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  • Please provide solution for RD Sharma class 12 chapter 7 Solution of Simultaneous Linear Exercise Fill in the blank Question 3 Maths Textbook solution.

Please provide solution for RD Sharma class 12 chapter 7 Solution of Simultaneous Linear  Exercise Fill in the blank  Question 3 Maths Textbook solution.


 

Answers (1)

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=1We 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.

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