Question

Matrices \(A\) and \(B\) are given below. Find \(X\) that satisfies the equation. $$ A=\left[\begin{array}{cc} 3 & -1 \\ 2 & 5 \end{array}\right] \quad B=\left[\begin{array}{cc} 1 & 7 \\ 3 & -4 \end{array}\right] $$ $$ A-\frac{1}{2} X=-B $$

Short answer

\( X = \left[ \begin{array}{cc} 8 & 12 \\ 10 & 2 \end{array} \right] \).

Step-by-step solution

Understand the Matrix Equation

The given equation is \( A - \frac{1}{2} X = -B \). To find \( X \), we need to isolate it on one side by simplifying this equation.

Rearrange the Equation

First, we rearrange the equation to express \( X \) as the subject. Start by multiplying both sides by \(-1\) to remove the negative sign on \( B \): \( -A + \frac{1}{2} X = B \). Next, add \( A \) to both sides: \( \frac{1}{2} X = A + B \).

Solve for \( X \) by Removing the Fraction

Multiply both sides of the equation \( \frac{1}{2} X = A + B \) by 2 to remove the fraction: \( X = 2(A + B) \).

Add Matrices A and B

Compute \( A + B \) by adding corresponding elements of matrices \( A \) and \( B \). \[ A + B = \left[ \begin{array}{cc} 3 & -1 \ 2 & 5 \end{array} \right] + \left[ \begin{array}{cc} 1 & 7 \ 3 & -4 \end{array} \right] = \left[ \begin{array}{cc} 4 & 6 \ 5 & 1 \end{array} \right] \].

Multiply the Result by 2

Multiply the resultant matrix from Step 4 by 2. \[ X = 2 \left( \left[ \begin{array}{cc} 4 & 6 \ 5 & 1 \end{array} \right] \right) = \left[ \begin{array}{cc} 8 & 12 \ 10 & 2 \end{array} \right] \].

Key concepts

Matrix Addition

Matrix addition is a straightforward process where you add corresponding elements from two matrices. For this operation, it is crucial that both matrices have the same dimensions. That means they must have the same number of rows and columns.

Here’s how you do matrix addition:

• Look at each element in a given position in the two matrices.
• Add these elements together to get a new element in the resulting matrix.
• Repeat the process for each position in the matrices.

In our example, matrices \(A\) and \(B\) are both 2x2 matrices. We add corresponding elements like this:

• Top-left elements: \(3 + 1 = 4\)
• Top-right elements: \(-1 + 7 = 6\)
• Bottom-left elements: \(2 + 3 = 5\)
• Bottom-right elements: \(5 - 4 = 1\)

So, the result of \(A + B\) is \[\left[ \begin{array}{cc} 4 & 6 \ 5 & 1 \end{array} \right]\], giving us a new matrix that forms part of the solution to the matrix equation.

Matrix Multiplication

Matrix multiplication involves more calculations than matrix addition. When performing matrix multiplication by a scalar (a single number), each element in the matrix is multiplied by that number.

In our problem, we need to multiply the resultant matrix from the addition by 2, which is a scalar multiplication.

• Take each element of the matrix \(\left[ \begin{array}{cc} 4 & 6 \ 5 & 1 \end{array} \right]\) and multiply it by 2.
• Top-left element: \(2 \times 4 = 8\)
• Top-right element: \(2 \times 6 = 12\)
• Bottom-left element: \(2 \times 5 = 10\)
• Bottom-right element: \(2 \times 1 = 2\)

The resulting matrix after this scalar multiplication is\[\left[ \begin{array}{cc} 8 & 12 \ 10 & 2 \end{array} \right]\]. Matrix multiplication in more complex scenarios may involve matrix/matrix multiplication where dimensions must be compatible, but our task focused on scalar multiplication.

Remember, scalar multiplication affects each matrix element individually.

Solving Linear Equations

Solving linear equations in the context of matrices often involves equating and simplifying matrix equations to find unknown matrix values. In this particular exercise, we were tasked with isolating matrix \(X\) from an equation involving \(A\) and \(B\).

The starting equation was \(A - \frac{1}{2} X = -B\). To solve for \(X\), one must perform the following steps:

• Rearrange the equation to get \(X\) on one side. This started with multiplying both sides by \(-1\) to make computation easier. Thus, \(-A + \frac{1}{2} X = B\).
• Next, add matrix \(A\) to both sides, yielding \(\frac{1}{2} X = A + B\).
• Finally, remove the fraction by multiplying both sides by 2, resulting in the equation \(X = 2(A + B)\).

Through these algebraic manipulations, we isolate \(X\) and solve the matrix equation, just like solving any standard linear equation. Each step is about methodically applying algebraic principles to simplify and solve for our target variable.