Question

* The voltage in an electric circuit are related by following equations :
V1 + V2 + V3 = 9; V1 – V2 + V3 = 3; V1 + V2 – V3 = 1 find V1, V2 and V3 by using
Cramer’s rule.

Answer 1

Given:

The given system of equations are

V₁ + V₂ + V₃ = 9

V₁ - V₂ + V₃ = 3

V₁ + V₂ - V₃ = 1

To find:

We have to find the values of V₁,V₂and V₃ by cramer's rule

Solution:

The coefficient matrix of given equations is  $\left[\begin{array}{ccc}1&1&1\\1&-1&1\\1&1&-1\end{array}\right]$

And Answer column is $\left[\begin{array}{c}9\\3\\1\end{array}\right]$

Now D= determinant of coefficient matrix = $1[(-1)(-1)-(1)(1)] -1[(1)(-1)-(1)(1)]+1[(1)(1)-(1)(-1)]$

⇒$1[1-1]-1[-1-1]+1[1+1]$

⇒$0+2+2$

⇒4

∴D = 4

$D_x$ = Determinant of coefficient matrix with answer column in V₁ column

$D_x$ = $\left|\begin{array}{ccc}9&1&1\\3&-1&1\\1&1&-1\end{array}\right|$ = $9(1-1)-1(-3-1)+1(3+1)$ = $4+4$ = $8$

$D_y$ = Determinant of coefficient matrix with answer column in V₂ column

$D_y$ =  $\left|\begin{array}{ccc}1&9&1\\1&3&1\\1&1&-1\end{array}\right|$ = $1(-3-1)-9(-1-1)+1(1-3)$ = $-4+18-2$ = $12$

$D_z$ =  Determinant of coefficient matrix with answer column in V₃ column

$D_z$ =  $\left|\begin{array}{ccc}1&1&9\\1&-1&3\\1&1&1\end{array}\right|$ = $1(-1-3)-1(1-3)+9(1-(-1))$ = $-4+2+18$ = $16$

So according to cramer's rule,

V₁ = $\frac{D_x}{D}$ = $\frac{8}{4}$ = $2$

V₂ = $\frac{D_y}{D}$ = $\frac{12}{4}$ = $3$

V₃ = $\frac{D_z}{D}$ = $\frac{16}{4}$ = $4$

∴V₁ = 2

  V₂ = 3

  V₃ = 4

Answer 2

Answer:

The given system of equations are

V₁ + V₂ + V₃ = 9

V₁ - V₂ + V₃ = 3

V₁ + V₂ - V₃ = 1

To find:

We have to find the values of V₁,V₂and V₃ by cramer's rule

Solution:

The coefficient matrix of given equations is \begin{gathered}\left[\begin{array}{ccc}1&1&1\\1&-1&1\\1&1&-1\end{array}\right]\end{gathered}

⎣

⎡

1

1

1

1

−1

1

1

1

−1

⎦

⎤

And Answer column is \begin{gathered}\left[\begin{array}{c}9\\3\\1\end{array}\right]\end{gathered}

⎣

⎡

9

3

1

⎦

⎤

Now D= determinant of coefficient matrix = 1[(-1)(-1)-(1)(1)] -1[(1)(-1)-(1)(1)]+1[(1)(1)-(1)(-1)]1[(−1)(−1)−(1)(1)]−1[(1)(−1)−(1)(1)]+1[(1)(1)−(1)(−1)]

⇒1[1-1]-1[-1-1]+1[1+1]1[1−1]−1[−1−1]+1[1+1]

⇒0+2+20+2+2

⇒4

∴D = 4

D_xD

x

= Determinant of coefficient matrix with answer column in V₁ column

D_xD

x

= \begin{gathered}\left|\begin{array}{ccc}9&1&1\\3&-1&1\\1&1&-1\end{array}\right|\end{gathered}

∣

∣

9

3

1

1

−1

1

1

1

−1

∣

∣

= 9(1-1)-1(-3-1)+1(3+1)9(1−1)−1(−3−1)+1(3+1) = 4+44+4 = 88

D_yD

y

= Determinant of coefficient matrix with answer column in V₂ column

D_yD

y

= \begin{gathered}\left|\begin{array}{ccc}1&9&1\\1&3&1\\1&1&-1\end{array}\right|\end{gathered}

∣

∣

1

1

1

9

3

1

1

1

−1

∣

∣

= 1(-3-1)-9(-1-1)+1(1-3)1(−3−1)−9(−1−1)+1(1−3) = -4+18-2−4+18−2 = 1212

D_zD

z

= Determinant of coefficient matrix with answer column in V₃ column

D_zD

z

= \begin{gathered}\left|\begin{array}{ccc}1&1&9\\1&-1&3\\1&1&1\end{array}\right|\end{gathered}

∣

∣

1

1

1

1

−1

1

9

3

1

∣

∣

= 1(-1-3)-1(1-3)+9(1-(-1))1(−1−3)−1(1−3)+9(1−(−1)) = -4+2+18−4+2+18 = 1616

So according to cramer's rule,

V₁ = \frac{D_x}{D}

D

D

x

= \frac{8}{4}

4

8

= 22

V₂ = \frac{D_y}{D}

D

D

y

= \frac{12}{4}

4

12

= 33

V₃ = \frac{D_z}{D}

D

D

z

= \frac{16}{4}

4

16

= 44

∴V₁ = 2

V₂ = 3

V₃ = 4