Problem 24 - Algebra and Trigonometry

Question

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ A-X=4 B $$

Step-by-Step Solution

Answer

The solution to the matrix equation is \(X = \left[\begin{array}{rr}17 & -3 \\ 2 & -9 \\ -7 & 16 \end{array}\right]\).

Step 1: Identifying the Equation

The matrix equation that needs to be solved is \(A - X = 4B\). Rearrange this equation to find \(X\).

Step 2: Rearranging the Equation

Rearrange equation to \(X = A - 4B\).

Step 3: Multiplying Matrix B with Scalar

Multiply each entry in matrix B by 4 to get \(4B = 4 \times \left[\begin{array}{rr}-5 & -1 \\ 0 & 0 \\ 3 & -4 \end{array}\right] = \left[\begin{array}{rr}-20 & -4 \\ 0 & 0 \\ 12 & -16 \end{array}\right]\).

Step 4: Subtracting Matrices

Subtract the resulting scalar-matrix multiplication from matrix A to get \(X = A - 4B = \left[\begin{array}{rr}-3 & -7 \\ 2 & -9 \\ 5 & 0 \end{array}\right] - \left[\begin{array}{rr}-20 & -4 \\ 0 & 0 \\ 12 & -16 \end{array}\right] = \left[\begin{array}{rr}17 & -3 \\ 2 & -9 \\ -7 & 16 \end{array}\right]\).

Key Concepts

Matrix Subtraction

Matrix subtraction is a straightforward yet important operation in linear algebra that involves subtracting corresponding elements of two matrices. For subtraction to be deemed valid, both matrices must have the same dimensions. In other words, they need to have the same number of rows and columns.

Here's how matrix subtraction works step-by-step:

• Set up both matrices side by side for easy comparison.
• Subtract each element in the first matrix by its corresponding element in the second matrix.
• Ensure that every subtraction follows the order of the matrices as the operation is not commutative (i.e., changing the order of matrices will lead to a different result).

Consider matrices A and B:
\[A = \begin{bmatrix} -3 & -7 \ 2 & -9 \ 5 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} -5 & -1 \ 0 & 0 \ 3 & -4 \end{bmatrix} \]

When subtracting matrix B from matrix A, compute \(-3 - (-5), -7 - (-1)\), and so on for all elements. The matrix subtraction process can change how matrices are manipulated and solved, emphasizing why correct configuration is essential.

Scalar Multiplication in Matrices

Scalar multiplication involves multiplying each entry of a matrix by the same scalar value, which is a single real number. It’s a simple yet powerful technique used often to modify matrix values across the board, making it crucial for many matrix-related computations.

To multiply a matrix by a scalar:

• Take the scalar (in this case, 4).
• Multiply the scalar value by each element of the matrix.
• Record the product in an organized, consistent manner to prevent errors.

For instance, if we take matrix B:
\[B = \begin{bmatrix} -5 & -1 \ 0 & 0 \ 3 & -4 \end{bmatrix} \]
and multiply by 4, the process would be:
\[4B = \begin{bmatrix} 4 \times (-5) & 4 \times (-1) \ 4 \times 0 & 4 \times 0 \ 4 \times 3 & 4 \times (-4) \end{bmatrix} = \begin{bmatrix} -20 & -4 \ 0 & 0 \ 12 & -16 \end{bmatrix} \]

Understanding scalar multiplication is a key foundational step when approaching more complex matrix operations. It allows for scaling matrices, which changes their size or influence without altering their structure.

Solving Matrix Equations

Solving a matrix equation involves finding a matrix variable that satisfies a given mathematical relationship between other matrices. Much like solving algebraic equations, matrix equations require understanding how to manipulate matrices with allowed operations. The same properties of equality and arithmetic apply here.

For solving the equation \( A - X = 4B \), the goal is to isolate matrix \( X \):

• First, clarify what each part of the equation means: \(A\) and \( B\) are known matrices, and \(X\) is the unknown matrix we need to find.
• Rearrange the equation to find \(X\) by adding \(X\) to both sides and then subtracting \(4B\) from both sides to derive \(X = A - 4B\).
• Calculate \(4B\) using scalar multiplication, then complete the subtraction \(A - 4B\).

After performing the calculations as outlined in earlier sections, the result is:
\[X = \begin{bmatrix} 17 & -3 \ 2 & -9 \ -7 & 16 \end{bmatrix} \]
This solution shows the matrix \(X\) that satisfies the original equation. Understanding these steps is essential for tackling more complex systems of equations and applying matrix operations in broader contexts.