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10. x+2y=1 3x+y=4
CBSE | Class 12 | Determinants | Exercise 6.4 | Maths | RD Sharma
10. x+2y=1 3x+y=4

Solution: Assume there be a system of n simultaneous linear equations and with ‘n’ unknown given by

{{a}_{11}}{{X}_{1}}+{{a}_{12}}{{X}_{2}}+...+{{a}_{1n}}{{X}_{n}}={{b}_{1}}

{{a}_{21}}{{X}_{1}}+{{a}_{22}}{{X}_{2}}+...+{{a}_{2n}}{{X}_{n}}={{b}_{2}}

{{a}_{n1}}{{X}_{1}}+{{a}_{n2}}{{X}_{2}}+...+{{a}_{nn}}{{X}_{n}}={{b}_{n}}

LetD=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & \cdots & {{a}_{1n}} \\ {{a}_{21}} & {{a}_{22}} & \cdots & {{a}_{2n}} \\ \vdots & \vdots & {} & \vdots \\ {{a}_{n1}} & {{a}_{n1}} & \cdots & {{a}_{nn}} \\ \end{matrix} \right|Let {{D}_{1}} be the determinant obtained from D after replacing the {{j}^{th}}column by

\left| \begin{matrix} {{b}_{1}} \\ {{b}_{2}} \\ \vdots \\ {{b}_{n}} \\ \end{matrix} \right|Then,

{{X}_{1}}=\frac{{{D}_{1}}}{D},{{X}_{2}}=\frac{{{D}_{2}}}{D},...{{X}_{n}}=\frac{{{D}_{n}}}{D}provided that D\ne 0

x+2y=1

3x+y=4

So by comparing with the theorem, let’s find D,{{D}_{1}}And{{D}_{2}}

\Rightarrow D=\left| \begin{matrix} 1 & 2 \\ 3 & 1 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

\Rightarrow D=1(1)-(3)(2)

\Rightarrow D=1-6

\Rightarrow D=-5

Again,

\Rightarrow {{D}_{1}}=\left| \begin{matrix} 1 & 2 \\ 4 & 1 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

\Rightarrow {{D}_{1}}=1(1)-(2)(4)

\Rightarrow {{D}_{1}}=1-8

\Rightarrow {{D}_{1}}=-7

\Rightarrow {{D}_{2}}=\left| \begin{matrix} 1 & 1 \\ 3 & 4 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

⇒ {{D}_{2}}\text{ }\!\!~\!\!\text{ }=1(4)-(1)(3)

⇒ {{D}_{2}}=4-3

⇒ {{D}_{2}}=1

Therefore, by Cramer’s Rule, we will get

\Rightarrow X=\frac{{{D}_{1}}}{D}

\Rightarrow X=\frac{-7}{-5}

\Rightarrow X=\frac{7}{5}

And

\Rightarrow Y=\frac{{{D}_{2}}}{D}

\Rightarrow Y=\frac{1}{-5}

\Rightarrow Y=-\frac{1}{5}

i

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