Problem 36 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the matrix \(A\) if $$\left[\begin{array}{rr}2 & 2 \\\\-1 & 3\end{array}\right] A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$$

Step-by-Step Solution

Answer

The short answer is: \(A = \begin{bmatrix} \frac{7}{8} & \frac{-1}{4} \\ \frac{5}{8} & \frac{5}{4}\end{bmatrix}\).

Step 1: Identify the matrix to invert

We are given the following equation: \(\begin{bmatrix}2 & 2 \\ -1 & 3 \end{bmatrix} A = \begin{bmatrix}3 & 2 \\ 1 & 4 \end{bmatrix}\) We want to find the matrix \(A\), so we must invert the matrix on the left side of the equation: \(\begin{bmatrix}2 & 2 \\ -1 & 3 \end{bmatrix}\)

Step 2: Compute the inverse of the matrix

To calculate the inverse of a 2x2 matrix, we use the formula: \((\begin{bmatrix}a & b \\ c & d\end{bmatrix})^{-1} = \frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}\) In our case, the given matrix is: \(\begin{bmatrix}2 & 2 \\ -1 & 3\end{bmatrix}\) So, we have \(a = 2\), \(b = 2\), \(c = -1\), and \(d = 3\). Now, compute the inverse matrix: \((\begin{bmatrix}2 & 2 \\ -1 & 3\end{bmatrix})^{-1} = \frac{1}{(2)(3) - (2)(-1)} \begin{bmatrix}3 & -2 \\ 1 & 2\end{bmatrix}\) \((\begin{bmatrix}2 & 2 \\ -1 & 3\end{bmatrix})^{-1} = \frac{1}{6 + 2} \begin{bmatrix}3 & -2 \\ 1 & 2\end{bmatrix}\) \((\begin{bmatrix}2 & 2 \\ -1 & 3\end{bmatrix})^{-1} = \frac{1}{8} \begin{bmatrix}3 & -2 \\ 1 & 2\end{bmatrix}\)

Step 3: Multiply both sides of the equation by the inverse matrix

Now that we have the inverse matrix, we can multiply both sides of the given equation to isolate matrix A: \((\begin{bmatrix}2 & 2 \\ -1 & 3\end{bmatrix})^{-1}(\begin{bmatrix}2 & 2 \\ -1 & 3 \end{bmatrix} A) = (\begin{bmatrix}2 & 2 \\ -1 & 3\end{bmatrix})^{-1} \begin{bmatrix}3 & 2 \\ 1 & 4 \end{bmatrix} \) By property of inverse matrices, the left side will simplify to matrix \(A\): \(A = \frac{1}{8} \begin{bmatrix}3 & -2 \\ 1 & 2\end{bmatrix} \begin{bmatrix}3 & 2 \\ 1 & 4 \end{bmatrix}\)

Step 4: Multiply the matrices to find matrix A

Now, we have to multiply both matrices on the right side of the equation: \(A = \frac{1}{8} \begin{bmatrix} (3)(3) + (-2)(1) & (3)(2) + (-2)(4) \\ (1)(3) + (2)(1) & (1)(2) + (2)(4)\end{bmatrix}\) \(A = \frac{1}{8} \begin{bmatrix} 7 & -2 \\ 5 & 10\end{bmatrix}\) Finally, divide each entry in the matrix by 8: \(A = \begin{bmatrix} \frac{7}{8} & \frac{-1}{4} \\ \frac{5}{8} & \frac{5}{4}\end{bmatrix}\) So, the matrix A is: \[A = \begin{bmatrix} \frac{7}{8} & \frac{-1}{4} \\ \frac{5}{8} & \frac{5}{4}\end{bmatrix}\]

Key Concepts

Matrix Inversion

In the world of matrix algebra, finding the inverse of a matrix is much like finding the reciprocal of a number. Not every matrix has an inverse, but those that do provide convenient solutions to certain equations. When dealing with a 2x2 matrix, the inverse can be found using a straightforward formula. The formula for the inverse of matrix \[\begin{bmatrix}a & b \ c & d\end{bmatrix}\]is given by:\[(\begin{bmatrix}a & b \ c & d\end{bmatrix})^{-1} = \frac{1}{ad - bc} \begin{bmatrix}d & -b \ -c & a \end{bmatrix}\].

• ad - bc is known as the determinant of the matrix.
• The determinant must be non-zero for the matrix to have an inverse.
• The elements of the original matrix are rearranged and signs are changed as per the formula to obtain the inverse.

Finding the inverse is crucial for solving matrix equations where you want to isolate a particular matrix variable, just as you would solve for x in linear equations. Remember, the determinant should not be zero; if it is, the inverse does not exist.

Matrix Multiplication

Matrix multiplication is quite different from regular multiplication. It involves combining rows and columns from two matrices to produce a new matrix. Let's say we have two matrices, \(\begin{bmatrix}a & b \ c & d\end{bmatrix}\)and \(\begin{bmatrix}e & f \ g & h\end{bmatrix}\).These matrices can be multiplied as follows:\[\begin{bmatrix}a & b \ c & d\end{bmatrix} \begin{bmatrix}e & f \ g & h\end{bmatrix} = \begin{bmatrix}ae + bg & af + bh \ ce + dg & cf + dh\end{bmatrix}\].

• Each element of the resulting matrix is a sum of products from the rows of the first matrix and the columns of the second matrix.
• The number of columns in the first matrix should match the number of rows in the second matrix for multiplication to be possible.
• This operation is not commutative, meaning \(AB\) does not necessarily equal \(BA\).

In our exercise, after determining the inverse of a matrix, we multiply to find the product needed to solve for matrix A. This highlights the importance of understanding the order and process of multiplication in matrix operations.

Linear Equations

Linear equations are foundational in mathematics, providing a method to express relationships using algebraic expressions. In matrix terms, systems of linear equations can be represented compactly using matrices. Consider a simple system represented as \(Ax = b\), where:

• A is a matrix representing coefficients.
• x is a column matrix representing variables.
• b is a column matrix representing constant terms.

To find x, you need to solve the matrix equation. If A is invertible, the solution is straightforward: multiply both sides by the inverse of A, giving \(x = A^{-1}b\). This approach is particularly useful when extending to larger systems of equations, as it leverages the power of matrix operations to handle several equations simultaneously. Recognizing how matrices can simplify and solve systems of linear equations is essential both in theoretical mathematics and practical applications such as engineering and economics.