Problem 15 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The solution is \( X = \begin{bmatrix} -1 & -0.5 \\ 1 & 2 \end{bmatrix} \).

Step 1: Understand the Equation

The given equation is \( 2X + A = B \), where \( A \) and \( B \) are matrices, and \( X \) is the unknown matrix we need to solve for. The goal is to isolate \( X \).

Step 2: Isolate the Matrix X

To isolate \( X \), first subtract matrix \( A \) from both sides of the equation. This gives:\[ 2X = B - A \]

Step 3: Calculate Matrix B - A

Subtract matrix \( A \) from matrix \( B \):\[B - A = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 - 4 & 5 - 6 \ 3 - 1 & 7 - 3 \end{bmatrix} = \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix}.\]

Step 4: Solve for X by Dividing by 2

Now divide the resulting matrix from Step 3 by 2:\[ X = \frac{1}{2} \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix} = \begin{bmatrix} -1 & -0.5 \ 1 & 2 \end{bmatrix}. \]

Step 5: Verify the Solution

Multiply the matrix \( X = \begin{bmatrix} -1 & -0.5 \ 1 & 2 \end{bmatrix} \) by 2 and add matrix \( A \) to check if the result is matrix \( B \):\[ 2X = \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix}, \]then \( 2X + A = \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} = B \). This verifies the solution is correct.

Key Concepts

Matrix Subtraction

Matrix subtraction is an essential operation in linear algebra. It's similar to regular subtraction, but instead of numbers, we subtract matrices element-by-element. For two matrices, say, \( A \) and \( B \), to be subtractable, they must have the same dimensions. This means they must have the same number of rows and columns.
To subtract matrix \( A \) from matrix \( B \), input the corresponding elements:

• Locate the same position in both matrices.
• Subtract the element of matrix \( A \) from the element of matrix \( B \).

For example, if you have \( B = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} \) and \( A = \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} \), the subtraction will be: \[ B - A = \begin{bmatrix} 2 - 4 & 5 - 6 \ 3 - 1 & 7 - 3 \end{bmatrix} = \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix} \].
Each element in matrix \( B \) is reduced by its corresponding element in matrix \( A \). This results in a new matrix that contains the differences of each element.

Matrix Division

Matrix division is a bit different from regular numerical division. Since matrices are multi-dimensional arrays, you cannot directly divide one matrix by another. Instead, we use a method that is analogous to division, called multiplication by the inverse.
In the context of solving matrix equations like \( 2X = B - A \), instead of dividing by 2, we multiply by the scalar inverse. This involves the following steps:

• Calculate the inverse of the scalar, which is quite straightforward. For a scalar 2, its inverse is simply \( \frac{1}{2} \).
• Multiply each element of the matrix \( B - A \) by this inverse.

In our exercise, the subtraction resulted in: \( \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix} \). When we multiply by \( \frac{1}{2} \), it becomes: \[ X = \begin{bmatrix} -1 & -0.5 \ 1 & 2 \end{bmatrix} \].
This process effectively "divides" the matrix by 2, giving us the required matrix \( X \).

Matrix Multiplication Verification

Verification through matrix multiplication is a critical step to ensure the correctness of a solution to a matrix equation. After computing what we believe is the correct solution for \( X \), we should check it by reversing the steps:
Start with your found solution \( X \), here: \( \begin{bmatrix} -1 & -0.5 \ 1 & 2 \end{bmatrix} \), and multiply it by 2 (the scalar originally used). The operation should be element-wise: \( 2X = \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix} \). We then add this product to matrix \( A \): \[ 2X + A = \begin{bmatrix} -2 & -1 \ 2 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} \].

• The final sum should match matrix \( B \) exactly.
• If the matrices are equal, your solution for \( X \) is verified to be correct.

This ensures that the calculations were done accurately and confirms the integrity of the solution.