Question

Solve each system of equations by calculator using the unit matrix method. Two Equations in Two Unknowns. $$\begin{aligned} &\frac{x}{2}-\frac{3 y}{4}=0\\\ &\frac{x}{3}+\frac{y}{4}=0 \end{aligned}$$

Short answer

By using a calculator to find the inverse of the given coefficient matrix and then multiplying it by the constants matrix, the solution to the system of equations is found.

Step-by-step solution

Write the system of equations in matrix form

Express the given system of equations in matrix form as AX = B, where A is the coefficient matrix, X is the matrix of variables, and B is the constants matrix. In this case:A = \[\begin{pmatrix}\frac{1}{2} & -\frac{3}{4} \frac{1}{3} & \frac{1}{4}\end{pmatrix}\], X = \[\begin{pmatrix}x \ y\end{pmatrix}\], and B = \[\begin{pmatrix}0 \ 0\end{pmatrix}\].

Find the inverse of matrix A

Calculate the inverse of matrix A, denoted by A^{-1}. To do this by calculator, enter the elements of matrix A and use the calculator's function to find the inverse.

Multiply the inverse of matrix A by matrix B

Once you have A^{-1}, multiply it by matrix B to find the values of X. This operation is done by using the calculator's matrix multiplication function. This will give you the solution to the system of equations.

Key concepts

Matrix Form Representation

To tackle the problem of solving systems of equations efficiently, converting them into matrix form is an excellent first step. This conversion streamlines the process, especially when using a calculator or computer software. Given a system of linear equations, the matrix form representation is denoted as \(AX = B\). Here, \(A\) is referred to as the coefficient matrix, representing the coefficients of the variables in the system. \(X\) stands for the matrix consisting of the variables, and \(B\) is the constants matrix, embodying the terms that remain on the other side of the equals sign.

In our example, the coefficient matrix \(A\) is constructed by taking the coefficients of \(x\) and \(y\) from each equation and placing them in a two-row matrix, as such:
\[A = \begin{pmatrix}\frac{1}{2} & -\frac{3}{4}\ \frac{1}{3} & \frac{1}{4}\end{pmatrix}\]
The variable matrix \(X\) consists of \(x\) and \(y\) placed in a column:
\[X = \begin{pmatrix}x\ y\end{pmatrix}\]
And the constants matrix \(B\) contains the constants from the right side of each equation, which are both zero in this case:
\[B = \begin{pmatrix}0\ 0\end{pmatrix}\]
Understanding this representation is foundational to grasping more advanced matrix operations and to using technological tools for solving systems of equations.

Inverse of a Matrix

The inverse of a matrix, denoted as \(A^{-1}\), plays a pivotal role in solving systems of equations. For a given square matrix \(A\), if an inverse exists, then \(A^{-1}A = AA^{-1} = I\), where \(I\) is the identity matrix. The inverse is akin to the reciprocal of a number, but for matrices.

To find the inverse of a matrix, one must ensure that the matrix is non-singular, which means it must have a non-zero determinant. If the determinant of \(A\) is zero, \(A\) has no inverse. For a 2x2 matrix, such as in our example, the inverse can be calculated using a formula. However, for larger matrices, the process can be more complex, often involving row operations or algorithms such as Gaussian elimination.

For the given matrix \(A\), finding \(A^{-1}\) would be the next logical step after establishing the matrix form. Using a calculator, this step is greatly simplified, as most scientific calculators have a built-in function to calculate the inverse of a matrix. Once the inverse is obtained, it can be used to isolate the variable matrix \(X\) in the matrix equation \(AX = B\) by multiplying both sides by \(A^{-1}\), leading us to \(X = A^{-1}B\).

Unit Matrix Method

The unit matrix method involves using the properties of the identity matrix, also known as the unit matrix, to find the solution set of a system of equations. The identity matrix \(I\) is a special type of diagonal matrix where all the diagonal elements are '1', and all off-diagonal elements are '0'. For any matrix \(A\), multiplying it by the identity matrix of compatible dimensions yields \(A\) itself, which is the fundamental feature exploited in this method.

Finding the Solution Using the Unit Matrix Method

Once we have the inverse \(A^{-1}\), we utilize the identity matrix to solve for \(X\) in the equation \(AX = B\), by performing the multiplication \(A^{-1}A = I\). Since \(A^{-1}AX = IX = X\), the result is the variable matrix we seek. Then, by multiplying both sides of the original equation by \(A^{-1}\), we derive \(A^{-1}AX = A^{-1}B\), which simplifies to \(X = A^{-1}B\). By performing this multiplication with a calculator, we retrieve the solution to our system of equations.

It is essential to note that the success of the unit matrix method hinges on the existence of the inverse of \(A\). Without it, we would have to explore other methods, such as row reduction or iterative approximations, to solve the system of equations.