Concept:

Zero matrices:

A zero matrix is a matrix with all entries are zero.

It is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns. 

Unit matrix: A unit matrix is a matrix whose diagonal entries are 1 i.e. all diagonal elements are same and remaining entries are zero

Calculations:

A zero matrix is a matrix with all entries are zero. It may be or may not a square matrix.

A matrix has a determinant not numerical value. It is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns. 

A unit matrix is a matrix whose diagonal entries are 1 i.e. all diagonal elements are same and remaining entries are zero.

Hence, a unit matrix is a diagonal matrix

Concept:

Diagonal Matrix:

Any square matrix in which all the elements are zero except those in the principal diagonal is called a diagonal matrix.

i.e A = [a_ij]_{n × n} is a diagonal matrix if a_ij = 0 for i not equal to j.

Identity Matrix:

A diagonal matrix in which all the principal diagonal elements are equal to 1 is called an identity matrix. It is also known as unit matrix whereas an identity matrix of order n is denoted by I or I_n

Scalar Matrix:

A diagonal matrix in which all the principal diagonal elements are equal is called a scalar matrix.

Calculation:

Given: $A=\left[\begin{array}{ccc}\sqrt{2}& 0& 0\\ 0& \sqrt{2}& 0\\ 0& 0& \sqrt{2}\end{array}\right],B=\left[\begin{array}{ccc}2& 0& 0\\ 0& 1& 0\\ 0& 0& -5\end{array}\right]$

A square matrix whose all the elements except the diagonal elements are zeroes is called a diagonal matrix.

$A={\left[a_{ij}\right]}_{m\times n}$ is a diagonal matrix if a_ij = 0 when i ≠ j.

A diagonal matrix is said to be a scalar matrix if its diagonal elements are same (non - zero).

$A={\left[a_{ij}\right]}_{m\times m}$ is a scalar matrix if a_ij = 0 when i ≠ j, a_ij = k when i = j.

From the above definitions we can clearly say that, matrix A is a scalar matrix and matrix B is a diagonal matrix.

Hence, option C is the correct answer.

Note: A scalar matrix is a diagonal matrix but a diagonal matrix may or may not be a scalar matrix.

Concept: 

Diagonal Matrix:

Any square matrix in which all the elements are zero except those in the principal diagonal is called a diagonal matrix.

i.e A = [aij]n × n is a diagonal matrix if aij = 0 for i not equal to j.

Identity Matrix:

A diagonal matrix in which all the principal diagonal elements are equal to 1 is called an identity matrix. It is also known as unit matrix whereas an identity matrix of order n is denoted by I or In

Calculation:

Given: $A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right]$

Here, we have to find the value of α such that A is an identity matrix.

i.e A = I

$\Rightarrow \left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

⇒ sin α = 1,

⇒ cos α = 0

⇒ α = 90°

∴ The required value is 90° .

Concept:

${\left(AB\right)}_{ij}=\sum _{k=1}^n{a}_{ik}\times b_{kj}\mathrm{\forall }i=1,2,\dots ,mandj=1,2,\dots .,p$

Calculation:

Given: A $=\left[\begin{array}{cc}1& 1\\ 4& 6\end{array}\right]$

Here, we have to find the value of k such that $A^2=kA-2I$

$$\begin{gathered} A^2=A.A=\left[\begin{array}{cc}1& 1\\ 4& 6\end{array}\right]\left[\begin{array}{cc}1& 1\\ 4& 6\end{array}\right] \\ =\left[\begin{array}{cc}1\left(1\right)+\left(1\right)\left(4\right)& 1\left(1\right)+\left(1\right)\left(6\right)\\ 4\left(1\right)+\left(6\right)\left(4\right)& 4\left(1\right)+\left(6\right)\left(6\right)\end{array}\right]=\left[\begin{array}{cc}5& 7\\ 28& 40\end{array}\right] \end{gathered}$$

$A^2=kA-2I$

$$\begin{gathered} \left[\begin{array}{cc}5& 7\\ 28& 40\end{array}\right]=k\left[\begin{array}{cc}1& 1\\ 4& 6\end{array}\right]-2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}k-2& k\\ 4k& 6k-2\end{array}\right] \end{gathered}$$

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements:

4k = 28 ⇒ k = 7

∴ The value of k is 7.

Concept Used:

A matrix A is said to be singular if | A | = 0

Calculation:

$A=\left[\begin{array}{ccc}2-x& 1& 1\\ 1& 3-x& 0\\ -1& -3& -x\end{array}\right]$

∣A∣ = 0 

⇒ $\left|\begin{array}{ccc}2-x& 1& 1\\ 1& 3-x& 0\\ -1& -3& -x\end{array}\right|$= 0

R₂ → R₂ + R₃

⇒ $\left|\begin{array}{ccc}2-x& 1& 1\\ 0& -x& -x\\ -1& -3& -x\end{array}\right|$ = 0

⇒ (2 - x) (x² - 3x) - 1[-x + x] = 0

⇒ (2 - x) x (x - 3) = 0

⇒ x = 0, 2, 3

⇒ S = {0, 2, 3}

∴  The solution set S = {0, 2, 3}

CONCEPT:

If A is a symmetric matrix and B is a skew-symmetric matrix then,

A = A^T, B = - B^T        ...(1)

By the properties of the transpose of the matrix,

(A + B)^T = AT + BT       ...(2)

CALCULATION

Given:

A + B = $\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]$        ...(3)

Using 2,

⇒ A' + B' = $\left[\begin{array}{cc}2& 5\\ 3& -1\end{array}\right]$

Using equation 1,

⇒ A - B = $\left[\begin{array}{cc}2& 5\\ 3& -1\end{array}\right]$          ...(4) 

After adding equations (i) and (ii)

A = $\left[\begin{array}{cc}2& 4\\ 4& -1\end{array}\right]$, B = $\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

⇒ AB = $\left[\begin{array}{cc}4& -2\\ -1& -4\end{array}\right]$

• So, the correct answer is option 3.

Concept:

• Symmetric Matrix: Any real square matrix A = (aij) is said to be symmetric matrix if and only if aij = aji, ∀ i and j or in other words we can say that if A is a real square matrix such that A = A’ then A is said to be a symmetric matrix.
• Skew-symmetric Matrix: Any real square matrix A = (a_ij) is said to be skew-symmetric matrix if and only if a_ij = - a_ji, ∀ i and j or in other words we can say that if A is a real square matrix such that A =- A’ then A is said to be a skew-symmetric matrix.
• (A ± B)' = A' ± B'
• (A ⋅ B)' = B' ⋅ A'

Calculation:

Given: A and B are two skew-symmetric matrices of order n

Statement 1:  A ⋅ B is a skew symmetric matrix when AB = - BA

Let's find out transpose of (A ⋅ B)

⇒ (A ⋅ B)' = B' ⋅ A'

∵ A and B are two skew - symmetric matrices of order n i.e A' = -A and B' = -B

⇒(A ⋅ B)' = -B ⋅ -A

⇒ (A ⋅ B)' = B ⋅ A

⇒ (A ⋅ B)' = - (A ⋅ B)--------------(∵ AB = - BA)

Hence, statement 1 is true.

Statement 2: A ⋅ B is a symmetric matrix when AB = BA

Let's find out transpose of (A ⋅ B)

⇒ (A ⋅ B)' = B' ⋅ A'

∵ A and B are two skew - symmetric matrices of order n i.e A' = -A and B' =- B

⇒(A ⋅ B)' = -B ⋅ -A

⇒ (A ⋅ B)' = B ⋅ A

⇒ (A ⋅ B)' = (A ⋅ B)--------------(∵ AB = BA)

Hence, statement 2 is also true.

CONCEPT:

Symmetric Matrix:

Any real square matrix A = (aij) is said to be symmetric matrix if and only if aij = aji, ∀ i and j or in other words we can say that if A is a real square matrix such that A = At then A is said to be a symmetric matrix.

Calculation:

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

A = At

 $A^t=\left[\begin{array}{ccc}1& 1-x& 2\\ 3+x& 2& 5-y\\ 2& y+1& 3\end{array}\right]=\left[\begin{array}{ccc}1& 3+x& 2\\ 1-x& 2& y+1\\ 2& 5-y& 3\end{array}\right]=A$

On comparing 

3 + x = 1 - x

⇒ x = - 1

And, y + 1 = 5 - y

⇒ y = 2

3x + y = 3(-1) + 2

∴ 3x + y = -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:

Transpose of a Matrix:

The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix.

For example: $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{x}& \mathrm{y}& \mathrm{z}\end{array}\right]\Rightarrow {\mathrm{A}}^{\prime }=\left[\begin{array}{cc}\mathrm{a}& \mathrm{x}\\ \mathrm{b}& \mathrm{y}\\ \mathrm{c}& \mathrm{z}\end{array}\right]$.

It is denoted by A' or A^T.

Properties of Transpose of a Matrix:

• The transpose of the product of two matrices is equivalent to the product of their transposes in reversed order:

  (AB)' = B'A'

• (ABC)' = C'B'A'

• (A')' = A

Calculation:

It is given that B is a skew-symmetric matrix.

∴ B' = -B

Now, consider the transpose of the product matrix A′BA.

(A′BA)' = A'B'(A')'

= A'(-B)A               [∵ B' = -B]

= -(A'BA)

Since the transpose is equal to its negative, A'BA is a Skew-Symmetric matrix.