Let us apply our knowledge of matrices to assist us in the analysis of an electric circuit. We consider the circuit shown below.

Figure 1: Electric circuit with two node voltages specified.

In this example, we wish to solve for the two node voltages v₁ and v₂. Since there are two unknowns in this problem, we must first establish two independent equations that reflect the operation of the circuit.

Kirchoff’s Current Law tells us that the sum of the currents that enter a node must equal the sum of the currents that leave a node. Let us focus first on node 1. The current that enters node 1 from the left can be stated mathematically as

The current that enters node 1 from the right can be stated as

The current that travels downward from node 1 is

We can arrange the expressions for each of the currents in terms of an equation via Kirchoff’s Current Law

We can combine and rearrange these terms into the equation

Now let us turn our attention to node 2. The current entering node 2 from the left is given by the expression

The current entering node 2 from the right is 2 A. The current leaving node 2 in a downward direction is

We proceed to combine these currents via Kirchoff’s Current Law

This equation can be rearranged as

So the pair of equations that we will use to solve for the two unknowns are

and

These equations may be expressed in matrix-vector form as

or

Let us find the inverse of the matrix A. The coefficients of this matrix are given by

The inverse can be found making use of the following formula

It should be noted that this formula works only with (2 x 2) matrices. For matrices of higher rank, other methods need to be applied.

For this example, the determinant of A is found as

We can incorporate this information to express the inverse matrix as

which can be written as

We can apply A^-1 to solve for the unknowns

Recognizing that A^-1A = I, we find that

So the node voltages are given as

and

Let us consider the electric circuit that is shown below.

Figure 2: Electric circuit with three unknown currents.

Suppose that we are interested in determining the value of the three unknown currents I₁, I₂ and I₃. In order to do so, we rely upon Ohm’s Law and Kirchoff’s Laws to develop a system of three independent, linear equations. We should note that because we have three unknowns (I₁, I₂ and I₃), we must have three independent, linear equations.

Let us define the matrix A=111−2300−36A=111−2300−36 size 12{A= left [ matrix { 1 {} # 1 {} # 1 {} ## - 2 {} # 3 {} # 0 {} ## 0 {} # - 3 {} # 6{} } right ]} {}

In order to find the unknowns, we must first find the inverse of the matrix A. This can be accomplished using elimination. To start the process, we adjoin the vector [0 24 0]^T to the matrix A.

Next, we wish to force the left-most constant of row 2 to take on a value of 0. We can do so by multiplying each value in the first row by (-2) and subtracting the result from the corresponding value in row 2. This process yields

Now, we divide each term in row 2 by (5) to yield

Next, we turn our attention to eliminating the (-3) term in row 3. We can do so by multiplying each term of row 2 by (-3) and subtracting the results from the corresponding terms in row 3. This produces the matrix

We can then divide the terms of row 3 by (36/5) to produce

Interpretation of the third row tells us that the value for the third unknown (I₃) is 2 A. We can use the coefficients from the second row along with the value for I₃ to solve for I₂.

which yields the result

Lastly, we may use the coefficients of the first row along with the previously determined values for I₂ and I₃ to produce the result for I₁.

I 1 + I 2 + I 3 = 0 I 1 + I 2 + I 3 = 0 size 12{I rSub { size 8{1} } +I rSub { size 8{2} } +I rSub { size 8{3} } =0} {}

Insertion of the previously found unknowns yields

So we find the value for I₁ to be -6 A.