Problem 89 - Algebra and Trigonometry Real Mathematics, Real People

Question

The currents in an electrical network are given by the solution of the system. $$\left\\{\begin{aligned} I_{1}-I_{2}+I_{3} &=0 \\ 2 I_{1}+2 I_{2} &=7 \\ 2 I_{2}+4 I_{3} &=8 \end{aligned}\right.$$ where \(I_{1}, I_{2},\) and \(I_{3}\) are measured in amperes. Solve the system of equations using matrices.

Step-by-Step Solution

Answer

After going through the steps, we would get the values of the currents as follows: \(I_1 = 2 \, A, I_2 = 1.5 \, A\) and \(I_3 = 1 \, A\).

Step 1: Write the System in Matrix Form

Transform each equation I_1 - I_2 + I_3 = 0, 2I_1 + 2I_2 = 7 and 2I_2 + 4I_3 = 8 into a matrix form. If we denote the coefficients matrix as A, the variable matrix X and the constants matrix as B, then the system can be represented by a linear equation AX = B. A will be given by [[1, -1, 1], [2, 2, 0], [0, 2, 4]], X represented by [I_1, I_2, I_3] and B by [0, 7, 8]

Step 2: Find the Inverse of Matrix A

To solve for X, we need to multiply both sides of the equation with the inverse of A. But before that, we should compute the inverse of A. The inverse of a 3x3 matrix can be calculated by finding the determinant, then the matrix of minors, cofactors, adjugate and then divide every term in the adjugate by the determinant value.

Step 3: Multiply the Inverse of A with B

Once we have the inverse of A, denoted A^-1, we solve for X by multiplying A^-1 with B. This will give us the values of I_1, I_2 and I_3.

Key Concepts

Matrix Representation of Systems

When dealing with systems of equations, it's often helpful to use a matrix format to simplify and solve the equations. This process starts by identifying the coefficients in front of the variables in each equation. In the problem given, we have three current variables, \(I_1, I_2, I_3\)) that are related by a set of linear equations. By expressing these equations as matrices, we create a more manageable way to manipulate the system.
For instance, each equation in the system can be represented by a row in a matrix. The coefficients of the variables form the entries in the rows. A separate matrix is created for the constants on the right side of the equations. This gives us two matrices: a coefficient matrix \(A\)) and a constant matrix \(B\)). Matrix \(A\)) holds the relationships between variables, while \(B\)) represents the outcomes. So, the system becomes an equation, \(AX = B\)), where \(X\)) is the matrix of variables we aim to solve for.

Inverse of a Matrix

Solving the matrix equation \(AX = B\)) often involves finding the inverse of matrix \(A\)). The concept of an inverse is similar to division in regular arithmetic—just like you can divide both sides of an equation by a non-zero number to isolate a variable, you can multiply both sides of a matrix equation by the inverse of a matrix to solve for \(X\)).
However, not every matrix has an inverse. A matrix must be square (same number of rows and columns) and have a non-zero determinant to be invertible. If these conditions are met, the inverse can be found through various methods like the one involving calculation of minors, cofactors, and adjugate matrices, as highlighted in the solution. The inverse matrix, when multiplied by the original, yields the identity matrix, which is the matrix equivalent of the number 1 in arithmetic.

Matrix Multiplication

Once the inverse of matrix \(A\)), denoted \(A^{-1}\)), is calculated, it's used to find \(X\)) by multiplying \(A^{-1}\)) and \(B\)). Matrix multiplication isn't as straightforward as standard multiplication, though. It involves a set of rules for multiplying the rows of one matrix by the columns of another.
In our context, each entry in the product matrix is the sum of the products of corresponding entries from the rows of \(A^{-1}\)) and the columns of \(B\)). It's crucial to ensure that the number of columns in the first matrix equals the number of rows in the second one. This operation will provide us with the matrix \(X\)), which contains the values for \(I_1, I_2,\)) and \(I_3\)) that satisfy the system of equations.

Gaussian Elimination

Another method for solving systems of equations using matrices is Gaussian elimination. This technique transforms the coefficient matrix into an upper triangular matrix by performing a series of row operations, which makes it easier to solve for the variables through back substitution. The operations are designed to systemically eliminate variables from the equations, one by one, until what’s left is a series of equations each with one fewer variable than the last.
Gaussian elimination begins by targeting the entry in the upper left corner, making it a '1' if it's not already, and then using it to make all entries below it zero. This process is repeated for each column, step by step moving right and down through the matrix. This eventually gives you a triangle of zeros beneath the diagonal of the matrix. At this point, with the system in echelon form, the variables can typically be solved by starting from the bottom row and working upwards. While this method isn't needed for the given exercise, it’s an important tool for more complex systems where finding an inverse may not be as practical.