Problem 16 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Equation

The given matrix equation is \(2A = B - 3X\). Our task is to solve for the matrix \(X\).

Step 2: Rearrange the Equation

To solve for \(X\), first isolate \(3X\) by moving \(B\) to the other side: \[3X = B - 2A\]

Step 3: Calculate Matrices B and 2A

Calculate \(2A\) by multiplying each element of matrix \(A\) by 2: \[2A = \begin{bmatrix}8 & 12 \2 & 6 \end{bmatrix}\] Matrix \(B\) is given as: \[B = \begin{bmatrix}2 & 5 \3 & 7 \end{bmatrix}\]

Step 4: Subtract 2A from B

Calculate \(B - 2A\): \[B - 2A = \begin{bmatrix}2 - 8 & 5 - 12 \3 - 2 & 7 - 6 \end{bmatrix} = \begin{bmatrix}-6 & -7 \1 & 1 \end{bmatrix}\]

Step 5: Solve for X

Now we have \(3X = \begin{bmatrix}-6 & -7 \1 & 1 \end{bmatrix}\). To find \(X\), divide each element of the right-hand side by 3:\[X = \begin{bmatrix}-2 & -\frac{7}{3} \\frac{1}{3} & \frac{1}{3} \end{bmatrix}\]

Step 6: Verify Dimensions of X

The dimensions of \(2A\) and \(B\) are both \(2 \times 2\), confirming that the resulting \(X\) is a \(2 \times 2\) matrix and our operations are valid.

Key Concepts

Matrix Multiplication

Matrix multiplication is a core operation in linear algebra, involving the combination of two matrices to produce a new matrix. Unlike arithmetic multiplication, not every pair of matrices can be multiplied together. For matrix multiplication to be valid, the number of columns in the first matrix must equal the number of rows in the second matrix.
In our step-by-step solution, multiplying a matrix by a scalar (2 in this case) is a specific application of matrix multiplication, often used to scale a matrix. This operation involves multiplying each element of the matrix by the scalar value. For example, multiplying matrix A by 2:

• Matrix A: \[ \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} \]
• Results in 2A: \[ \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix} \]

This simple operation scales the magnitude of the matrix without changing its dimensions.

Matrix Subtraction

Matrix subtraction is similar to matrix addition and involves subtracting the elements of one matrix from the corresponding elements of another. For subtraction to be valid, both matrices must have the same dimensions. Simply put, each element from the first matrix is subtracted from the corresponding element in the second matrix.
In our example problem, we calculated \( B - 2A \) as part of solving for the unknown matrix \( X \). To perform this operation, each element from matrix 2A is subtracted from the corresponding element in matrix B:

• Matrix B: \[ \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} \]
• 2A: \[ \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix} \]
• Result of subtraction \( B - 2A \): \[ \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix} \]

Subtracting matrices element-wise is a straightforward way to find the difference between two matrices, and each element of the matrices plays a role in the final result.

Matrix Division

Matrix division is conceptually different from standard arithmetic division. In matrix algebra, dividing one matrix by another isn't directly possible. Instead, division is done using the concept of the inverse. To "divide" a matrix by another, we typically multiply by the inverse of the matrix.
In the step-by-step solution provided, division was demonstrated by solving the equation \( 3X = B - 2A \) by dividing each element of the resulting matrix by 3. Since matrices themselves do not support direct division, breaking down the solution in a simpler way like scalar division can aid understanding:

• Matrix: \[ \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix} \]
• By dividing each element by 3: \[ \begin{bmatrix} -2 & -\frac{7}{3} \ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \]

In practical terms, solving equations like this often involves decomposing the matrix equation using inverses or scalars.