RD Sharma Class 12 Exercise 7.2 Simultaneous Linear Equation Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 7.2 Simultaneous Linear Equation Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 03:26 PM IST

RD Sharma Class 12 Solutions Chapter 7 Exercise 7.2 (Solution of Simultaneous Linear Equations) consists of problems on finding the homogeneous system of linear equations. These problems are solved by CAREER360 experts in a basic manner to accelerate the test readiness of students. It essentially helps during the update and further develops certainty among understudies before showing up for the board test.

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

# Solution of simultaneous linear equation exercise 7.2 question 1

$x=y=z=0$
Hint:
If $\left | A \right |\neq 0$, then $x=y=z=0$ is the only solution of the homogeneous system.
Given:
$2x-y+z=0$
$3x+2y-z=0$
$x+4y+3z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 2 & -1 &1 \\ 3& 2 &-1 \\ 1& 4 &3 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
$A= \begin{bmatrix} 2 & -1 &1 \\ 3& 2& -1\\ 1& 4 & 3 \end{bmatrix}$
Let
$A= \begin{vmatrix} 2 & -1 &1 \\ 3& 2& -1\\ 1& 4 & 3 \end{vmatrix}$
$=2(6+4)+1(9+1)+1(12-2)$
$=20+10+10$
$=40$
$\left | A \right |\neq 0$
So the given system of equations has a trivial solution.
$x=0, y=0, z=0$

Solution of simultaneous linear equation exercise 7.2 question 2

$x=\frac{-5k}{11}, y=\frac{12k}{11}, z=k$
Hint:
If $\left | A \right |= 0$, then the system of equation has non-trivial solution.
Given:
$2x-y+2z=0$
$5x+3y-z=0$
$x+5y-5z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
Let $A=\begin{bmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{bmatrix},X=\begin{bmatrix} x\\ y\\ z\end{bmatrix},B=\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
i.e. $AX=B$
$\left | A \right |=\begin{vmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{vmatrix}$
$=2(-15+5)+1(-25+1)+2(25-3)$
$=-20-24+ 44$
$=0$
So, the given system has non-trivial solution.
To find the solutions we write
$2x-y=-2z$
$5x+3y=z$
let $z=k$
$\begin{bmatrix} 2 &-1 \\ 5 & 3 \end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} -2k\\ k\end{bmatrix}$

$\left | A \right |=\begin{vmatrix} 2 &-1 \\ 5 & 3 \end{vmatrix}=11\neq 0$
So A-1 exists
$adjA=\begin{bmatrix} 3 & 1\\ -5 & 2\end{bmatrix}$
$A^{-1}=\frac{1}{\left | A \right |}adjA$
$A^{-1}=\frac{1}{11}\begin{bmatrix} 3 & 1\\ -5 &2 \end{bmatrix}$
Now, $X=A^{-1}B$
$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{11}\begin{bmatrix} 3 & 1\\ -5 &2 \end{bmatrix}\begin{bmatrix} -2k\\ k\end{bmatrix}$
$=\frac{1}{11}\begin{bmatrix} -5k\\ 12k\end{bmatrix}$
$x=\frac{-5k}{11},y=\frac{12k}{11}$
x,z and y satisfy the third equation
$\frac{-5k}{11}+5\left ( \frac{12k}{11} \right )-5z=0$
$5z=\frac{55k}{11}$
$z=k$

Hence $x=\frac{-5k}{11}, y=\frac{12k}{11}, z=k$, where $k\in R$

Solution of simultaneous linear equation exercise 7.2 question 3

$x= \frac{-9k}{13},y= \frac{-k}{13},z= k$
Hint:
If $\left | A \right |= 0$, then the system of equation has non-trivial solution.
Given:
$3x-y+2z=0$
$4x+3y+3z=0$
$5x+7y+4z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 3 & -1 & 2\\ 4& 3 & 3\\ 5& 7 & 4 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$
$A=\begin{bmatrix} 3 & -1 & 2\\ 4& 3 & 3\\ 5& 7 & 4 \end{bmatrix}$
let
$\left | A \right |=\begin{vmatrix} 3 & -1 & 2\\ 4& 3 & 3\\ 5& 7 & 4 \end{vmatrix}$
$= 3(12-21)+1(16-15)+2(28-15)$
$= -27+1+26$
$= 0$
So, the given system of equation has a non-trivial solution.
To find these solutions, we write the first two equations as
$3x-y=-2z$
$4x+3y=-3z$
Let $z=k$, then we have
$\begin{bmatrix} 3 & -1\\ 4 & 3\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} -2k\\ -3k\end{bmatrix}$
$A=\begin{bmatrix} 3 & -1\\ 4 & 3\end{bmatrix},X=\begin{bmatrix} x\\ y\end{bmatrix},B=\begin{bmatrix} -2k\\ -3k\end{bmatrix}$
Where,
$\left | A \right |=\begin{vmatrix} 3 & -1\\ 4 & 3\end{vmatrix}=13\neq 0$
So $A^{-1}$ exists
$adjA=\begin{bmatrix} 3 & 1\\ -4 & 3\end{bmatrix}$
$A^{-1}=\frac{1}{\left | A \right |}adjA$
$A^{-1}=\frac{1}{13}\begin{bmatrix} 3 & 1\\ -4& 3\end{bmatrix}$
Now, $X=A^{-1}B$
$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{13}\begin{bmatrix} 3 & 1\\ -4& 3\end{bmatrix}\begin{bmatrix} -2k\\ -3k\end{bmatrix}$
$=\frac{1}{13}\begin{bmatrix} -9k\\ -k\end{bmatrix}$
$x=\frac{-9k}{13}, y=\frac{-k}{13}$
Put in third equation, it satisfies it too.
Hence $x=\frac{-9k}{13}, y=\frac{-k}{13},z=k$, where $k\in R$

Solution of simultaneous linear equation exercise 7.2 question 4

$x= 2k,y= 4k,z= k$
Hint:
If $\left | A \right |= 0$, then the system of equation has non-trivial solution.
Given:
$x+y-6z=0$
$x-y+2z=0$
$-3x+y+2z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$
$A=\begin{bmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{bmatrix}$
let
$\left | A \right |=\begin{vmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{vmatrix}$
$= 1(-2-2)-1(2+6)-6(1-3)$
$= -4-8+12$
$= 0$
So, the given system of equation has a non-trivial solution.
To find these solutions, we write the first two equations as
$x+y=6z$
$x-y=-2z$
Let $z=k$, then we have
$\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 6k\\ -2k\end{bmatrix}$
$A=\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix},X=\begin{bmatrix} x\\ y\end{bmatrix},B=\begin{bmatrix} 6k\\ -2k\end{bmatrix}$
Where,
$\left | A \right |=\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}=-2\neq 0$
So $A^{-1}$ exists
$adjA=\begin{bmatrix} -1 & -1\\ -1 & 1\end{bmatrix}$
$A^{-1}=\frac{1}{\left | A \right |}adjA$
$A^{-1}=\frac{1}{-2}\begin{bmatrix} -1 & -1\\ -1& 1\end{bmatrix}$
Now, $X=A^{-1}B$
$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{-2}\begin{bmatrix} -1 & -1\\ -1& 1\end{bmatrix}\begin{bmatrix} 6k\\ -2k\end{bmatrix}$
$\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 2k\\ 4k\end{bmatrix}$
$x=2k, y=4k$
Put in third equation, so these values $x=2k, y=4k,z=k$ satisfies it.
Hence ,$x=2k, y=4k,z=k$, where $k\in R$

Solution of simultaneous linear equation exercise 7.2 question 5

$x= 2k,y= -3k,z= k$
Hint:
If $\left | A \right |= 0$, then the system of equation has non-trivial solution.
Given:
$x+y+z=0$
$x-y-5z=0$
$x+2y+4z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 1 & 1 & 1\\ 1& -1 & -5\\ 1& 2 & 4\end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$
$A=\begin{bmatrix} 1 & 1 & 1\\ 1& -1 & -5\\ 1& 2& 4 \end{bmatrix}$
let
$\left | A \right |=\begin{vmatrix} 1 & 1 & 1\\ 1& -1 & -5\\ 1& 2 & 4 \end{vmatrix}$
$= 1(-4+10)-1(4+5)+1(2+1)$
$= 6-9+3$
$= 0$
So, the given system of equation has a non-trivial solution.
To find these solutions, we write the first two equations as
$x+y=6z$
$x-y=-2z$
Let $z=k$, then we have
$\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} -k\\ 5k\end{bmatrix}$
$A=\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix},X=\begin{bmatrix} x\\ y\end{bmatrix},B=\begin{bmatrix} -k\\ -5k\end{bmatrix}$
Where,
$\left | A \right |=\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}=-2\neq 0$
So $A^{-1}$ exists
$adjA=\begin{bmatrix} -1 & -1\\ -1 & 1\end{bmatrix}$
$A^{-1}=\frac{1}{\left | A \right |}adjA$
$A^{-1}=\frac{1}{-2}\begin{bmatrix} -1 & -1\\ -1& 1\end{bmatrix}$
Now, $X=A^{-1}B$
$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{-2}\begin{bmatrix} -1 & -1\\ -1& 1\end{bmatrix}\begin{bmatrix} -k\\ 5k\end{bmatrix}$
$\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 2k\\ -3k\end{bmatrix}$
$x=2k, y=-3k$
Put in third equation, so these values $x=2k, y=-3k,z=k$ satisfies it.
Hence ,$x=2k, y=-3k,z=k$, where $k\in R$

Solution of simultaneous linear equation exercise 7.2 question 6

$x= \frac{k}{3},y= \frac{2k}{3},z= k$
Hint:
If $\left | A \right |= 0$, then the system of equation has non-trivial solution.
Given:
$x+y-z=0$
$x-2y+z=0$
$3x+6y-5z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 1 & 1 & -1\\ 1& -2 & 1\\ 3& 6 & -5 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$
$A=\begin{bmatrix} 1 & 1 & -1\\ 1& -2 & 1\\ 3& 6 & -5 \end{bmatrix}$
let
$\left | A \right |=\begin{vmatrix} 1 & 1 & -1\\ 1& -2 & 1\\ 3& 6 & -5 \end{vmatrix}$
$= 1(10-6)-1(-5-3)-1(6+6)$
$= 4+8-12$
$= 0$
So, the given system of equation has a non-trivial solution.
To find these solutions, we write the first two equations as
$x+y=z$
$x-2y=-z$
Let $z=k$, then we have
$\begin{bmatrix} 1 & 1\\ 1 & -2\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} k\\ -k\end{bmatrix}$
$A=\begin{bmatrix} 1 & 1\\ 1 & -2\end{bmatrix},X=\begin{bmatrix} x\\ y\end{bmatrix},B=\begin{bmatrix} k\\ -k\end{bmatrix}$
Where,
$\left | A \right |=\begin{vmatrix} 1 & 1\\ 1 & -2\end{vmatrix}=-3\neq 0$
So $A^{-1}$ exists
$adjA=\begin{bmatrix} -2 & -1\\ -1 & 1\end{bmatrix}$
$A^{-1}=\frac{1}{\left | A \right |}adjA$
$A^{-1}=\frac{1}{-3}\begin{bmatrix} -2 & -1\\ -1& 1\end{bmatrix}$
Now, $X=A^{-1}B$
$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{-3}\begin{bmatrix} -2 & -1\\ -1& 1\end{bmatrix}\begin{bmatrix} k\\ -k\end{bmatrix}$
$\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} \frac{k}{3}\\ \frac{2k}{3} \end{bmatrix}$
$x=\frac{k}{3}, y=\frac{2k}{3}$
Put in third equation, so these values$x=\frac{k}{3}, y=\frac{2k}{3},z=k$ satisfies it.
Hence ,$x=\frac{k}{3}, y=\frac{2k}{3},z=k$ where $k\in R$

Solution of simultaneous linear equation exercise 7.2 question 7

$x=y=z=0$
Hint:
If $\left | A \right |\neq 0$, then $x=y=z=0$
Given:
$3x+y-2z=0$
$x+y+z=0$
$x-2y+z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 3 & 1 &-2 \\ 1& 1 &1 \\ 1& -2 &1 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
$A= \begin{bmatrix} 3 & 1 &-2 \\ 1& 1& 1\\ 1& -2 & 1 \end{bmatrix}$
Let
$\left | A \right |= \begin{vmatrix} 3& 1 &-2 \\ 1& 1& 1\\ 1& -2 & 1 \end{vmatrix}$
$=3(1+2)-1(1-1)-2(-2-1)$
$=9+0+6$
$=15$
$\left | A \right |\neq 0$
Hence, $x=0, y=0, z=0$

Solution of simultaneous linear equation exercise 7.2 question 8

$x=y=z=0$
Hint:
If $\left | A \right |\neq 0$, then $x=y=z=0$
Given:
$2x+3y-z=0$
$x-y-2z=0$
$3x+y+3z=0$
Explanation:
The given homogeneous system can be written in matrix form
i.e. $\begin{bmatrix} 2 & 3 &-1 \\ 1& -1 &-2 \\ 3& 1 &3 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$
$A=\begin{bmatrix} 2 & 3 &-1 \\ 1& -1 &-2 \\ 3& 1 &3 \end{bmatrix}$
Let
$\left | A \right |=\begin{vmatrix} 2 & 3 &-1 \\ 1& -1 &-2 \\ 3& 1 &3 \end{vmatrix}$
$=2(-3+2)-3(3+6)-1(1+3)$
$=-2-27-4$
$=-33$
$\left | A \right |\neq 0$
Hence, $x=y=z=0$

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