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Unformatted text preview: use Kirchoff’s First Law to get the three equations i 1 + E + 2( i 1 i 3 ) = 0 2 i 2 + ( i 2 i 3 ) E = 0 2( i 3 i 1 ) + 1( i 3 i 2 ) 12 + 4 i 3 = 0 Kirchoff’s Second Law shows that the current from the source must be equal to the current given by the loop currents, i.e. i 2 i 1 = 1 . Simplifying yields the following system: 3 i 1 2 i 3 + E = 0 3 i 2 i 3 E = 0 2 i 1 i 2 + 7 i 3 = 12 i 1 + i 2 = 0 Then solving this either with Gaussian elimination or substitution yields the unique solution i 1 = 12 11 , i 2 = 12 11 , i 3 = 24 11 , and E = 12 11 . So, the current through the 4Ω resistor is i 3 = 24 11 to the left, and the voltage across the current source is E = 12 11 . 2. Suppose that the reduced row echelon form of a matrix A is 1 2 0 0 0 0 1 0 0 0 0 1 . Were the columns of A linearly dependent or independent? (HINT: consider the solution set to A~x = ~ 0.) Solution: Solving A~x = ~ 0 is done by doing Gaussian elimination on the augmented matrix [ A  ~ 0]. We know what the result will be once this is done; we will get the augmented matrix 1 2 0 0 0 0 1 0...
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This note was uploaded on 03/30/2011 for the course MATH 152 taught by Professor Caddmen during the Spring '08 term at The University of British Columbia.
 Spring '08
 Caddmen
 Equations

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