RD Sharma Class 12 Exercise 4.4 Algebra of Matrices Solutions Maths

RD Sharma Class 12 Exercise 4.4 Algebra of Matrices Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 01:54 PM IST

The RD Sharma solution books are used by every student who is preparing for their public examinations. When it comes to mathematics, many students struggle to find solutions while doing their homework. Even though Algebra of Matrices is an easy chapter, many students find it challenging to recheck their answer if it is correct or not. The RD Sharma Class 12th exercise 4.4 book will be a great help for them.

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

Algebra of Matrices Excercise: 4.4

Algebra of Matrices Excercise 4.4 Question 1 (i).

Given:
$A=\begin{bmatrix} 2 &-3 \\ -7 &5 \end{bmatrix}, B=\begin{bmatrix} 1 &0 \\ 2 & -4 \end{bmatrix}$
To prove: $\left ( 2A \right )^{T}=2A^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along its main diagonal
Solution:
$\! \! \! \! \! \! \! \! A=\begin{bmatrix} 2 &-3 \\ -7 & 5 \end{bmatrix}, A^{T}=\begin{bmatrix} 2 &-7 \\ - 3& 5 \end{bmatrix}\\\\ 2A=2\begin{bmatrix} 2 &-3 \\ -7 & 5 \end{bmatrix}\\\\ 2A=\begin{bmatrix} 4 & -6\\ -14 & 10 \end{bmatrix}\\\\$
$\left ( 2A \right )^{T}=\begin{bmatrix} 4 &-14 \\ -6& 10 \end{bmatrix}$ …… (1)
$2A ^{T}=2\begin{bmatrix} 2 & -7\\ -3 & 5 \end{bmatrix}=\begin{bmatrix} 4 &-14 \\ -6& 10 \end{bmatrix}$ …… (2)
From 1 & 2
$\left ( 2A \right ) ^{T}=2A^{T}$

Algebra of Matrices Excercise 4.4 Question 1 (ii).

Answer: $\left ( A+B\right )^{T}=A^{T}+B^{T}$
Given:
$A=\begin{bmatrix} 2 &-3 \\ -7& 5 \end{bmatrix}, B=\begin{bmatrix} 1 &0 \\ 2& -4 \end{bmatrix}$
To prove: $\left ( A+B\right )^{T}=A^{T}+B^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along its main diagonal
Solution:
$\left ( A+B\right )^{T}=A^{T}+B^{T}$
R.H.S:
$A^{T}=\begin{bmatrix} 2 &-7 \\ 3& 5 \end{bmatrix}, B^{T}=\begin{bmatrix} 1 &2 \\ 0& -4 \end{bmatrix}$
$A^{T}B^{T}=\begin{bmatrix} 2 & -7\\ 3 & 5 \end{bmatrix}+\begin{bmatrix} 1 &2\\ 0 &-4 \end{bmatrix}$
$=\begin{bmatrix} 3 & -5\\ -3& -1 \end{bmatrix}$ …… (1)
$\left ( A+B \right )^{T}=\left ( \begin{bmatrix} 2 & -3\\ -7 & 5 \end{bmatrix}+\begin{bmatrix} 1 &0\\ 2 &-4 \end{bmatrix} \right )$
$=\left ( \begin{bmatrix} 2+1 & -3+0\\ -7+2 & 5-4 \end{bmatrix} \right )$
$=\left ( \begin{bmatrix} 3 & -3\\ -5 & 1 \end{bmatrix} \right )^{T}$
$= \begin{bmatrix} 3 & -5\\ -3 & 1 \end{bmatrix} -\left ( 2 \right )$
From 1 & 2
$\left ( A+B\right )^{T}=A^{T}+B^{T}$

Algebra of Matrices Excercise 4.4 Question 1 (iii).

Answer: $\left ( A-B \right )^{T}=A^{T}-B^{T}$
Given: $A=\begin{bmatrix} 2 &-3 \\ -7 &-5 \end{bmatrix}, B=\begin{bmatrix} 1 &0 \\ 2& -4 \end{bmatrix}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$A^{T}=\begin{bmatrix} 2 &-7 \\ -3 &-5 \end{bmatrix}, B^{T}=\begin{bmatrix} 1 &2 \\ 0& -4 \end{bmatrix}$
$\left ( A-B \right )^{T}=A^{T}-B^{T}$
$\left ( \begin{bmatrix} 2 &-3 \\ -7 &5 \end{bmatrix}-\begin{bmatrix} 1 &0 \\ 2& -4 \end{bmatrix} \right )=\begin{bmatrix} 2 &-7 \\ -3 & 5 \end{bmatrix}-\begin{bmatrix} 1 &2 \\ 0&-4 \end{bmatrix}$
$\begin{bmatrix} 1 &-3 \\ -9 &9 \end{bmatrix}^{T} =\begin{bmatrix} 1 &-9 \\ -3 & 9 \end{bmatrix}$
$\begin{bmatrix} 1 &-9 \\ -3 &9 \end{bmatrix}^{T} =\begin{bmatrix} 1 &-9 \\ -3 & 9 \end{bmatrix}$
∴ LHS=RHS

Algebra of Matrices Excercise 4.4 Question 1 (iv).

Answer: $\left ( AB \right )^{T}=B^{T}A^{T}$
Given: $A=\begin{bmatrix} 2 &-3 \\ -7 & 5 \end{bmatrix},B=\begin{bmatrix} 1 &0 \\ 2 &-4 \end{bmatrix}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$B^{T}=\begin{bmatrix} 1 &2 \\ 0& -4 \end{bmatrix}, A^{T}=\begin{bmatrix} 2 & -7\\ -3& 5 \end{bmatrix}$
$\left ( AB \right )^{T}=B^{T}A^{T}$
$\left ( \begin{bmatrix} 2 &-3 \\ -7& 5 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 2& -4 \end{bmatrix}\right )=\begin{bmatrix} 1 &2 \\ 0& -4 \end{bmatrix}\begin{bmatrix} 2 & -7\\ -3& 5 \end{bmatrix}$
$\left ( \begin{bmatrix} 2-6 &0+12\\ -7+10& 0-20 \end{bmatrix} \right )^{T}=\begin{bmatrix} 2-6 &-7+10\\ 0+12& 0-20 \end{bmatrix}$
$\left ( \begin{bmatrix} -4 &12\\ -3& -20 \end{bmatrix} \right )^{T}=\begin{bmatrix} -4 &3\\ 12& -20 \end{bmatrix}$
$\begin{bmatrix} -4 &3\\ 12& -20 \end{bmatrix} =\begin{bmatrix} -4 &3\\ 12& -20 \end{bmatrix}$
∴LHS=RHS

Algebra of Matrices Excercise 4.4 Question 2

Answer: $\left ( AB \right )^{T}=B^{T}A^{T}$
Given: $A=\begin{bmatrix} 3\\ 5\\ 1 \end{bmatrix}, B=\begin{bmatrix} 1 &0 &4 \end{bmatrix}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$A^{T}=\begin{bmatrix} 3\\ 5\\ 2 \end{bmatrix}, B^{T}=\begin{bmatrix} 1\\ 0\\ 4 \end{bmatrix}$
$A=\begin{bmatrix} 3\\ 5\\ 2 \end{bmatrix}, B=\begin{bmatrix} 1 &0 &4 \end{bmatrix}$
$\left ( AB \right )^{T}=\begin{bmatrix} 3 &5 &5 \\ 0& 0 & 0\\ 12 & 20 & 8 \end{bmatrix}$ … (1)
$B^{T}A^{T}=\begin{bmatrix} 1\\ 0\\ 4 \end{bmatrix}\begin{bmatrix} 3 & 5 &2 \end{bmatrix}$
$=\begin{bmatrix} 3 &5 &2 \\ 0& 0 & 0\\ 12 & 20 & 8 \end{bmatrix}$ ….. (2)
(1) &( 2)
$\left ( AB \right )^{T}=B^{T}A^{T}$

Algebra of Matrices Excercise 4.4 Question 3 (i).

Answer: $\left ( A+B\right )^{T}=A^{T}+B^{T}$
Given: $A=\begin{bmatrix} 1 & -1 &0 \\ 2&1 & 3\\ 1 &2 & 1 \end{bmatrix}, B=\begin{bmatrix} 1 & 2 &3 \\ 2 & 1&3 \\ 0 & 1 & 1 \end{bmatrix}$
To prove: $\left ( A+B\right )^{T}=A^{T}+B^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$\left ( A+B\right )^{T}=A^{T}+B^{T}$
$\left ( \begin{bmatrix} 1 &-1 &0 \\ 2 & 1 &3 \\ 1 & 2 & 1 \end{bmatrix}+\begin{bmatrix} 1 & 2 &3 \\ 2 &1 &3 \\ 0 & 1 & 1 \end{bmatrix} \right )^{T}= \begin{bmatrix} 1 &-1 &0 \\ 2 &1 &3 \\ 1 &2 &1 \end{bmatrix}^{T}+\begin{bmatrix} 1 & 2 & 3\\ 2 & 1 &3 \\ 0 & 1 & 1 \end{bmatrix}^{T}$
$\begin{bmatrix} 1+1 &-1+2 &0+3 \\ 2+2 & 1+1 &3+3 \\ 1+0 & 2+1 & 1+1 \end{bmatrix} ^{T}= \begin{bmatrix} 1 &2 &1 \\ -1 &1 &2 \\ 0 &3 &1 \end{bmatrix}+\begin{bmatrix} 1 & 2 & 0\\ 2 & 1 &1 \\ 3 & 3 & 1 \end{bmatrix}$
$\begin{bmatrix} 2 &1 &3 \\ 4 & 2 &6 \\ 1 & 3 & 2 \end{bmatrix} ^{T}= \begin{bmatrix} 2 &4 &1 \\ 1 &2 &3 \\ 3&6 &2 \end{bmatrix}$
$\begin{bmatrix} 2 &4 &1 \\ 1 &2 &3 \\ 3&6 &2 \end{bmatrix}= \begin{bmatrix} 2 &4 &1 \\ 1 &2 &3 \\ 3&6 &2 \end{bmatrix}$
LHS=RHS

Algebra of Matrices Excercise 4.4 Question 3 (ii).

Answer: $\left ( AB \right )^{T}=B^{T}A^{T}$
Given: $A=\begin{bmatrix} 1 & -1 &0 \\ 2&1 & 3\\ 1 &2 & 1 \end{bmatrix}, B=\begin{bmatrix} 1 & 2 &3 \\ 2 & 1&3 \\ 0 & 1 & 1 \end{bmatrix}$
To prove: $\left ( AB \right )^{T}=B^{T}A^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$B^{T}=\begin{bmatrix} 1 &2 & 0\\ 2 & 1 & 1\\ 3 &3 &1 \end{bmatrix}, A^{T}=\begin{bmatrix} 1 & 2 &1 \\ -1 & 1 & 2\\ 0 & 3 & 1 \end{bmatrix}$
$\left ( AB \right )^{T}=B^{T}A^{T}$
$\left ( \begin{bmatrix} 1 & -1 &0 \\ 2 & 1 &1 \\ 3 &3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 2 & 1 & 3\\ 0 & 1 & 1 \end{bmatrix}\right )^{T}=\begin{bmatrix} 1 & 2 &0 \\ 2 & 1 & 1\\ 3 & 3 &1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1\\ -1 & 1 & 2\\ 0 &3 & 1 \end{bmatrix}$
$\begin{bmatrix} 1-2+0 & 2-1+0 &3-3+0 \\ 2+2+0 & 4+1+3 &6+3+3 \\ 1+4+0 & 2+2+1 & 3+6+1 \end{bmatrix}^{T}=\begin{bmatrix} 1-2+0 & 2+2+0 &1+4+0 \\ 2-1+0 &4+1+3 &2+2+1 \\ 3-3+0 &6+3+3 & 3+6+1 \end{bmatrix}$
$\begin{bmatrix} -1 & 1 &0 \\ 4 & 8 &12\\ 5 & 5 & 10 \end{bmatrix}^{T}=\begin{bmatrix} -1 & 4 &5 \\ 1 &8 &5 \\ 0&12 & 10 \end{bmatrix}$
$\begin{bmatrix} -1 & 4 &5 \\ 1 &8 &5 \\ 0&12 & 10 \end{bmatrix}=\begin{bmatrix} -1 & 4 &5 \\ 1 &8 &5 \\ 0&12 & 10 \end{bmatrix}$
∴LHS=RHS

Algebra of Matrices Excercise 4.4 Question 3 (iii).

Answer: $\left ( 2A^{T} \right )=2A^{T}$
Given: $A=\begin{bmatrix} 1 & -1 &0 \\ 2&1 & 3\\ 1 &2 & 1 \end{bmatrix}, B=\begin{bmatrix} 1 & 2 &3 \\ 2 & 1&3 \\ 0 & 1 & 1 \end{bmatrix}$
To prove: $\left ( 2A^{T} \right )=2A^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$A^{T}=\begin{bmatrix} 1 & 2 &1 \\ -1 & 1& 2\\ 0 &3 & 1 \end{bmatrix}$
$\left ( 2A^{T} \right )=2A^{T}$
$\left ( 2\begin{bmatrix} 1 & -1 & 0\\ 2& 1 &3 \\ 1 & 2 & 1 \end{bmatrix} \right )^{T}=2\begin{bmatrix} 1 &2 &-1 \\ -1 &1 & 2\\ 0 &3 & 1 \end{bmatrix}$
$\begin{bmatrix} 2 & -2 & 0\\ 4& 2 &6 \\ 2 & 4 & 2 \end{bmatrix}^{T}=\begin{bmatrix} 2 &4 &2 \\ -2 &2 & 4\\ 0 &6 & 2 \end{bmatrix}$
$\begin{bmatrix} 2 &4 &2 \\ -2 &2 & 4\\ 0 &6 & 2 \end{bmatrix}=\begin{bmatrix} 2 &4 &2 \\ -2 &2 & 4\\ 0 &6 & 2 \end{bmatrix}$
∴LHS=RHS
$\left ( 2A^{T} \right )=2A^{T}$

Algebra of Matrices Excercise 4.4 Question 4

Answer: $\left ( AB \right )^{T}=B^{T}A^{T}$
Given: $A=-\begin{bmatrix} -2\\ 4\\ 5 \end{bmatrix}, B=\begin{bmatrix} 1 & 3 &- 6 \end{bmatrix}$
To prove: $\left ( AB \right )^{T}=B^{T}A^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$A^{T}=\begin{bmatrix} -2 & 4 &5 \end{bmatrix}, B^{T}=\begin{bmatrix} 1\\ 3\\ -6 \end{bmatrix}$
$\left ( AB \right )^{T}=B^{T}A^{T}$
$\left ( \begin{bmatrix} -2\\ 4\\ 5 \end{bmatrix}\begin{bmatrix} 1 &3 & -6 \end{bmatrix} \right )^{T}=\begin{bmatrix} 1\\ 3\\ -6 \end{bmatrix}\begin{bmatrix} -2 &4 & 5 \end{bmatrix}$
$\begin{bmatrix} -2 & -6 &12 \\ 4& 12 & -24\\ 5 & 15 & -30 \end{bmatrix}=\begin{bmatrix} -2 & 4 &5 \\ -6 &12 & 15\\ 12 &-24 & -30 \end{bmatrix}$
$\begin{bmatrix} -2 & 4 &5 \\ -6 &12 & 15\\ 12 &-24 & -30 \end{bmatrix}=\begin{bmatrix} -2 & 4 &5 \\ -6 &12 & 15\\ 12 &-24 & -30 \end{bmatrix}$
∴LHS=RHS
Hence, $\left ( AB \right )^{T}=B^{T}A^{T}$ is proved.

Question:5

Algebra of Matrices Excercise 4.4 Question 5

Answer: $\left ( AB\right )^{T}=\begin{bmatrix} 0 &1 \\ 15 & -2 \end{bmatrix}$
Given: $A=\begin{bmatrix} 2 & 4 & -1\\ -1 &0 & 2 \end{bmatrix},B=\begin{bmatrix} 3 &4 \\ -1 &2 \\ 2 & 1 \end{bmatrix}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$\left ( AB \right )=\begin{bmatrix} 2 & 4 &-1 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 3 & 4\\ -1 & 2\\ 2 & 1 \end{bmatrix}$
$=\begin{bmatrix} 6-4-2 & 8+8-1\\ -3-0+4 & -4+0+2 \end{bmatrix}$
$\left [ AB \right ]=\begin{bmatrix} 0 & 15\\ 1 & -2 \end{bmatrix}$
$\left [ AB \right ]^{T}=\begin{bmatrix} 0 & 1\\ 15 & -2 \end{bmatrix}$

Algebra of Matrices Excercise 4.4 Question 6 (i).

Answer: $\left (AB \right )^{T}=B^{T}A^{T}$
Given: $A=\begin{bmatrix} 2 & 1 &3 \\ 4 & 1 & 0 \end{bmatrix}, B=\begin{bmatrix} 1 &-1 \\ 0 & 2\\ 5 & 0 \end{bmatrix}$
To prove: $\left ( AB \right )^{T}=B^{T}A^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$A^{T}=\begin{bmatrix} 2 &4 \\ 1 &1 \\ 3 & 0 \end{bmatrix}, B^{T}=\begin{bmatrix} 1 & 0 &5 \\ -1 & 2& 0 \end{bmatrix}$
$\left ( AB \right )^{T}=B^{T}A^{T}$
$\left ( \begin{bmatrix} 2 & 1 &3 \\ 4 &1 & 0 \end{bmatrix}\begin{bmatrix} 1 &-1 \\ 0 &2 \\ 5 & 0 \end{bmatrix} \right )^{T}=\begin{bmatrix} 1& 0 &5 \\ -1 &2 & 0 \end{bmatrix}\begin{bmatrix} 2 &4 \\ 1 &1 \\ 3 & 0 \end{bmatrix}$
$\left ( \begin{bmatrix} 2+0+15& -2+2+0\\ 4+0+0 & -4+2+0 \end{bmatrix} \right )^{T}=\begin{bmatrix} 2+0+15 &4+0+0 \\ -2+2+0 & -4+2+0 \end{bmatrix}$
$\begin{bmatrix} 17 &0 \\ 4 & -2 \end{bmatrix}^{T}=\begin{bmatrix} 17 &4 \\ 0& -2 \end{bmatrix}$
$\begin{bmatrix} 17 &0 \\ 4 & -2 \end{bmatrix}^{T}=\begin{bmatrix} 17 &4 \\ 0& -2 \end{bmatrix}$
∴LHS=RHS
Hence, $\left ( AB \right )^{T}=B^{T}A^{T}$ , is proved.

Algebra of Matrices Excercise 4.4 Question 6 (ii).

Answer: $\left (AB \right )^{T}=B^{T}A^{T}$
Given: $A=\begin{bmatrix} 1 &3 \\ 2 & 4 \end{bmatrix},B=\begin{bmatrix} 1 &4 \\ 2& 5 \end{bmatrix}$
To prove: $\left (AB \right )^{T}=B^{T}A^{T}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$A^{T}=\begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix},B^{T}=\begin{bmatrix} 1 &2\\ 4& 5 \end{bmatrix}$
$\left (AB \right )^{T}=B^{T}A^{T}$
$\left ( \begin{bmatrix} 1 &3 \\ 2& 4 \end{bmatrix}\begin{bmatrix} 1 &4 \\ 2 & 5 \end{bmatrix} \right )^{T}=\begin{bmatrix} 1 &2 \\ 4& 5 \end{bmatrix}\begin{bmatrix} 1 &2 \\ 3& 4 \end{bmatrix}$
$\begin{bmatrix} 1+6 &4+15 \\ 2+8 & 8+20 \end{bmatrix}^{T}=\begin{bmatrix} 1+6 & 2+8\\ 4+15 &8+20 \end{bmatrix}$
$\begin{bmatrix} 7 & 19\\ 10 & 28 \end{bmatrix}^{T}=\begin{bmatrix} 7 &10 \\ 19& 28 \end{bmatrix}$
$\begin{bmatrix} 7 & 19\\ 10 & 28 \end{bmatrix}=\begin{bmatrix} 7 &19 \\ 19& 28 \end{bmatrix}$
∴LHS=RHS
Hence, $\left (AB \right )^{T}=B^{T}A^{T}$ is proved.

Algebra of Matrices Excercise 4.4 Question 7

Answer: $A^{T}-B^{T}=\begin{bmatrix} 4 &3 \\ -3& 0\\ -1 & -2 \end{bmatrix}$
Given: $\begin{bmatrix} 3 &4 \\ -1 & 2\\ 0 & 1 \end{bmatrix}=A^{T},B=\begin{bmatrix} -1 &2 &1 \\ 1& 2 & 3 \end{bmatrix}$
Hint: The $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.
Solution:
$A^{T}=\begin{bmatrix} 3 &4 \\ -1 & 2\\ 0 & 1 \end{bmatrix},B^{T}=\begin{bmatrix} -1 &1 \\ 2 &2 \\ 1 &3 \end{bmatrix}$
$A^{T}-B^{T}=\begin{bmatrix} 3 &4 \\ -1 & 2\\ 0 & 1 \end{bmatrix}-\begin{bmatrix} -1 &1 \\ 2 &2 \\ 1 &3 \end{bmatrix}$
$\begin{bmatrix} 3+1 &4-1 \\ -1-2& 2-2\\0-1&1-3 \end{bmatrix}=\begin{bmatrix} 4 &3 \\ -3& 0\\ -1 &-2 \end{bmatrix}$

Algebra of Matrices Excercise 4.4 Question 8

Answer: $A^{T}A=I_{2}$
Given: $=\begin{bmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$
Hint: $I_{2}$ refers to an identity matrix with two rows and two columns.
Solution:
$A^{T}A=I_{2}$
Consider: $LHS=A^{T}A$
$=\begin{bmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha &\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$
$=\begin{bmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha &\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$
$\begin{bmatrix} \cos ^{2 }\alpha +\sin^{2} \alpha &\cos \alpha \sin \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\cos \alpha \sin \alpha & \sin ^{2}\alpha +\cos ^{2}\alpha \end{bmatrix}$
$= \begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}=RHS$
∴LHS=RHS
Hence, $A^{T}A=I_{2}$ is proved.

Algebra of Matrices Excercise 4.4 Question 9

Answer: $A^{T}A=I_{2}$
Given: $A=\begin{bmatrix} \sin \alpha &\cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$
Hint: $I_{2}$ refers to an identity matrix with two rows and two columns.
Try to multiply $A^{T}$ with $A$ .
Solution:
$A^{T}A=I_{2}$
Consider: $LHS=A^{T}A$
$=\begin{bmatrix} \sin \alpha &\cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}^{T}\begin{bmatrix} \sin \alpha &\cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$
$=\begin{bmatrix} \sin \alpha &-\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}\begin{bmatrix} \sin \alpha &\cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$
$\begin{bmatrix} \sin ^{2 }\alpha +\cos^{2} \alpha &\sin \alpha \cos \alpha -sin \alpha \cos \alpha \\ \sin \alpha \cos\alpha -\sin \alpha \cos\alpha & \cos ^{2}\alpha +\sin^{2}\alpha \end{bmatrix}$
$= \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}= I$
LHS=RHS
$A^{T}A=I_{2}$ is proved

Algebra of Matrices Excercise 4.4 Question 10

Answer: $AA^{T}=I$
Given: If $l_{i},m_{i},n_{i},i=1,2,3$ denotes direction cosines of 3 mutually ⊥ vertices
$A=\begin{bmatrix} l_{1} &m_{1} &n_{1} \\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3} \end{bmatrix}$
Hint: I refers to identity function, the $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along its main diagonal.
Solution:
Given $\left ( l_{1}, m_{1}, n_{1} \right ),\left ( l_{2}, m_{2}, n_{2}\right ), (l_{3}, m_{3}, n_{3})$ are the direction cosines of 3 mutually + vectors in space.
$\begin{Bmatrix} l_{1}^{2}+ m_{1}^{2}+ n_{1}^{2}=1 \\ l_{2}^{2}+ m_{2}^{2}+ n_{2}^{2}=1\\ l_{3}^{2}+m_{3}^{2}+ n_{3}^{2} =1 \end{Bmatrix}$ ……. (1)
$\begin{Bmatrix} l_{1}l_{2}+ m_{1}m_{2}+ n_{1}n_{2}=0 \\ l_{2}l_{3}+ m_{2}m_{3}+ n_{2}n_{3}=0\\ l_{3}l_{1}+m_{3}m_{1}+ n_{3}n_{1} =0 \end{Bmatrix}$ ……….. (2)
Let, $A=\begin{bmatrix} l_{1} &m_{1} &n_{1} \\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3} \end{bmatrix}$ $,A^{T}=\begin{bmatrix} l_{1} &l_{2} &l_{3} \\ m_{1} & m_{2} & m_{3}\\ n_{1} & n_{2} & n_{3} \end{bmatrix}$
$AA^{T}=\begin{bmatrix} l_{1} &m_{1} &n_{1} \\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3} \end{bmatrix}\begin{bmatrix} l_{1} &l_{2} &l_{3} \\ m_{1} & m_{2} & m_{3}\\ n_{1} & n_{2} & n_{3} \end{bmatrix}$
$AA^{T}=\begin{bmatrix} l_{1}^{2}+m_{1}^{2} +n_{1}^{2} & l_{1} l_{2}+m_{1}m_{2}+n_{1}n_{2} & l_{3}l_{1}+m_{3}m_{1}+n_{3}n_{1}\\ l_{1} l_{2}+m_{1}m_{2}+n_{1}n_{2} & l_{2}^{2}+m_{2}^{2} +n_{2}^{2} &l_{2}l_{3}+m_{2}m_{3}+n_{2}n_{3} \\ l_{3}l_{1}+m_{3}m_{1}+n_{3}n_{1} &l_{2}l_{3}+m_{2}m_{3}+n_{2}n_{3} & l_{3}^{2}+m_{3}^{2}+n_{3}^{2} \end{bmatrix}$
From (1) & (2) we get
$AA^{T}=\begin{bmatrix} 1 & 0 &0 \\ 0& 1 & 0\\ 0 &0 &1 \end{bmatrix}=I$
Hence, proved.

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