Mathematics Test - 27

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: \(A = \left[ {\begin{array}{*{20}{c}} {\sqrt 2 }&0&0\\ 0&{\sqrt 2 }&0\\ 0&0&{\sqrt 2 } \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} 2&0&0\\ 0&1&0\\ 0&0&{ - 5} \end{array}} \right]\)

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

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

i.e A = I

\(⇒ \left[ {\begin{array}{*{20}{c}} {\sin α }&{ - \cos α }\\ {\cosα }&{\sin α } \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

⇒ sin α = 1,

⇒ cos α = 0

⇒ α = 90°

∴ The required value is 90° .

Concept:

 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 = [aij] is a m × n matrix and B = [bij] 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 \(= \;\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right]\)

Here, we have to find the value of k such that \({A^2} = kA - 2I\)

\({A^2}\; = \;A.A\; = \;\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right]\;\\ = \;\left[ {\begin{array}{*{20}{c}} {1\left( 1 \right)\; + \;\left( { 1} \right)\left( 4 \right)}&{1\left( { 1} \right)\; + \;\left( { 1} \right)\left( { 6} \right)}\\ {4\left( 1 \right)\; + \;\left( { 6} \right)\left( 4 \right)}&{4\left( { 1} \right)\; + \;\left( { 6} \right)\left( { 6} \right)} \end{array}} \right]\; = \;\left[ {\begin{array}{*{20}{c}} 5&{ 7}\\ 28&{ 40} \end{array}} \right]\)

\({A^2}\; = \;kA - 2I\)

\(\left[ {\begin{array}{*{20}{c}} 5&{ 7}\\ 28&{ 40} \end{array}} \right]\; = \;k\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\\= \;\left[ {\begin{array}{*{20}{c}} {k - 2}&{ k}\\ {4k}&{ 6k - 2} \end{array}} \right]\)

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:

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

∣A∣ = 0 

⇒ \(\begin{vmatrix}2-x&1&1\\\ 1&3-x&0\\\ -1&-3&-x\end{vmatrix}\)= 0

R₂ → R₂ + R₃

⇒ \(\begin{vmatrix}2-x&1&1\\\ 0&-x&-x\\\ -1&-3&-x\end{vmatrix}\) = 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 = \(\left[ {\begin{array}{*{20}{c}} 2&3\\ 5&{ - 1} \end{array}} \right]\)        ...(3)

Using 2,

⇒ A' + B' = \(\left[ {\begin{array}{*{20}{c}} 2&5\\ 3&{ - 1} \end{array}} \right]\)

Using equation 1,

⇒ A - B = \(\left[ {\begin{array}{*{20}{c}} 2&5\\ 3&{ - 1} \end{array}} \right]\)          ...(4) 

After adding equations (i) and (ii)

A = \(\left[ {\begin{array}{*{20}{c}} 2&4\\ 4&{ - 1} \end{array}} \right]\), B = \(\left[ {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right]\)

⇒ AB = \(\left[ {\begin{array}{*{20}{c}} 4&{ - 2}\\ { - 1}&{ - 4} \end{array}} \right]\)

• 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:

\(A = \left[ {\begin{array}{*{20}{c}} 1&{3 + x}&2\\ {1 - x}&2&{y + 1}\\ 2&{5 - y}&3 \end{array}} \right]\)

A = At

 \( A^t=\left[ {\begin{array}{*{20}{c}} 1&{1 - x}&2\\ {3 + x}&2&{5 - y}\\ 2&{y + 1}&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{3 + x}&2\\ {1 - x}&2&{y + 1}\\ 2&{5 - y}&3 \end{array}} \right] = A\)

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| = \(\rm 1\over {|A^{-1}|}\)

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:

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: \(\rm A=\begin{bmatrix} \rm a & \rm b &\rm c \\ \rm x & \rm y & \rm z \end{bmatrix}\Rightarrow A'=\begin{bmatrix} \rm a & \rm x \\ \rm b & \rm y \\ \rm c & \rm z \end{bmatrix}\).

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.