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

Calculation:

Given,

Matrix A is defined as:

$A=\left[\begin{array}{ccc}x& y& z\\ y& z& x\\ z& x& y\end{array}\right]$

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

$A^{T}A=I$

Now, we calculate A²:

$A^2=A\cdot A$

$A^2=\left[\begin{array}{ccc}x& y& z\\ y& z& x\\ z& x& y\end{array}\right]\cdot \left[\begin{array}{ccc}x& y& z\\ y& z& x\\ z& x& y\end{array}\right]$

Since A is orthogonal, $A^{T}A=I$, and hence:

$A^2=I$

∴ A² = Identity matrix (I).

Calculation:

Given,

$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$

$A^2$ $\Rightarrow A^2=A\times A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]$

Now 4A $4A$

$\Rightarrow 4A=4\times \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$

$\Rightarrow 4A=\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

Also A² - 4A $A^2 - 4A$

$\Rightarrow A^2-4A=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

$\Rightarrow A^2-4A=\left[\begin{array}{ccc}9-4& 8-8& 8-8\\ 8-8& 9-4& 8-8\\ 8-8& 8-8& 9-4\end{array}\right]$

$\Rightarrow A^2-4A=\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]$

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">I3$I_3$ 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: $f(\theta )=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
• 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: $f(\pi )=\left[\begin{array}{cc}\cos \pi & \sin \pi \\ -\sin \pi & \cos \pi \end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$

Calculation:

Given,

Rotation matrix at θ = π:

$f(\pi )=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$

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

$f(\pi )\times f(\pi )=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\times \left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$

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:

$f(\pi {)}^2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

∴ (f(π))² is equal to the identity matrix, which is $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.

Calculation:

Given,

The matrix A is:

$A=\left[\begin{array}{ccc}y& x& x\\ z& x& y\\ x& y& z\end{array}\right]$

Since A  is an orthogonal matrix, we know that:

$A^T=A^{-1}\Rightarrow A^{T}A=I$

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:

$A^{T}A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Now, let’s calculate $A^{T}A$ step by step. The transpose of matrix A , denoted $A^T$ is:

$A^T=\left[\begin{array}{ccc}y& z& x\\ x& x& y\\ x& y& z\end{array}\right]$

Now, we perform matrix multiplication between $A^T$ and A:

$A^{T}A=\left[\begin{array}{ccc}y& z& x\\ x& x& y\\ x& y& z\end{array}\right]\left[\begin{array}{ccc}y& x& x\\ z& x& y\\ x& y& z\end{array}\right]$

Performing this multiplication, we get the following matrix:

$A^{T}A=\left[\begin{array}{ccc}{y}^2+z^2+x^2& xy+zx+xy& xz+yz+x^2\\ xy+zx+xy& x^2+x^2+y^2& xy+xz+yz\\ xz+yz+x^2& xy+xz+yz& x^2+y^2+z^2\end{array}\right]$

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

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

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:

$\left[\begin{array}{ccc}x& 1& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]=\left[\begin{array}{ccc}x+4+7& 2x+5+8& 3x+6+9\end{array}\right]$

⇒ $\left[\begin{array}{ccc}x+11& 2x+13& 3x+15\end{array}\right]$

Multiply the resulting matrix with Matrix 3:

$\left[\begin{array}{ccc}x+11& 2x+13& 3x+15\end{array}\right]\times \left[\begin{array}{c}1\\ 1\\ x\end{array}\right]$

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

⇒ $(x+11)+(2x+13)+(3x^2+15x)$

⇒ $(3x^2+18x+24)$

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 = $\left[\begin{array}{ccc}2& x-3& x-2\\ 3& -2& -1\\ 4& -1& -5\end{array}\right]$

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 - $\mathrm{A}=\left(\begin{array}{cc}4& \mathrm{x}+2\\ 2\mathrm{x}-3& \mathrm{x}+1\end{array}\right)$

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

Where A^T = transpose of matrix A

${\mathrm{A}}^{\mathrm{T}}=\left(\begin{array}{cc}4& 2\mathrm{x}-3\\ \mathrm{x}+2& \mathrm{x}+1\end{array}\right)$

∴ A = A^T

$\Rightarrow \left[\begin{array}{cc}4& \mathrm{x}+2\\ 2\mathrm{x}-3& \mathrm{x}+1\end{array}\right]=\left[\begin{array}{cc}4& 2\mathrm{x}-3\\ \mathrm{x}+2& \mathrm{x}+1\end{array}\right]$

Compare A₂₁ element.

⇒ x + 2 =2x - 3 

⇒ x = 5

Concept:

For an invertible matrix A:

• A^-1 = $\frac{\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})}{|\mathrm{A}|}$.
• |A-1| = |A|^-1 = $\frac{1}{|\mathrm{A}|}$.

Calculation:

${\mathrm{A}}^{-1}=\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 3\\ 3& 1& 6\end{array}\right]=\frac{\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})}{\mathrm{k}}$         -----(1)

From the definition of the inverse of a matrix, 

A-1 = $\frac{\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})}{|\mathrm{A}|}$              -----(2)

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

k = |A|  

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

k = |A| = $\frac{1}{|{\mathrm{A}}^{-1}|}$        ----(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 = $-\frac{1}{15}$.

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| = $\frac{1}{|{\mathrm{A}}^{-1}|}$

Calculation:

Given: $\mathrm{A}=\left[\begin{array}{cc}x& 2\\ 4& 3\end{array}\right]$ and ${\mathrm{A}}^{-1}=\left[\begin{array}{cc}\frac{1}{8}& \frac{-1}{12}\\ \frac{-1}{6}& \frac{4}{9}\end{array}\right]$

|A^-1| = $\frac{4}{72}-\frac{1}{72}=\frac{3}{72}=\frac{1}{24}$

|A| = $\frac{1}{|{\mathrm{A}}^{-1}|}$ = 24

⇒ 3x - 8 = 24

∴ x = $\frac{32}{3}$

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. 

$\mathrm{t}\mathrm{r}\left(\mathrm{A}\right)=\sum _{\mathrm{n}=1}^{\mathrm{n}}{\mathrm{A}}_{\mathrm{n}\mathrm{n}}$

Calculation:

Given: $A=\left[\begin{array}{cc}-1& 4\\ 5& 8\end{array}\right]$

Trace of matrix = sum of elements on the main diagonal

= -1 + 8

= 7

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:

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∶

$A=\left[\begin{array}{cc}n& 9\\ 2& 1\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

We cannot find anything from this statement.

Combining statement 1 and 2∶

$AB=\left[\begin{array}{cc}(n\times 1+9\times 0)& (n\times 0+9\times 1)\\ (2\times 1+1\times 0)& (2\times 0+1\times 1)\end{array}\right]$

$AB=\left[\begin{array}{cc}n& 9\\ 2& 1\end{array}\right]$

Also, $A=\left[\begin{array}{cc}n& 9\\ 2& 1\end{array}\right]$

∴ 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