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16. 5x-7y+z=11 6x-8y-z=15 3x+2y-6z=7
CBSE | Class 12 | Determinants | Exercise 6.4 | Maths | RD Sharma
16. 5x-7y+z=11 6x-8y-z=15 3x+2y-6z=7

Solution:

As per the question it is given that,

5x-7y+z=11

6x-8y-z=15

3x+2y-6z=7

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

Then, here we get

5x-7y+z=11

6x-8y-z=15

3x+2y-6z=7

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

\Rightarrow D=\left| \begin{matrix} 5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & -6 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

\Rightarrow D=5\left[ (-8)(-6)-(-1)(2) \right]-7\left[ (-6)(6)-3(-1) \right]+1\left[ 2(6)-3(-8) \right]

\Rightarrow D=5[48+2]-7[-36+3]+1[12+24]

\Rightarrow D=250-231+36

\Rightarrow D=55

Now, solve {{D}_{1}}formed by replacing {{1}^{st}} Colum by B matrices

Here,B=\left| \begin{matrix} 11 \\ 15 \\ 7 \\ \end{matrix} \right|

\Rightarrow {{D}_{1}}=\left| \begin{matrix} 11 & -7 & 1 \\ 15 & -8 & -1 \\ 7 & 2 & -6 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

\Rightarrow {{D}_{1}}=11\left[ (-8)(-6)-(2)(-1) \right]-(-7)\left[ (15)(-6)-(-1)(7) \right]+1\left[ (15)2-(7)(-8) \right]

\Rightarrow {{D}_{1}}=11[48+2]+7[-90+7]+1[30+56]

\Rightarrow {{D}_{1}}=11[50]+7[-83]+86

\Rightarrow {{D}_{1}}=550-581+86

\Rightarrow {{D}_{1}}=55

Then, solve {{D}_{2}} formed by replacing {{1}^{st}} column by B matrices

Here,B=\left| \begin{matrix} 11 \\ 15 \\ 7 \\ \end{matrix} \right|

\Rightarrow {{D}_{2}}=\left| \begin{matrix} 5 & -7 & 11 \\ 6 & -8 & 15 \\ 3 & 2 & 7 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

\Rightarrow {{D}_{2}}=5\left[ (15)(-6)-(7)(-1) \right]-11\left[ (6)(-6)-(-1)(3) \right]+1\left[ (6)7-(15)(3) \right]

⇒ {{D}_{2}}=5[-90+7]-11[-36+3]+1[42-45]

\Rightarrow {{D}_{2}}=5[-83]-11(-33)-3

\Rightarrow {{D}_{2}}=-415+363-3

\Rightarrow {{D}_{2}}=-55`

Then, solve {{D}_{3}} formed by replacing {{1}^{st}} column by B matrices

Here, B=\left| \begin{matrix} 11 \\ 15 \\ 7 \\ \end{matrix} \right|

\Rightarrow {{D}_{2}}=\left| \begin{matrix} 5 & -7 & 11 \\ 6 & -8 & 15 \\ 3 & 2 & 7 \\ \end{matrix} \right|

Solving for the determinant, expanding along 1st row

\Rightarrow {{D}_{3}}=5\left[ (-8)(7)-(15)(2) \right]-(-7)\left[ (6)(7)-(15)(3) \right]+11\left[ (6)2-(-8)(3) \right]

⇒ {{D}_{3}}=5\left[ -56-30 \right]-(-7)[42-45]+11[12+24]

\Rightarrow {{D}_{3}}=5\left[ -86 \right]+7[-3]+11[36]

⇒ {{D}_{3}}=-430-21+396

\Rightarrow {{D}_{3}}=-55

Therefore, by Cramer’s Rule, we will get

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

\Rightarrow X=\frac{55}{55}

\Rightarrow X=1

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

\Rightarrow Y=\frac{-55}{55}

\Rightarrow Y=-1

\Rightarrow Z=\frac{{{D}_{3}}}{D}

\Rightarrow Z=\frac{-55}{55}

\Rightarrow Z=-1

i

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