Problem 13 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\(X = \begin{bmatrix} -3 & -5 \\ -2 & 7 \end{bmatrix}\) is the solution.

Step 1: Understanding the Equation

The equation to solve is \(2(B - X) = D\) for the matrix \(X\). This can be rewritten to find \(X\) as \(X = B - \frac{1}{2}D\).

Step 2: Determine Matrix Dimensions

Check if matrix subtraction is possible by verifying that matrices \(B\) and \(\frac{1}{2}D\) have the same dimensions. Both \(B\) and \(D\) are 2x2 matrices, which makes the subtraction feasible.

Step 3: Simplify \(\frac{1}{2}D\)

Calculate \(\frac{1}{2}D\):\[\frac{1}{2} \begin{bmatrix} 10 & 20 \ 10 & 0 \end{bmatrix} = \begin{bmatrix} 5 & 10 \ 5 & 0 \end{bmatrix}\]

Step 4: Perform Subtraction \(B - \frac{1}{2}D\)

Subtract \(\frac{1}{2}D\) from \(B\):\[\begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - \begin{bmatrix} 5 & 10 \ 5 & 0 \end{bmatrix} = \begin{bmatrix} 2-5 & 5-10 \ 3-5 & 7-0 \end{bmatrix} = \begin{bmatrix} -3 & -5 \ -2 & 7 \end{bmatrix}\]

Step 5: Write the Solution

The solution for \(X\) is \[\begin{bmatrix} -3 & -5 \ -2 & 7 \end{bmatrix}\]. Since the subtraction was successful and \(X\) has been calculated, a solution exists.

Key Concepts

Matrix Subtraction

Matrix subtraction is a fundamental operation in linear algebra, much like subtraction with regular numbers. However, there is a key rule: you can only subtract matrices of the same dimensions. In the given exercise, matrices \(B\) and \(\frac{1}{2}D\) are both 2x2 matrices. This means they can be subtracted from each other.

• Step one involves checking if the matrices are of the same dimensions, which they are in this case.
• Next, subtraction involves taking each element from the first matrix and subtracting the corresponding element in the second. For example, the top-left element of \(B - \frac{1}{2}D\) would be \(2 - 5\), resulting in \(-3\).

Ensuring matrices have compatible dimensions before performing any operations is vital for maintaining accuracy in calculations.

Matrix Dimensions

Understanding matrix dimensions is crucial when working with matrices as it dictates the operations you can perform. A matrix's dimensions are described by rows and columns. For instance, a 2x2 matrix has two rows and two columns.

• In the exercise, both matrices \(B\) and \(D\) have dimensions of 2x2.
• This means they can be directly compared and combined through operations like addition or subtraction.

Being clear on the dimensions helps to avoid errors in basic matrix operations, as mismatched dimensions can render operations invalid.

Matrix Simplification

Matrix simplification involves reducing a matrix to a simpler form while retaining its essential properties. This is often achieved through scaling, where each element in the matrix is multiplied by a common factor.

• In this exercise, the matrix \(D\) is scaled by \(\frac{1}{2}\). This involves multiplying every element of \(D\) by 0.5.
• Through simplification, \(D\) becomes \(\begin{bmatrix} 5 & 10 \ 5 & 0 \end{bmatrix}\).

Simplification aids in further operations, ensuring calculations become more straightforward and manageable.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and systems of linear equations. It is a powerful tool used in various scientific disciplines.

• This exercise requires an understanding of matrix laws, such as how to rearrange equations to isolate unknowns like the matrix \(X\).
• In the example provided, manipulating the given equation to find \(X\) involves rearranging \(2(B - X) = D\) to \(X = B - \frac{1}{2}D\).

Linear algebra provides a framework for solving complex problems and is foundational in disciplines such as physics, engineering, computer science, and more.