I will begin by saying hello! I will provide a picture and my objectives with the question.

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I am wanting to present the following loops in the matrix for $\displaystyle Ax=B$ We will call that "a)". Additionally, I want to solve for b) $\displaystyle I_1$, $\displaystyle I_2$ and $\displaystyle I_3$, providing their sense. Lastly, I want to determine c) the power dissipated in the $\displaystyle 6.8k\Omega$ resitor.

a)

$\displaystyle I_1: (2.2k\Omega+9.1k\Omega)I_1-(9.1k\Omega)I_2=18V$

$\displaystyle I_2: (9.1k\Omega+7.5k\Omega+6.8k\Omega)I_2-(9.1k\Omega)I_1-(6.8k\Omega)I_3=-18V$

$\displaystyle I_3: (6.8k\Omega+3.3k\Omega)I_3-(6.8k\Omega)I_2=-3V$

and hence, we have our values for putting it in the form of Ax=B

____11.3 -9.1 0

A= -9.1 23.4 -6.8

____0 -6.8 10.1

___ I1

x = I2

___ I3

____18

B= -18

____-3

I apologize but I don't yet know how to execute matrices via latex code. Regardless, that's part a).

b)

First, I will determine the determinate for A, since it applies to all of the currents.

det(A)=(11.3*23.4*10.1)-(11.3*-6.8*-6.8)+(-9.1*-9.1*10.1)-(-9.1*-6.8*0)+0-0

det(A)=2984.511

For I1, det(A1)=(18*23.4*10.1)-(18*-6.8*-6.8)+(-9.1*-18*10.1)-(-9.1*-6.8*-3)=5261.82

and hence I1=5261.82/2984.511=1.76A clockwise.

I2= -3939.24/2984.511=-1.32A= 1.32 A counter-clockwise.

I3= -1310.97/2984.511=-0.439A = 0.439 A counter-clockwise.

c) The power dissipated in the 6.8-kolm resistor will be affected by currents I2 and I3, which are I=1.32A+0.439A=1.76A

P=I^2*R=(1.76A)^2*(6.8*10^3 olm) =21.0 kJ