RD Sharma Solutions Class 12 Mathematics Chapter 7 FBQ

Updated on Jan 20, 2022 03:10 PM IST

The Class 12 RD Sharma chapter 7 exercise FBQ solution tops all the NCERT solutions as per the opinion of students as well as teachers. Since every CBSE school requires their students to practise and master their NCERT maths book, they are required to answer all the questions that are important for their exam. The RD Sharma class 12th exercise FBQ is said to be quite an efficient guide for the students to help them score high in exams and ace the maths paper.

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

RD Sharma Class 12 Solutions Chapter 7 FBQ Solution of Simultaneous Linear Equation - Other Exercise

Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 1

Answer $\rightarrow a=-1$
Given $\rightarrow$ Here given that $x+a y=0, a z+y=0, a x+z-0$ has infinitely many solutions.
To find $\rightarrow$ The value of $a$.
Hint $\rightarrow$ Given system of the equation can be written as $AX=0$ then solve to find $a$.
Solution$\rightarrow$ We know that $x+a y=0, a z+y=0, a x+z-0$
Given system of linear equation can be written as $AX=0$ where,$A=\left[\begin{array}{lll} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } 0=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]$

We know that for infinitely many solution $D=0$

$\Rightarrow\left|\begin{array}{lll} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{array}\right|=0$
Expanding along row $1$ we get

$\begin{aligned} &\Rightarrow 1(1-0)-a\left(0-a^{2}\right)+0(0-a)=0 \\ &\Rightarrow 1+a^{3}=0 \end{aligned}$

$\begin{aligned} &\Rightarrow a^{3}+1=0 \\ &\Rightarrow a^{3}=-1 \\ &\Rightarrow a=-1 \end{aligned}$
This is required value of $a$

Solution of Simultaneous Linear Exercise Fill in the blank Question 2

Answer $\rightarrow \lambda =3$
Given$\rightarrow$ We have, the given system of equations $x+y+z=6, x+2 y+3 z=10, x+2 y+\lambda z=12$ is inconsistent.
To find $\rightarrow$ The value of $\lambda$
Hint $\rightarrow$ For inconsistent, determinant of the given system of equation will be zero i.e.$D=0$
Solution$\rightarrow$ We know that
$\begin{aligned} &x+y+z=6 \\ &x+2 y+3 z=10 \\ &x+2 y+\lambda z=12 \end{aligned}$
Here, we know that for inconsistent then $D=0$
$\Rightarrow\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{array}\right|=0$
Expanding along row $1$ we get
$\begin{aligned} &\Rightarrow 1(2 \lambda-6)-1(\lambda-3)+1(2-2)=0\\ &\Rightarrow 2 \lambda-6-\lambda+3=0\\ &\Rightarrow \lambda-3=0\\ &\Rightarrow \lambda=3 \text { is required solution. } \end{aligned}$

Solution of Simultaneous Linear Exercise Fill in the blank Question 3

Answer$\rightarrow$ No solution.
Given$\rightarrow$ Here given the system of equations $x+2 y+z=3,2 x+3 y+z=3,3 x+5 y+2 z=1$We have to find the number of solutions of given linear equations.
Hint $\rightarrow$ First we have to check $|A| \neq 0$ or not then we check $(a d j A) B=0$ or not
Solution $\rightarrow$ Here, we have,
$\begin{aligned} &x+2 y+z=3 \\ &2 x+3 y+z=3 \\ &3 x+5 y+2 z=1 \end{aligned}$
First we check $|A| \neq 0$ or not

$\Rightarrow A=\left|\begin{array}{lll} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{array}\right|$

$\begin{aligned} &\Rightarrow|A|=1(6-5)-2(4-3)+1(10-9) \\ &\Rightarrow 1-2+1=0 \\ &\Rightarrow|A|=0 \end{aligned}$
Again we have to find $(a d j A) B$ where $B=\left[\begin{array}{l} 3 \\ 3 \\ 1 \end{array}\right], A=\left[\begin{array}{lll} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{array}\right]$
For $a d j \: A$

$\text { Here, } A=\left[\begin{array}{lll} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{array}\right]$
The cofactor of the elements of $\left | A \right |$ are given by
$A_{11}=\left|\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right|=1 \quad \quad A_{12}=-\left|\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right|-1$

$\begin{aligned} &A_{13}=\left|\begin{array}{ll} 2 & 3 \\ 3 & 5 \end{array}\right|=1 \quad \quad A_{21}=-\left|\begin{array}{ll} 2 & 1 \\ 5 & 2 \end{array}\right|=1 \\\\ &A_{22}=\left|\begin{array}{ll} 1 & 1 \\ 3 & 2 \end{array}\right|=-1 \quad \quad A_{23}=-\left|\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right|=1 \end{aligned}$
$A_{31}=\left|\begin{array}{ll} 2 & 1 \\ 3 & 1 \end{array}\right|=-1 \quad A_{32}=-\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=1$
$A_{33}=\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|=-1$
$\text { Then } a d j A=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 1 & -1 & 1 \\ -1 & 1 & -1 \end{array}\right]^{T}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]$
$\text { Then }(a d j A) B=\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\left[\begin{array}{l} 3 \\ 3 \\ 1 \end{array}\right]$
$\Rightarrow\left[\begin{array}{c} 3+3-1 \\ -3-3+1 \\ 3+3-1 \end{array}\right]=\left[\begin{array}{c} 5 \\ -5 \\ 5 \end{array}\right] \neq 0$
Here, $(a d j A) B\neq 0$
Hence, the system of equation has no solution.

Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 4

Answer$\rightarrow \lambda=-\frac{5}{3}$
Given$\rightarrow$ Given that the system of equations $2 x-y-z=12, x-2 y+z=-4, x+y+\lambda z=4$ has no solution.
To find$\rightarrow$ We have to find out the value of$\lambda$
Hint$\rightarrow$ If system has no solution then $(\operatorname{adj} A) B \neq 0 \text { and }|A|=0$
Solution$\rightarrow$ We have
$\begin{aligned} &2 x-y-z=12 \\ &x-2 y+z=-4 \\ &x+y+\lambda z=4 \end{aligned}$
$A=\left|\begin{array}{ccc} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{array}\right|$
We know that the system of given equation has no solution when $\left | A \right |=0$
$\Rightarrow\left|\begin{array}{ccc} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{array}\right|=0$
$\begin{aligned} &\Rightarrow 2(-2 \lambda-1)+1(\lambda-1)-1(1+2)=0 \\\\ &\Rightarrow-4 \lambda-2+\lambda-1-2=0 \\ &\Rightarrow-3 \lambda=5 \\ &\Rightarrow \lambda=-\frac{5}{3} \end{aligned}$

Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 5

Answer$\rightarrow k=\pm 1$
Given$\rightarrow$ The system of equations $x-k y-z=0, k x-y-z=0 \text { and } x+y-z=0$ has non-zero solution
To find$\rightarrow$ We have to find out the value of $k$.
Hint$\rightarrow$ The system has a non-zero solution if $\left | A \right |=0$
Solution$\rightarrow$ Given system of linear equation
$\begin{aligned} &x-k y-z=0 \\ &k x-y-z=0 \\ &x+y-z=0 \end{aligned}$
For the given system of equations we have,
$\Rightarrow A=\left|\begin{array}{ccc} 1 & -k & -1 \\ k & -1 & -1 \\ 1 & 1 & -1 \end{array}\right|$
We know, if system of equation has a non-zero solution then $\left | A \right |=0$
$\text { Now, }|A|=\left|\begin{array}{ccc} 1 & -k & -1 \\ k & -1 & -1 \\ 1 & 1 & -1 \end{array}\right|=0$
$\begin{aligned} &\Rightarrow 1(1+1)+k(-k+1)-1(k+1)=0 \\ &\Rightarrow 2-k^{2}+k-k-1=0 \\ &\Rightarrow k^{2}=1 \\ &\Rightarrow k=\pm 1 \end{aligned}$

Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 6

Answer $\rightarrow$
Given$\rightarrow$ The given system of equations $\lambda x+y+z=0,-x+\lambda y+z=0 \text { and }-x-y+\lambda z=0$has a non-zero solution.
To find $\rightarrow$ We have to find the value of $\lambda$.
Hint $\rightarrow$ The system has a non-zero solution if $|A|=0$
Solution $\rightarrow$ Here system of equations are
$\begin{aligned} &\lambda x+y+z=0 \\ &-x+\lambda y+z=0 \\ &-x-y+\lambda z=0 \end{aligned}$
Then,
$\Rightarrow A=\left|\begin{array}{ccc} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{array}\right|$
We know, if system of equation has a non-zero solution then $\left | A \right |=0$
$\text { Now, }|A|=\left|\begin{array}{ccc} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{array}\right|=0$
$\begin{aligned} &\Rightarrow \lambda\left(\lambda^{2}+1\right)-1(-\lambda+1)+1(1+\lambda)=0 \\ &\Rightarrow \lambda^{3}+\lambda+\lambda-1+1+\lambda=0 \\ &\Rightarrow \lambda^{3}+3 \lambda=0 \\ &\Rightarrow \lambda\left(\lambda^{2}+3\right)=0 \\ &\Rightarrow \lambda^{2}+3=0 \text { or } \quad \lambda=0 \end{aligned}$
In this case $\lambda$ is an imaginary number which is non-existent.
$\Rightarrow \lambda=0$
Hence, $\lambda=0$ is required answer.

Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 7

Answer $\rightarrow$ $k \neq 0, k=\pm 1, \pm 2, \pm 3, \pm 4, \ldots \ldots . \pm n$
Given $\rightarrow$ Given that the system of equations $x+y+z=2,2 x+y-z=3 \text { and } 3 x+2 y+k z=4$ has a unique solution.
To find $\rightarrow$ We have to find the real value of $k$
Hint $\rightarrow$ If the system of equations has a unique solution $\Rightarrow|A| \neq 0$
Solution $\rightarrow$ We have system of equation,
$\begin{aligned} &x+y+z=2 \\ &2 x+y-z=3 \\ &3 x+2 y+k z=4 \end{aligned}$
Then,
$\Rightarrow A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{array}\right]$
We know that the system of equations has a unique solution.
$\Rightarrow \left | A \right |$ should not be zero
$\text { i.e. }|A| \neq 0$
$\Rightarrow\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{array}\right| \neq 0$
$\begin{aligned} &\Rightarrow 1(k+2)-1(2 k+3)+1(4-3) \neq 0 \\ &\Rightarrow k+2-2 k-3+1 \neq 0 \\ &\Rightarrow-k \neq 0 \\ &\Rightarrow k=\pm 1, \pm 2, \ldots \ldots \pm n \end{aligned}$

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The RD Sharma class 12th exercise FBQ solution includes the questions from simultaneous linear equations, where students need to find out the system of equations using real numbers. This exercise FBQ has only 7 questions having a system of equations related to a unique solution, no solution, and infinitely many solutions makes it short and concise to solve. The FBQ section is specifically very essential as it covers the entire chapter’s concept.

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