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3. 2x-y=17 3x+5y=6
CBSE | Class 12 | Determinants | Exercise 6.4 | Maths | RD Sharma
3. 2x-y=17 3x+5y=6

Solution:

According to the question it is given that, 2x-y=17

And 3x+5y=6

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

Now, here we have

2x-y=17

3x+5y=6

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

\Rightarrow D=\left| \begin{matrix} 2 & -1 \\ 3 & 5 \\ \end{matrix} \right|Solving for the determinant, expanding along 1st row

\Rightarrow {{D}_{1}}=17(5)-(6)(-1)\text{ }~\text{ }

\Rightarrow {{D}_{2}}=85+6

⇒ {{D}_{1}}=91

\Rightarrow {{D}_{2}}=\left| \begin{matrix} 2 & 17 \\ 3 & 6 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

⇒ D2 = 2(6) – (17) (3)

⇒ D2 = 12 – 51

⇒ D2 = – 39

So, by Cramer’s Rule, we will get

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

\Rightarrow X=\frac{91}{13}

\Rightarrow X=7

And

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

\Rightarrow Y=\frac{-39}{13}

\Rightarrow Y=-3

i

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