Question #350832

The following were obtained by applying Kirchoff’s laws to an electric circuit


2IA +IB −IC = −8


−IA +IB +IC = 3


−2IA +4IC = 18


1
Expert's answer
2022-06-16T09:49:51-0400

Augmented matrix


(2118111320418)\begin{pmatrix} 2 & 1 & -1 & & -8 \\ -1 & 1 & 1 & & 3 \\ -2 & 0 & 4 & & 18 \\ \end{pmatrix}

R1=R1/2R_1=R_1/2


(11/21/24111320418)\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ -1 & 1 & 1 & & 3 \\ -2 & 0 & 4 & & 18 \\ \end{pmatrix}

R2=R2+R1R_2=R_2+R_1


(11/21/2403/21/2120418)\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ 0 & 3/2 & 1/2 & & -1 \\ -2 & 0 & 4 & & 18 \\ \end{pmatrix}

R3=R3+2R1R_3=R_3+2R_1


(11/21/2403/21/2101310)\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ 0 & 3/2 & 1/2 & & -1 \\ 0 & 1 & 3 & & 10 \\ \end{pmatrix}

R2=2R2/3R_2=2R_2/3


(11/21/24011/32/301310)\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 1 & 3 & & 10 \\ \end{pmatrix}

R1=R1R2/2R_1=R_1-R_2/2


(102/311/3011/32/301310)\begin{pmatrix} 1 & 0 & -2/3 & & -11/3 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 1 & 3 & & 10 \\ \end{pmatrix}

R3=R3R2R_3=R_3-R_2


(102/311/3011/32/3008/332/3)\begin{pmatrix} 1 & 0 & -2/3 & & -11/3 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 0 & 8/3 & & 32/3 \\ \end{pmatrix}

R3=3R3/8R_3=3R_3/8


(102/311/3011/32/30014)\begin{pmatrix} 1 & 0 & -2/3 & & -11/3 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 0 & 1 & & 4 \\ \end{pmatrix}

R1=R1+2R3/3R_1=R_1+2R_3/3


(1001011/32/30014)\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 0 & 1 & & 4 \\ \end{pmatrix}

R2=R2R3/3R_2=R_2-R_3/3


(100101020014)\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & 0 & & -2\\ 0 & 0 & 1 & & 4 \\ \end{pmatrix}

IA=1,IB=2,IC=4I_A=-1, I_B=-2, I_C=4

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS