Problem 21 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

Matrix X is \[ X = \left[ \begin{array}{cc} 0 & -5 \\ -25 & -20 \\ -10 & 10 \end{array} \right] \]

Step 1: Understand the equation

The equation given is \( \frac{1}{5}(X + D) = C \). This involves aligning matrix operations and solving for the unknown matrix \( X \). First, let's eliminate the scalar fraction by multiplying both sides of the equation by 5.

Step 2: Eliminate the fraction

Multiply the entire equation by 5 to remove the fraction: \[ X + D = 5C \] This makes our task simpler by getting rid of the fraction.

Step 3: Find 5C

Calculate \( 5C \) to substitute into the equation. Since \( C = \left[ \begin{array}{cc} 2 & 3 \ 1 & 0 \ 0 & 2 \end{array} \right] \), then:\[ 5C = \left[ \begin{array}{cc} 5 \times 2 & 5 \times 3 \ 5 \times 1 & 5 \times 0 \ 5 \times 0 & 5 \times 2 \end{array} \right] = \left[ \begin{array}{cc} 10 & 15 \ 5 & 0 \ 0 & 10 \end{array} \right] \]

Step 4: Substitute 5C

Substitute \( 5C \) into the equation:\[ X + D = \left[ \begin{array}{cc} 10 & 15 \ 5 & 0 \ 0 & 10 \end{array} \right] \]

Step 5: Rearrange for X

We rearrange to solve for \( X \):\[ X = \left[ \begin{array}{cc} 10 & 15 \ 5 & 0 \ 0 & 10 \end{array} \right] - D \]

Step 6: Calculate X

Subtract matrix \( D \) from \( 5C \): Given \( D = \left[ \begin{array}{cc} 10 & 20 \ 30 & 20 \ 10 & 0 \end{array} \right] \), calculate:\[ X = \left[ \begin{array}{cc} 10 & 15 \ 5 & 0 \ 0 & 10 \end{array} \right] - \left[ \begin{array}{cc} 10 & 20 \ 30 & 20 \ 10 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & -5 \ -25 & -20 \ -10 & 10 \end{array} \right] \]

Step 7: Conclusion: Matrix X

The resulting matrix \( X \) after solving the given equation is \[ X = \left[ \begin{array}{cc} 0 & -5 \ -25 & -20 \ -10 & 10 \end{array} \right] \]

Key Concepts

Matrix Subtraction

Matrix subtraction is a fundamental operation in algebra involving matrices. It is similar to the subtraction of regular numbers or scalars, but it operates element-wise between two matrices. For matrix subtraction to be valid, the matrices involved must have the same dimensions.

Given two matrices, let's say matrix A and matrix B, both of size \(m \times n\), the subtraction is performed by subtracting the corresponding elements in the matrices. The result is another \(m \times n\) matrix, where each element is calculated as follows:

• \( C_{ij} = A_{ij} - B_{ij} \)

In our step-by-step solution, we subtracted matrix D from matrix \(5C\). Since both matrices are \(3 \times 2\), the subtraction can be performed without issues.

We calculated: \[ X = \left[ \begin{array}{cc} 10 & 15 \ 5 & 0 \ 0 & 10 \end{array} \right] - \left[ \begin{array}{cc} 10 & 20 \ 30 & 20 \ 10 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & -5 \ -25 & -20 \ -10 & 10 \end{array} \right] \]

Scalar Multiplication

Scalar multiplication in the context of matrices involves multiplying each entry of a matrix by a scalar. This operation is straightforward and easy to perform. If you have a matrix, say matrix \(A\), and a scalar \(k\), the scalar multiplication results in another matrix of the same size as follows:

• \(B_{ij} = k \times A_{ij} \)

In the original problem, we encountered scalar multiplication when eliminating the fraction \(\frac{1}{5}\). We multiplied the entire matrix C by 5, transforming \(C\) to \(5C\):

• First row: \(5 \times 2 = 10\), \(5 \times 3 = 15\)
  
• Second row: \(5 \times 1 = 5\), \(5 \times 0 = 0\)
  
• Third row: \(5 \times 0 = 0\), \(5 \times 2 = 10\)
  

The resulting matrix is \[ \left[ \begin{array}{cc} 10 & 15 \ 5 & 0 \ 0 & 10 \end{array} \right] \] according to the solution.

Matrix Operations

Matrix operations encompass a range of techniques used to manipulate matrices, and these include operations such as addition, subtraction, multiplication, and scalar multiplication. Understanding these operations is vital for solving matrix equations.

Each operation follows specific rules:

• Addition and Subtraction: Same dimensions are required, and operations are done element-wise.
• Multiplication: A matrix of \( m \times n\) can multiply a matrix of \( n \times p\), resulting in a new matrix of size \( m \times p\).
• Scalar Multiplication: Every element in the matrix is multiplied by the scalar.

In our example, we specifically dealt with subtraction and scalar multiplication, where we managed the equation by individually applying these operation rules to achieve the desired results. Understanding and applying the correct rule for each operation ensures the successful manipulation of matrices.

Solving Matrix Equations

Solving matrix equations involves isolating an unknown matrix, just as you would isolate an unknown variable in algebra. The goal is to express the unknown matrix solely in terms of known matrices or scalars.

In our task, we started with the equation \( \frac{1}{5}(X + D) = C \). The initial step is to clear out scalars and fractions to simplify the equation. We do this by multiplying both sides by a scalar, which in this case was 5, to get rid of the fraction. Our equation thus transforms into:
\[ X + D = 5C \]

Next, we solve for the unknown matrix \(X\) by subtracting matrix \(D\):\[ X = 5C - D \]

The subtraction is computed element-wise due to matrix subtraction rules, ensuring that \(X\) is correctly isolated based on the given condition. This process of gradually eliminating other terms until the unknown matrix stands alone is central to solving these equations.