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

Calculation:

Given,

Matrix A is defined as:

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

It is mentioned that A is an orthogonal matrix. Therefore,

\(A^T A = I\)

Now, we calculate A²:

\(A^2 = A \cdot A\)

\(A^2 = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix} \cdot \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}\)

Since A is orthogonal, \(A^T A = I\), and hence:

\(A^2 = I\)

∴ A² = Identity matrix (I).

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

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:

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.

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

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

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:

Properties of identity matrix:

If A is the square matrix of order n × n

• AI = IA = A
• Iⁿ = I           (Where n ∈ N)

Calculation:

Given

A² = I

Now, A3 + (A + I)2 - 9A - I² - A²

= A². A + A² + I² + 2AI - 9A - I² - A²

= I. A + I + I + 2AI - 9A - I - I           [∵ A2 = I and AI = IA = A]

= AI + 2AI - 9A 

= 3AI - 9A

= 3A - 9A

= - 6A

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