Problem 22 - Precalculus: Mathematics for Calculus

Question

Solve the matrix equation for the unknown matrix \(X\), or explain why no solution exists. $$\begin{array}{c}A=\left[\begin{array}{ll}4 & 6 \\\1 & 3\end{array}\right] & B=\left[\begin{array}{ll}2 & 5 \\\3 & 7\end{array}\right] \\\C=\left[\begin{array}{ll}2 & 3 \\\1 & 0 \\\0 & 2\end{array}\right] & D=\left[\begin{array}{lr}10 & 20 \\\30 & 20 \\\10 & 0\end{array}\right]\end{array}$$ $$2 A=B-3 X$$

Step-by-Step Solution

Answer

The solution is \(X = \begin{bmatrix} 2 & \frac{7}{3} \\ -\frac{1}{3} & -\frac{1}{3} \end{bmatrix}\).

Step 1: Simplify the Matrix Equation

The given matrix equation is \(2A = B - 3X\). Start by simplifying this equation to express it in terms of the unknown matrix \(X\). By isolating \(-3X\) on one side, the equation becomes \(-3X = B - 2A\). To simplify, our goal is to express \(X\). We will need to do further operations to isolate \(X\).

Step 2: Substitute Matrices A and B

Substitute matrices \(A\) and \(B\) into the equation \(-3X = B - 2A\):\[2 \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - 3X\] Calculate \(2A\):\[2 \cdot \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix}\] Substitute back:\[-3X = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix}\]

Step 3: Perform Matrix Subtraction

Perform the subtraction of matrices \( B - 2A \):\[-3X = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix}\]\[= \begin{bmatrix} 2 - 8 & 5 - 12 \ 3 - 2 & 7 - 6 \end{bmatrix}\]\[= \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\] Now, the equation becomes:\[-3X = \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\]

Step 4: Solve for X

Since we have \(-3X = \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\), we solve for \(X\) by dividing both sides by \(-3\).\[X = \frac{1}{-3} \cdot \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\]This results in:\[X = \begin{bmatrix} -6/-3 & -7/-3 \ 1/-3 & 1/-3 \end{bmatrix}\]Simplify:\[X = \begin{bmatrix} 2 & rac{7}{3} \ -rac{1}{3} & -rac{1}{3} \end{bmatrix}\]

Key Concepts

Matrix Subtraction

Matrix subtraction is a fundamental operation that involves element-wise subtraction between two matrices of the same dimensions. In this exercise, matrix subtraction is used to simplify the expression \(B - 2A\). Both matrices \(B\) and \(2A\) must have the same number of rows and columns to be subtracted, which they do in this problem (each having dimensions 2x2).

To perform matrix subtraction:

• Subtract each corresponding element in the matrices.
• Write down the result as a new matrix.

For example, in the subtraction \(-3X = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix}\):
- The element in the first row, first column is \(2 - 8 = -6\).
- The element in the first row, second column is \(5 - 12 = -7\).
Continue this process for each element. This results in a new matrix, \(\begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\), which is what we see in the step-by-step solution.

Matrix Multiplication

Matrix multiplication is another crucial operation in solving matrix equations. In our problem, multiplication is seen in the context of scalar multiplication and the operation \(-3X\). Scalar multiplication involves multiplying every element of the matrix by the scalar value. Here, the equation presents this when modifying \(\begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\) with the scalar \(-1/3\).

To perform scalar multiplication:

• Multiply each element of the matrix by the scalar. For example, if multiplying by \(-1/3\), multiply each element of \(-3X\) by \(-1/3\).

For instance, \(-6\) becomes \(-6 \times (-1/3) = 2\), and similarly, \(-7\) becomes \(-7 \times (-1/3) = 7/3\).

Matrix multiplication is a more complex operation when not dealing with scalars and requires compatible matrices, where the number of columns in the first matrix matches the number of rows in the second matrix. Always check compatibility before performing matrix multiplication in a broader context.

Solving Linear Equations

Solving linear equations in matrix form is a powerful method for finding unknown variables arranged in matrices. The given exercise illustrates this method by solving the equation \(2A = B - 3X\) for the unknown matrix \(X\).

Steps to solve matrix equations:

• First, isolate the term containing the unknown matrix. Start by rearranging the equation to collect like terms on appropriate sides. For example, moving \(-3X\) to one side in our exercise gave \(-3X = B - 2A\).
• Perform necessary matrix operations like addition, subtraction, or multiplication on each side to simplify the matrices involved.
• If needed, apply scalar multiplicative division to each element within the matrix to further simplify and isolate the variable as shown when dividing by \(-3\).

Finally, compute the transformations to derive the unknown matrix. Here, dividing the matrix \(\begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\) by \(-3\) yields \(X\), resulting in \(\begin{bmatrix} 2 & \frac{7}{3} \ -\frac{1}{3} & -\frac{1}{3} \end{bmatrix}\). Ensure steps are presented clearly and operations validated for accurate solutions.