Given:

The first system equation is:

  16I1−9I2=110 ..... (1)

The second system equation is:

  20I2−9I1+110=0 ..... (2)

Calculation:

Solve the system equation (1) for I1

  16I1−9I2=11016I1=110+9I2

Simplify further.

I1=110+9I216 ..... (3)

Substitute 110+9I216 for I1 in equation (2).

  20I2−9( 110+9 I 2 16)+110=020I2−61.875−5.0625I2+110=014.9375I2+48.125=0

Rearrange for I2.

  14.9375I2=−48.125I2=−48.12514.9375=−3.22 A

Substitute −3.22 A for I2 in equation (3).

  I1=110+9( −3.22)16=5.06 A

Conclusion:

Thus, the value of I1 is 5.06 A and I2 is −3.22 A.

Concept used:

The system of two simultaneous equations of variables in the form a1x+b1y=c1 and a2x+b2y=c2 can be written in the form of a matrix equation.

  AX=b as follows:

  [ a 1 b 1 a 2 b 2][xy]=[ c 1 c 2]

Here,

  A=[ a 1 b 1 a 2 b 2] is the 2×2 coefficient matrix.

  X=[xy] is the 2×1 matrix of unknown variables and b=[ c 1 c 2] is the 2×1 constant matrix on the right-hand side of the equation.

Calculation:

The system of equations represented by equation (1) and equation (2) can be written in the matrix form as follows:

  [16−9−920][ I 1 I 2]=[110−110]

Conclusion:

Thus, the matrix form of the equation is [16−9−920][ I 1 I 2]=[110−110].

Concept used:

Write the expression to calculate the value of the current using matrix method.

I=A−1b ...... (4)

Here, A−1 is the inverse of the coefficient matrix A, I is the matrix of unknown currents and b is the coefficient matrix on the right-hand side of the equation.

Write the inverse of the matrix.

A−1=1Δ[adjA] ...... (5)

Here, A is the coefficient matrix, Δ is the determinant of the matrix and [adjA] is the adjoint of the matrix A.

Calculation:

The equation in the form AI=b where A=[16−9−920] is the 2×2 coefficient matrix.

  I=[ I 1 I 2] is the 2×1 matrix of unknown currents and b=[110−110] is the 2×1 coefficient matrix on the right-hand side of the equation.

The determinant of matrix A is given by:

  Δ=| 16 −9 −9 20|=(16)(20)−(−9)(−9)=320−81=239

The adjoint of matrix A is given by:

  [adjA]=[209916]

Substitute 239 for Δ and [209916] for [adjA] in equation (5).

  A−1=1239[209916]

Substitute 1239[209916] for A−1, [ I 1 I 2] for I and [110−110] for b in equation (4).

   [ I 1 I 2 ]=1239[ 20 9 9 16][ 110 −110]=1239[ 2200−990 990−1760]=1239[ 1210 −770]=[ 5.06 −3.22]

Conclusion:

Thus, the value of I1 is 5.06 A and I2 is −3.22 A.

Concept used:

Write the expression for I1 using Cramer's rule.

  I1=| b 1 a 12 b 2 a 22 ||A| ...... (6)

Here, I1 is the one solution of the equation and |A| is the determinant of the matrix.

Write the expression for I2 using Cramer's rule.

  I2=| a 11 b 1 a 21 b 2 ||A| ...... (7)

Here, I2 is the other solution of the equation.

Calculation:

Substitute 110 for b1, −110 for b2, −9 for a12, 20 for a22 and 239 for |A| in equation (6).

  I1=| 110 −9 −110 20 |239=( 2200−990)239=1210239=5.06 A

Substitute 110 for b1, −110 for b2, 16 for a11, −9 for a21 and 239 for |A| in equation (7).

  I2=| 16 110 −9 −110 |239=( −1760+990)239=770239=3.22 A

Conclusion:

Thus, the value of I1 is 5.06 A and I2 is −3.22 A.