Problem 81 - Algebra 2

Question

Solve each matrix equation. If the coefficient matrix has no inverse, write \(n o\) unique solution. $$ \left[\begin{array}{rr}{2} & {-1} \\ {-3} & {2}\end{array}\right]\left[\begin{array}{l}{x} \\\ {y}\end{array}\right]=\left[\begin{array}{r}{5} \\ {-10}\end{array}\right] $$

Step-by-Step Solution

Answer

The solution is x = 0 and y = -5.

Step 1: Write down the equation with given matrices

Set up the matrix equation in the form of AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. $$A = \begin{bmatrix}2 & -1\-3 & 2d{bmatrix}, X = \begin{bmatrix}x\yd{bmatrix}, B = \begin{bmatrix}5\-10d{bmatrix}$$

Step 2: Find the determinant of the coefficient matrix

Calculate the determinant of matrix A. If the determinant is 0, the inverse does not exist. Otherwise, proceed to find the inverse. $$det(A) = (2)(2) - (-1)(-3) = 4 - 3 = 1$$

Step 3: Find the inverse of the coefficient matrix

Since the determinant is not zero, matrix A has an inverse. Find the inverse by applying the formula for a 2x2 matrix. $$A^{-1} = \frac{1}{det(A)} \begin{bmatrix} d & -b\ -c & a\ \end{bmatrix} = \begin{bmatrix} 2 & 1\ 3 & 2\end{bmatrix}$$

Step 4: Multiply the inverse of the coefficient matrix by the constant matrix

Multiply matrix A^{-1} by matrix B to find the values of x and y. $$X = A^{-1}B = \begin{bmatrix} 2 & 1\ 3 & 2\end{bmatrix} \begin{bmatrix}5\-10\end{bmatrix} = \begin{bmatrix} (2)(5) + (1)(-10)\ (3)(5) + (2)(-10) \end{bmatrix} = \begin{bmatrix} 10 -10\ 15 - 20\ \end{bmatrix} = \begin{bmatrix} 0\ -5\end{bmatrix}$$

Step 5: Write the solution for the variables

The solution to the matrix equation is the values of x and y that we have obtained. $$x = 0, y = -5$$

Key Concepts

Coefficient Matrix

In the context of matrix equations, a coefficient matrix is a matrix that contains the coefficients of the variables for a system of linear equations. It’s essentially the numerical component of the equation that interacts with the variable matrix.

Let’s look at the setup of a matrix equation, which is commonly noted as AX = B. Here, A represents the coefficient matrix. This matrix is key to understanding the system as it encodes the relationships between different variables.

For example, in the matrix equation provided, the coefficient matrix is: \[ A = \begin{bmatrix}2 & -1\ -3 & 2\end{bmatrix} \]. Each row of the matrix A represents a linear equation, and each column corresponds to a particular variable of the system. In the end, we are aiming to find the values of the variable matrix X which consists of the variables x and y.

Matrix Inverse

A matrix inverse is integral to solving matrix equations. Not every matrix has an inverse; only square matrices who have a non-zero determinant do. If a matrix A has an inverse, it is denoted by A^-1.

The inverse of a matrix is like the reciprocal of a number—it allows us to 'cancel out' the matrix in an equation and isolate the variable matrix X. When we multiply a matrix by its inverse, we get the identity matrix (which acts like the number 1 in matrix arithmetic).

To find the inverse of a 2x2 matrix, like our coefficient matrix A, we use the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b\ -c & a\end{bmatrix} \]. In this problem, the inverse of matrix A was found to be \[ A^{-1} = \begin{bmatrix} 2 & 1\ 3 & 2\end{bmatrix} \], which is then used to solve for X.

Determinant of a Matrix

The determinant is a scalar attribute of a square matrix. It provides crucial information about the matrix, such as whether an inverse exists, and is a factor in solving systems of linear equations.

For a 2x2 matrix, the determinant is calculated by the difference between the products of its diagonals. The formula is: \[ \text{det}(A) = ad - bc \]. In the given exercise, the determinant of A is 1, which confirms that the matrix A does indeed have an inverse, since the determinant is not zero.

The determinant can also inform us about the nature of solutions. If the determinant is zero, the system may not have a unique solution, meaning it could either have no solution or an infinite number of solutions. In this case, because det(A) is 1, we are assured of a unique solution.

Systems of Linear Equations

Systems of linear equations comprise two or more linear equations that share several variables. We can solve these systems using various methods, such as substitution, elimination, or matrix operations.

In matrix form, systems can be succinctly represented and conveniently manipulated. The goal is to find the variable matrix X that satisfies AX = B. With matrix methods, particularly when we employ the inverse of the coefficient matrix, we streamline the solution process. This method is efficient and less error-prone, as it reduces the system to a straightforward matrix multiplication once the inverse has been found.

In the given exercise, the system is represented by two linear equations with two variables, x and y, and the solution, obtained by multiplying the inverted coefficient matrix by the constants matrix, gives us the values of the variables that satisfy both equations simultaneously: \[ x = 0, y = -5 \]. This neat, elegant solution tells us exactly where the lines would intersect on a graph, representing the system visually.