 = 0.

⇒ a = -1 and b = 4.

Concept:

• cos² θ – sin² θ = cos 2θ
• 2 sin θ cos θ = sin 2θ
• cos (x + y) = cos x cos y – sin x sin y
• sin (x + y) = sin x cos y + cos x sin y

Calculation:

Given:

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

\({{\rm{A}}^2} = {\rm{AA}} = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

\( \Rightarrow {{\rm{A}}^2} = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}{\rm{\theta }} - {\rm{\;}}{{\sin }^2}{\rm{\theta }}}&{\cos {\rm{\theta }}\sin {\rm{\theta }} + {\rm{\;}}\sin {\rm{\theta }}\cos {\rm{\theta }}}\\ { - \cos {\rm{\theta }}\sin {\rm{\theta }} - {\rm{\;}}\cos {\rm{\theta }}\sin {\rm{\theta }}}&{ - {{\sin }^2}{\rm{\theta }} + {\rm{\;}}{{\cos }^2}{\rm{\theta }}} \end{array}} \right]\)

\( \Rightarrow {{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}}&{\sin 2{\rm{\theta }}}\\ { - \sin 2{\rm{\theta }}}&{\cos 2{\rm{\theta }}} \end{array}} \right]\)

Now,

\(\Rightarrow {{\rm{A}}^3} = {{\rm{A}}^2}{\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}}&{\sin 2{\rm{\theta }}}\\ { - \sin 2{\rm{\theta }}}&{\cos 2{\rm{\theta }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

\(\Rightarrow {{\rm{A}}^3} = \left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}\cos {\rm{\theta }} - \sin 2{\rm{\theta }}\sin {\rm{\theta \;}}}&{\cos 2{\rm{\theta }}\sin {\rm{\theta }} + \sin 2{\rm{\theta }}\cos {\rm{\theta }}}\\ { - \sin 2{\rm{\theta }}\cos {\rm{\theta }} - {\rm{\;}}\cos 2{\rm{\theta }}\sin {\rm{\theta }}}&{ - \sin 2{\rm{\theta }}\sin {\rm{\theta }} + \cos 2{\rm{\theta }}\cos {\rm{\theta }}} \end{array}} \right]{\rm{\;}}\)

\(\Rightarrow {{\rm{A}}^3} = \left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}\cos {\rm{\theta }} - \sin 2{\rm{\theta }}\sin {\rm{\theta \;}}}&{\cos 2{\rm{\theta }}\sin {\rm{\theta }} + \sin 2{\rm{\theta }}\cos {\rm{\theta }}}\\ { - \left( {\cos 2{\rm{\theta }}\sin {\rm{\theta }} + \sin 2{\rm{\theta }}\cos {\rm{\theta }}} \right)}&{\cos 2{\rm{\theta }}\cos {\rm{\theta }} - \sin 2{\rm{\theta }}\sin {\rm{\theta }}} \end{array}} \right]\)

\(\Rightarrow {{\rm{A}}^3} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos (2{\rm{\theta }} + {\rm{\theta }}){\rm{\;}}}&{\sin (2{\rm{\theta }} + {\rm{\theta }}){\rm{\;}}}\\ { - \sin (2{\rm{\theta }} + {\rm{\theta }})}&{\cos (2{\rm{\theta }} + {\rm{\theta }}){\rm{\;}}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos 3{\rm{\theta }}}&{\sin 3{\rm{\theta }}}\\ { - \sin 3{\rm{\theta }}}&{\cos 3{\rm{\theta }}} \end{array}} \right]{\rm{\;}}\)

∴ Option 1^st is correct answer.