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

Calculation:

Given,

\(A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\)

$A^2$ \(\Rightarrow A^2 = A \times A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\)

 

\(\Rightarrow A^2 = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}\)

Now 4A $4A$

\(\Rightarrow 4A = 4 \times \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\)

\(\Rightarrow 4A = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}\)

Also A² - 4A $A^2 - 4A$

\(\Rightarrow A^2 - 4A = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}\)

\(\Rightarrow A^2 - 4A = \begin{bmatrix} 9-4 & 8-8 & 8-8 \\ 8-8 & 9-4 & 8-8 \\ 8-8 & 8-8 & 9-4 \end{bmatrix}\)

\(\Rightarrow A^2 - 4A = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}\)

Relate the result to the identity matrix I3" id="MathJax-Element-594-Frame" role="presentation" style="position: relative;" tabindex="0">I3I3" id="MathJax-Element-39-Frame" role="presentation" style="position: relative;" tabindex="0">I3I3 I3 $I_3$

\(\Rightarrow A^2 - 4A = 5I_3\)

Hence, the correct answer is option 4.

Concept:

Rotation Matrix:

• A rotation matrix is used to perform a rotation in a Euclidean space. It is a square matrix that describes the rotation of a vector space.
• For a 2D rotation, the matrix is given by: \(f(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\)
• Here, θ is the angle of rotation in radians.
  • cos θ: Represents the cosine of the rotation angle.
  • sin θ: Represents the sine of the rotation angle.
• Key property of a rotation matrix:
  • The transpose of the matrix is equal to its inverse.
  • The determinant of the matrix is always equal to 1.
• When θ = π, the rotation matrix becomes: \(f(\pi) = \begin{bmatrix} \cos \pi & \sin \pi \\ -\sin \pi & \cos \pi \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\)

 

Calculation:

Given,

Rotation matrix at θ = π:

\(f(\pi) = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\)

To find (f(π))², multiply the matrix by itself:

\(f(\pi) × f(\pi) = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} × \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\)

Using matrix multiplication:

Top-left element: (-1)(-1) + (0)(0) = 1

Top-right element: (-1)(0) + (0)(-1) = 0

Bottom-left element: (0)(-1) + (-1)(0) = 0

Bottom-right element: (0)(0) + (-1)(-1) = 1

Resulting matrix:

\(f(\pi)^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

∴ (f(π))² is equal to the identity matrix, which is \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).

Hence, the correct answer is Option 4.

Calculation:

Given,

The matrix A is:

\( A = \begin{bmatrix} y & x & x \\ z & x & y \\ x & y & z \end{bmatrix} \)

Since A  is an orthogonal matrix, we know that:

\( A^T = A^{-1} \quad \Rightarrow \quad A^T A = I \)

This property tells us that A  is orthogonal, and it implies that \(A^T A \) (the product of A's transpose and A is equal to the identity matrix I , which is:

\( A^T A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Now, let’s calculate \(A^T A \) step by step. The transpose of matrix A , denoted \(A^T \) is:

\( A^T = \begin{bmatrix} y & z & x \\ x & x & y \\ x & y & z \end{bmatrix} \)

Now, we perform matrix multiplication between \(A^T\) and A:

\( A^T A = \begin{bmatrix} y & z & x \\ x & x & y \\ x & y & z \end{bmatrix} \begin{bmatrix} y & x & x \\ z & x & y \\ x & y & z \end{bmatrix} \)

Performing this multiplication, we get the following matrix:

\( A^T A = \begin{bmatrix} y^2 + z^2 + x^2 & xy + zx + xy & xz + yz + x^2 \\ xy + zx + xy & x^2 + x^2 + y^2 & xy + xz + yz \\ xz + yz + x^2 & xy + xz + yz & x^2 + y^2 + z^2 \end{bmatrix} \)

This matrix must be equal to the identity matrix I , which is:

\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

By comparing the elements of the matrices, we get the following system of equations:

1. \( y^2 + z^2 + x^2 = 1 \) 2. \(xy + zx + xy = 0 \) 3. \(xz + yz + x^2 = 1 \)

Thus, the key result from the orthogonality condition is:

\( x^2 + y^2 + z^2 = 1 \)

Hence, the correct answer is Option 2. 

Calculation:

Multiply Matrix 1 and Matrix 2:

 

\(\begin{bmatrix} x & 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} x+4+7 & 2x+5+8 & 3x+6+9 \end{bmatrix}\)

⇒ \(\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix}\)

Multiply the resulting matrix with Matrix 3:

\(\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix} \times \begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix} \)

\(= (x+11) \cdot 1 + (2x+13) \cdot 1 + (3x+15) \cdot x\)

⇒ \((x+11) + (2x+13) + (3x^2+15x)\)

⇒ \((3x^2 + 18x + 24)\)

Equate the result to 45:

\((3x^2 + 18x + 24 = 45)\)

⇒ \((3x^2 + 18x - 21 = 0)\)

\((x^2 + 6x - 7 = 0)\)

 

\((x^2 + 7x - x - 7 = 0)\)

 

⇒ \((x = 1 \text{ or } x = -7)\)

Step 5: Verify:

For \(x = 1\), substitute back:

\((3(1)^2 + 18(1) + 24 = 45)\)

45 =45

∴ The correct value of x is 1.

Hence, the correct answer is Option 4.

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

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:

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.

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:

Orthogonal matrix: When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

AA^T = I

Calculation:

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

Then according to the definition;

AA^T = I

Pre multiplication by A^-1

A^-1 AA^T = A^-1 I

IA^T = A^-1

A^T = A^-1 or A’ = A^-1
then A is orthogonal matrix.

∴ Option 2 is correct

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 \(\rm -\dfrac{23}{2}\).

Calculation:

Given:

x + 2y = \(\begin{bmatrix} 2 & -3\\ 1 & 5 \end{bmatrix}\)                    .... (1)

2x + 5y = \(\begin{bmatrix} 7 & 5\\ 2 & 3 \end{bmatrix}\)                     .... (2)

Multiplying by 2 in the equation (1), we get

⇒ 2x + 4y = \(\begin{bmatrix} 4 & -6\\ 2 & 10 \end{bmatrix}\)             .... (3)

Subtracting equation (3) from equation (2), we get

⇒ (2x + 5y) - (2x + 4y) = \(\begin{bmatrix} 7 & 5\\ 2 & 3 \end{bmatrix}-\begin{bmatrix} 4 & -6\\ 2 & 10 \end{bmatrix}\)

∴ y = \( \begin{bmatrix} 3 & 11\\ 0 & -7 \end{bmatrix}\)

 

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:

If two matrices A and B are said to be equal if the following conditions hold true:

• Order of matrix A = Order of matrix B
• Corresponding element of matrix A = Corresponding element of matrix B

 

Calculation:

Given: \(\begin{bmatrix} \rm 2x & 5\\ 7 & \rm -y \end{bmatrix} = \begin{bmatrix} 8 & 5\\7 & 3 \end{bmatrix}\)

As we know that, if two matrices A and B are equal then their corresponding elements are also the same.

⇒ 2x = 8 

∴ x = 4

Now,

⇒ -y = 3

∴ y = -3

We have to find the value of x + y

So,  x + y = 4 - 3 = 1

Hence,  option 2 is the correct answer.