RD Sharma Class 12 Exercise 4.5 Algebra of Matrices Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 4.5 Algebra of Matrices Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 01:58 PM IST

Students in class 12 have a vast syllabus when it comes to mathematics. This is why students are recommended to use RD Sharma class 12th exercise 4.5 solutions to practice at home and improve their performance on the subject. The 4th chapter of the Mathematics Textbook has Algebra of Matrices, which talks about types of Matrices like skew-symmetric matrix and symmetric matrix, which are used in various Algebra calculations. RD Sharma Solutions The exercise in the book and in the RD Sharma class 12 chapter 4 exercise 4.5 solutions contain ten questions based on the chapter.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## RD Sharma Class 12 Solutions Chapter 4 Algebra of Matrices - Other Exercise

Answer:$A - A^{T}$ is a skew-symmetric matrix.
Hint: Matrix$B$ is said to be skew-symmetric if transpose of matrix $A$ is equal to negative of matrix $A$ i.e $(B^{T} = -B)$
Find AT and prove that $(A -A^{T}) = - (A - A^{T})$
Given:$A =\begin{bmatrix} 2 &3 \\ 4& 5 \end{bmatrix}$
Solution:
$A^T= \begin{bmatrix} 2 & 4\\ 3 &5 \end{bmatrix}$
$A - A^{T} = \begin{bmatrix} 2 &3 \\ 4& 5 \end{bmatrix}-\begin{bmatrix} 2 &4 \\ 3& 5 \end{bmatrix}$
$= \begin{bmatrix} 0 &-1 \\ 1& 0 \end{bmatrix}$
$= B$
$B^{T} = \begin{bmatrix} 0 &1 \\ -1 &0 \end{bmatrix}$
$-B^{T} = \begin{bmatrix} 0 &-1 \\ 1 &0 \end{bmatrix}$
$- B^{T}= B$
Hence,
$B$ is a skew-symmetric matrix.

Algebra of Matrices Exercise 4.5- Question:2

Answer:$A - A^{T}$ is a skew-symmetric matrix.
Hint: Matrix$B$ is said to be skew-symmetric if transpose of matrix $A$ is equal to negative of matrix $A$ i.e $(B^{T} = -B)$
Prove that : $(A - A^{T}) = - (A - A^{T})$
Given:
$A=\begin{bmatrix} 3 &-4 \\ 1 & 1 \end{bmatrix}$
Solution:
$A^{T}=\begin{bmatrix} 3 &1 \\ -4 & 1 \end{bmatrix}$
$A - A^{T} =\begin{bmatrix} 3 &-4 \\ 1& 1 \end{bmatrix} - \begin{bmatrix} 3& 1\\ -4 &1 \end{bmatrix}$
$=\begin{bmatrix} 0 & -5\\ 5& 0 \end{bmatrix}$
$= B$
$B^{T} =\begin{bmatrix} 0 &5 \\ -5& 0 \end{bmatrix}$
$-B^{T} =\begin{bmatrix} 0 &-5 \\ 5& 0 \end{bmatrix}$
$- B^{T} = B$
Hence,
$B$ is a skew-symmetric matrix.

### Question:3

Algebra of Matrices Exercise 4.5

Answer: $x = 4, y = 2, z\: \epsilon \: c, t = -3$
Hint: If $A$ is symmetric matrix, then $A = A^{T}$
Given:$A = \begin{bmatrix} 5 &2 &x \\ y& z &-3 \\ 4& t &-7 \end{bmatrix}$
Solution:
$A^{T} = \begin{bmatrix} 5 &y &4 \\ 2& z &t \\ x& -3 &-7 \end{bmatrix}$
We know that, $A = A^{T}$
$\Rightarrow \begin{bmatrix} 5 &2 & x\\ y &z &-3 \\ 4& t &-7 \end{bmatrix}= \begin{bmatrix} 5 & y &4 \\ 2 &z & t\\ x& -3 &-7 \end{bmatrix}$
Hence, we get
$x = 4, y = 2, z = z, t = - 3$
$x = 4, y = 2, z\: \epsilon \: c, t = - 3$

## Question:4

Algebra of Matrices Exercise 4.5

$X =\begin{bmatrix} 3 &\frac{3}{2} &\frac{5}{2} \\\\ \frac{3}{2} &4 & 4\\ \\\frac{5}{2} & 4 & 8 \end{bmatrix} and$$Y = \left [ 0 \frac{1}{2} \frac{9}{2} \frac{-1}{2} 0 -1 \frac{-9}{2} 1 0 \right ]$
Hint: Find $X =\frac{1}{2} (A + A^{T}) and \: Y=\frac{1}{2} (A - A^{T})$
Given:$A = \begin{bmatrix} 3 &2 &7 \\ 1 &4 &3 \\ -2 &5 &8 \end{bmatrix}$
Solution:
$Step - 1\rightarrow A =\begin{bmatrix} 3 &2 &7 \\ 1& 4 &3 \\ -2 & 5 &8 \end{bmatrix} and AT =\begin{bmatrix} 3& 1 &-2 \\ 2 &4 &5 \\ 7& 3 &8 \end{bmatrix}$$\rightarrow A =\begin{bmatrix} 3 &2 &7 \\ 1& 4 &3 \\ -2 & 5 &8 \end{bmatrix} and AT =\begin{bmatrix} 3& 1 &-2 \\ 2 &4 &5 \\ 7& 3 &8 \end{bmatrix}$
$Step - 2\rightarrow (A + A^{T}) = \left [ 6\, 3\, 5\, 3\, 8\, 8 \,5\,8 \,16 \right ]$
$Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 &1 &9 \\ -1& 0 & -2\\ -9& 2 &0 \end{bmatrix}$
$Step - 4 \rightarrow X= (A + A^{T}) = \left [ 3\,\frac{3}{2}\,\frac{5}{2}\,\frac{3}{2}\,4\,4\,\frac{5}{2}\,4\,8 \right ]$
$Y =\frac{1}{2} (A - A^{T})= \begin{bmatrix} 0 &\frac{1}{2} &\frac{9}{2} \\ \\ \frac{-1}{2} & 0 & -1\\ \\ \frac{-9}{2} &1 & 0 \end{bmatrix}$
$Step - 5\rightarrow X^{T} =\begin{bmatrix} 3 &\frac{3}{2} &\frac{5}{2} \\ \frac{3}{2} & 4 & 4\\ \frac{5}{2} &4 & 8 \end{bmatrix}= X and \, Y^{T} =\left [ 0 \frac{1}{2} \frac{9}{2} \frac{-1}{2} 0 -1 \frac{-9}{2} 1 0 \right ]= -Y$

$X$ is symmetric and $Y$ is skew-symmetric. Also, $X + Y = A.$
$\left [ 3\, \frac{3}{2}\, \frac{5}{2}\, \frac{3}{2}\, 4\, 4\, \frac{5}{2} \,4\, 8 \right ] + \left [ 0\, \frac{1}{2} \,\frac{9}{2}\, \frac{-1}{2} -1\, \frac{-9}{2}\, 1\, 0 \right ] = \begin{bmatrix} 3 &2 &7 \\ 1& 4 &3 \\ -2 &5 &8 \end{bmatrix}$
Thus, $A$ is expressed as a sum of symmetric and skew symmetric matrix.

### Question:5

Algebra of Matrices Exercise 4.5

Answer: Symmetric matrix $= \begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}$ and Skew-symmetric matrix $= \begin{bmatrix} 0 &\frac{-1}{2} &-1 \\ \\ \frac{1}{2} & 0 &\frac{9}{2} \\ \\ 1 & \frac{-9}{2} & 0 \end{bmatrix}$
Hint: Find$P= \frac{1}{2} (A + A^{T}) and\: Q= \frac{1}{2}(A - A^{T})$
Given:$A = \begin{bmatrix} 4 & 2 &-1 \\ 3 &5 &7 \\ 1 &-2 &1 \end{bmatrix}$
Solution:
$Step - 1\rightarrow A\begin{bmatrix} 4 & 2&-1 \\ 3& 5& 7\\ 1&-2 & 1 \end{bmatrix} = and\: AT^{T}=\begin{bmatrix} 4 & 3 & 1\\ 2 &5 & -2\\ -1& 7 &1 \end{bmatrix}$
$Step - 2\rightarrow (A + A^{T}) = \left [ 8\, 5\, 0\, 5\, 10\, 5\, 0\, 5\, 2 \right ]$
$Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 &-1 &-2 \\ 1 &0 &9 \\ 2& -9 &0 \end{bmatrix}$
$Step - 4\rightarrow P=\frac{1}{2} (A + A^{T}) = \begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}$
$Q =\frac{1}{2} (A - A^{T}) = \begin{bmatrix} 0 &\frac{-1}{2} &-1 \\ \\ \frac{1}{2} & 0 &\frac{9}{2} \\ \\ 1 & \frac{-9}{2} & 0 \end{bmatrix}$
$Step - 5\rightarrow P^{T} = \begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}= P \: and \: Q^{T} = \begin{bmatrix} 0 &\frac{-1}{2} &-1 \\ \\ \frac{1}{2} & 0 &\frac{9}{2} \\ \\ 1 & \frac{-9}{2} & 0 \end{bmatrix}= -Q$

Now,$P+ Q = A$ where $P$ is symmetric and $Q$is skew-symmetric.
$\begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}$$+ \begin{bmatrix} 0 &\frac{-1}{2} &-1 \\ \\ \frac{1}{2} & 0 &\frac{9}{2} \\ \\ 1 & \frac{-9}{2} & 0 \end{bmatrix}$$= \begin{bmatrix} 4 & 2 &-1 \\ 3 &5 &7 \\ 1 &-2 &1 \end{bmatrix}$
Thus,$A$ is expressed as a sum of symmetric and skew symmetric matrix.

### Question:6

Algebra of Matrices Exercise 4.5

Answer:$A + A^{T}$is a symmetric matrix.
Hint: Find$A + A^{T}$and prove that$(A + A^{T}) = (A + A^{T})^{T}$
Given:$A = \begin{bmatrix} 2 &4 \\ 5 &6 \end{bmatrix}$
Solution:$A.$ is a symmetric matrix if and only if$A + A^{T}$ where AT is the transpose of matrix $A.$
$AT = ^{T}= \begin{bmatrix} 2 &5 \\ 4 &6 \end{bmatrix}$
$A + A^{T} =\begin{bmatrix} 2 &4 \\ 5& 6 \end{bmatrix} + \begin{bmatrix} 2 & 5\\ 4& 6 \end{bmatrix}$
$= \begin{bmatrix} 4 &9 \\ 9& 12 \end{bmatrix}$
$(A + A^{T})^{T} =\begin{bmatrix} 4 &9 \\ 9 &12 \end{bmatrix}= A + A^{T}$
Hence,
$A + A^{T}$ is a symmetric matrix.

### Question:6

Algebra of Matrices Exercise 4.5

Answer:$A + A^{T}$is a symmetric matrix.
Hint: Find$A + A^{T}$and prove that$(A + A^{T}) = (A + A^{T})^{T}$
Given:$A = \begin{bmatrix} 2 &4 \\ 5 &6 \end{bmatrix}$
Solution:$A.$ is a symmetric matrix if and only if$A + A^{T}$ where AT is the transpose of matrix $A.$
$AT = ^{T}= \begin{bmatrix} 2 &5 \\ 4 &6 \end{bmatrix}$
$A + A^{T} =\begin{bmatrix} 2 &4 \\ 5& 6 \end{bmatrix} + \begin{bmatrix} 2 & 5\\ 4& 6 \end{bmatrix}$
$= \begin{bmatrix} 4 &9 \\ 9& 12 \end{bmatrix}$
$(A + A^{T})^{T} =\begin{bmatrix} 4 &9 \\ 9 &12 \end{bmatrix}= A + A^{T}$
Hence,
$A + A^{T}$ is a symmetric matrix.

### Question:7

Algebra of Matrices Exercise 4.5

Answer: Symmetric matrix $\begin{bmatrix} 3 &\frac{-3}{2} \\ \frac{-3}{2} & -1 \end{bmatrix}$ and Skew-symmetric matrix $=\begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$
Hint: Find$P= \frac{1}{2} (A + A^{T}) and\: Q=\frac{1}{2} (A - A^{T})$
Given:$A = \begin{bmatrix} 3 &-4 \\ 1 & -1 \end{bmatrix}$
Solution:
$Step - 1\rightarrow A^{T} =\begin{bmatrix} 3 &1 \\ -4 &-1 \end{bmatrix}$
$Step - 2\rightarrow (A + A^{T}) = \begin{bmatrix} 6 &-3 \\ -3& -2 \end{bmatrix}$
$Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 & -5\\ 5& 0 \end{bmatrix}$
$Step - 4\rightarrow P=\frac{1}{2} (A + A^{T}) = \begin{bmatrix} 3 & \frac{-3}{2}\\ \frac{-3}{2} & -1 \end{bmatrix}$
$Q = \frac{1}{2} (A - A^{T}) = \begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2}& 0 \end{bmatrix}$
$Step - 5\rightarrow P^{T} =\begin{bmatrix} 3 & \frac{-3}{2}\\ \frac{-3}{2}& -1 \end{bmatrix} = P \: and\: Q^{T} = \begin{bmatrix} 0 &\frac{5}{2} \\ \frac{-5}{2}& 0 \end{bmatrix}= -Q$
Now, $P+ Q = A$ where$P$ is symmetric and $Q$ is skew-symmetric.
$\begin{bmatrix} 3 &\frac{-3}{2} \\ \frac{-3}{2} & -1 \end{bmatrix}$$+\begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$$=\begin{bmatrix} 3 &-4\\ \1 & -1\end{bmatrix}$
Thus, $A$is expressed as a sum of symmetric and skew symmetric matrix

### Question:8

Algebra of Matrices Exercise 4.5

Answer: Symmetric matrix $= \begin{bmatrix} 3 &\frac{1}{2} &\frac{-5}{2} \\ \frac{1}{2} &-2 &-2 \\ \frac{-5}{2} & -2 &-2 \end{bmatrix}$ and Skew-symmetric matrix $= \begin{bmatrix} 0 &\frac{-5}{2} &\frac{-3}{2} \\ \frac{5}{2} &0 &-3 \\ \frac{3}{2} & 3 &0 \end{bmatrix}$
Hint: Find $P= \frac{1}{2} (A + A^{T}) and \: Q=\frac{1}{2} (A - A^{T})$
Given:$A = \begin{bmatrix} 3 &-2 &-4 \\ 3& -2 &-5 \\ -1 &1 & 2 \end{bmatrix}$
Solution:
$Step - 1\rightarrow A^{T} = \begin{bmatrix} 3 &3 &-1 \\ -2& -2 &1 \\ -4 &-5 & 2 \end{bmatrix}$
$Step - 2\rightarrow (A + A^{T}) = \begin{bmatrix} 6 &-1 &-5 \\ 1& -4 & -4\\ -5& -4 &4 \end{bmatrix}$
$Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 & -5 & -3\\ 5 &0 &-6 \\ 3& 6 & 0 \end{bmatrix}$
$Step - 4\rightarrow P=\frac{1}{2} (A + A^{T}) =\begin{bmatrix} 3 &\frac{1}{2} &\frac{-5}{2} \\ \frac{1}{2} & -2 &-2 \\ \frac{-5}{2} & -2 &2 \end{bmatrix}$
$Q =\frac{1}{2} (A - A^{T}) = \begin{bmatrix} 0 &\frac{-5}{2} & \frac{-3}{2}\\ \frac{5}{2} & 0 & -3\\ \frac{3}{2} &3 &0 \end{bmatrix}$
$Step-5\rightarrow P^{T}=\begin{bmatrix} 3 &\frac{1}{2} &\frac{-5}{2} \\ \frac{1}{2} &-2 &-2 \\ \frac{-5}{2} &-2 &2 \end{bmatrix}= P\: and\: Q^{T} = \begin{bmatrix} 0 &\frac{-5}{2} &\frac{-3}{2} \\ \frac{5}{2} &0 &-3 \\ \frac{3}{2} & 3 &0 \end{bmatrix}= -Q$
Verify:$P+ Q = A$ where$P$ is symmetric and $Q$ is skew-symmetric.
$\begin{bmatrix} 3 &\frac{1}{2} &\frac{-5}{2} \\ \frac{1}{2} &-2 &-2 \\ \frac{-5}{2} & -2 &-2 \end{bmatrix}$$+ \begin{bmatrix} 0 &\frac{-5}{2} &\frac{-3}{2} \\ \frac{5}{2} &0 &-3 \\ \frac{3}{2} & 3 &0 \end{bmatrix}$$=\begin{bmatrix} 3 & -2 & -4\\ 3 &-2 &-5 \\ -1& 1 &2 \end{bmatrix}$
Thus, $A$ is expressed as a sum of the symmetric and skew-symmetric matrix

### Question:9

Algebra of Matrices Exercise 4.5

Answer:$A + A^{T}$ is a symmetric matrix.
Hint: Find$A + A^{T}$and prove that $(A + A^{T}) = (A + A^{T})^{T}$
Given:$A = \begin{bmatrix} 2 &3 \\ 5& 7 \end{bmatrix}$
Solution: $A$ is a symmetric matrix if and only if $A + A^{T}$ where $A^{T}$ is the transpose of the matrix $A$
$A^{T} = \begin{bmatrix} 2 & 5\\ 3 &7 \end{bmatrix}$
$A + A^{T} = \begin{bmatrix} 2 &3 \\ 5& 7 \end{bmatrix}+\begin{bmatrix} 2 & 5\\ 3& 7 \end{bmatrix}$
$= \begin{bmatrix} 4 &8 \\ 8& 14 \end{bmatrix}$
$(A + A^{T})^{T} =\begin{bmatrix} 4 &8 \\ 8& 14 \end{bmatrix} = A + A^{T}$
Hence,
$A + A^{T}$is a symmetric matrix.

### Question:10

Algebra of Matrices Exercise 4.5.

Answer:$5$
Hint:$A$ is symmetric matrix $A = A^{T}$
Given:$A=\begin{bmatrix} 4 & x+2\\ 2x-3& x+1 \end{bmatrix}$
Solution:
$A^{T}=\begin{bmatrix} 4 & 2x-3\\ x+2& x+1 \end{bmatrix}$
$A = A^{T}$
$\begin{bmatrix} 4 & x+2\\ 2x-3 & x+1 \end{bmatrix}= \begin{bmatrix} 4 & 2x-3\\ x+2& x+1 \end{bmatrix}$
$2x - 3 = x + 2$
$x = 5$

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