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I'm not sure if this even belongs in the calculus forum because it is rather basic, but here it goes.

I'm working on this circuit problem and I could NOT figure out what I was doing wrong.

My solution has led me to the following system of equations:

(1) [tex] (2+j2)\bar I_1 +(j1)\bar I_2 = 10 [/tex]

(2) [tex] (j)\bar I_1 +(2+j2)\bar I_2 = 0 [/tex]

I then bought a cramster account since they have the solutions for my book. Their solution is the same system of equations (1) and (2).

HOWEVER. When the book AND the cramster solution solves for [itex] \bar I_2 [/itex] they yield the following:

(by the way I just need [itex] \bar I_2 [/itex] to solve the problem)

[tex] \bar I_2 = \frac{10}{-2-j} =-4+2j[/tex]

Now when I solve the following matrix (with the help of a calculator) I have:

[tex] \left( \begin{array}{cc} 2+j2 & j \\ j & 2+j2 \end{array} \right) \left( \begin{array}{c} \bar I_1 \\ \bar I_2 \end{array} \left) = \left( \begin{array}{c} 10 \\ 0 \end{array} \right) [/tex]

Solving yields the following result:

(1*) [tex] \bar I_1 = \frac{36}{13} - \frac{28}{13}j [/tex]

(2*) [tex] \bar I_2 = -\frac{16}{13} - \frac{2}{13}j [/tex]

Now if I plug (1*) and (2*) back into (1) and (2) I get [itex] 10 [/itex] and [itex] 0 [/itex] respectively.

So what gives? Is the book and cramster wrong?

I haven't taken complex analysis, so maybe there are more then one solutions to [itex] \bar I_2 [/itex] How do I know which one is right?

grr... I was tearing my hair out thinking I did something wrong.

I'm working on this circuit problem and I could NOT figure out what I was doing wrong.

My solution has led me to the following system of equations:

(1) [tex] (2+j2)\bar I_1 +(j1)\bar I_2 = 10 [/tex]

(2) [tex] (j)\bar I_1 +(2+j2)\bar I_2 = 0 [/tex]

I then bought a cramster account since they have the solutions for my book. Their solution is the same system of equations (1) and (2).

HOWEVER. When the book AND the cramster solution solves for [itex] \bar I_2 [/itex] they yield the following:

(by the way I just need [itex] \bar I_2 [/itex] to solve the problem)

[tex] \bar I_2 = \frac{10}{-2-j} =-4+2j[/tex]

Now when I solve the following matrix (with the help of a calculator) I have:

[tex] \left( \begin{array}{cc} 2+j2 & j \\ j & 2+j2 \end{array} \right) \left( \begin{array}{c} \bar I_1 \\ \bar I_2 \end{array} \left) = \left( \begin{array}{c} 10 \\ 0 \end{array} \right) [/tex]

Solving yields the following result:

(1*) [tex] \bar I_1 = \frac{36}{13} - \frac{28}{13}j [/tex]

(2*) [tex] \bar I_2 = -\frac{16}{13} - \frac{2}{13}j [/tex]

Now if I plug (1*) and (2*) back into (1) and (2) I get [itex] 10 [/itex] and [itex] 0 [/itex] respectively.

So what gives? Is the book and cramster wrong?

I haven't taken complex analysis, so maybe there are more then one solutions to [itex] \bar I_2 [/itex] How do I know which one is right?

grr... I was tearing my hair out thinking I did something wrong.

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