Problem 19 - College Algebra

Question

In Exercises \(17-26,\) 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 \(X\). $$ 2 X+A=B $$

Step-by-Step Solution

Answer

The solution matrix X is \[X=\left[\begin{array}{rr} -1 & 3 \ -1 & 4.5 \ -1 & -2 \end{array}\right]\]

Step 1: Analyze the given equation

The given equation is \(2X + A = B\). The objective is to solve for X, which implies isolating X on one side of the equation.

Step 2: Initialize the given matrices

Matrix A is \[A=\left[\begin{array}{rr} -3 & -7 \ 2 & -9 \ 5 & 0 \end{array}\right]\]. And Matrix B is \[B=\left[\begin{array}{rr} -5 & -1 \ 0 & 0 \ 3 & -4 \end{array}\right]\]

Step 3: Isolate X

Solving for X, we get \(X = (B - A) / 2\). This is achieved by moving matrix A to the other side of the equation (which changes the sign), and then dividing both sides of the equation by 2.

Step 4: Perform matrix subtraction

First, subtract matrix A from matrix B element-wise to get another matrix (let’s say C). So, \(C = B - A\). This gives: \[C=\left[\begin{array}{rr} -5 - (-3) & -1 - (-7) \ 0 - 2 & 0 - (-9) \ 3 - 5 & -4 - 0 \end{array}\right] = \left[\begin{array}{rr} -2 & 6 \ -2 & 9 \ -2 & -4 \end{array}\right]\]

Step 5: Divide by the scalar

Finally, divide every element of the resulting matrix C by 2, since \(2X = C\) according to the formulation of our problem. Now, \(X = C / 2 = \left[\begin{array}{rr} -2 / 2 & 6 / 2 \ -2 / 2 & 9 / 2 \ -2 / 2 & -4 / 2 \end{array}\right]\). This gives us the matrix X: \[X=\left[\begin{array}{rr} -1 & 3 \ -1 & 4.5 \ -1 & -2 \end{array}\right]\]

Key Concepts

Matrix Subtraction

Matrix subtraction involves subtracting one matrix from another. It is an element-wise operation, meaning each element of the first matrix is subtracted from the corresponding element of the second matrix. To perform matrix subtraction, both matrices must have the same dimensions.

Here's how it works using the matrices from our exercise:

• Matrix A: \[\begin{bmatrix} -3 & -7 \2 & -9 \5 & 0 \end{bmatrix}\]
• Matrix B: \[\begin{bmatrix} -5 & -1 \0 & 0 \3 & -4 \end{bmatrix}\]
• To find \( B - A \), subtract each element of A from the corresponding element of B:\[\begin{bmatrix} -5 - (-3) & -1 - (-7) \0 - 2 & 0 - (-9) \3 - 5 & -4 - 0 \end{bmatrix} = \begin{bmatrix} -2 & 6 \-2 & 9 \-2 & -4 \end{bmatrix}\]

This resulting matrix is the difference, calculated by subtracting the elements of A from the corresponding elements of B.

Matrix Division

Matrix division, unlike matrix subtraction, is not a straightforward element-wise operation. Instead, it refers to processes like matrix inversion or scalar division. In our exercise, we essentially perform scalar division on each element.

It's important to note:

• Matrix division does not mean dividing one matrix by another.
• Handling matrices with division often involves manipulating the entire matrix by a scalar value.

In the context of our exercise, after performing matrix subtraction to get matrix \( C \), we divided each element of \( C \) by 2 to solve for matrix \( X \). This process is often referred to as scalar division.

Scalar Multiplication

Scalar multiplication involves multiplying each element of a matrix by a scalar value (a single number, not another matrix). This is a simple operation, yet powerful when transforming matrices.

In our exercise, we rearranged the equation from \( 2X + A = B \) to find \( C = 2X \). Hence, every element in the matrix C was divided by 2 to find the original matrix X:

• Consider matrix C:\[\begin{bmatrix} -2 & 6 \-2 & 9 \-2 & -4 \end{bmatrix}\]
• Perform scalar multiplication by dividing each element by 2:\[\frac{1}{2} \times \begin{bmatrix} -2 & 6 \-2 & 9 \-2 & -4 \end{bmatrix} = \begin{bmatrix} -1 & 3 \-1 & 4.5 \-1 & -2 \end{bmatrix}\]

This operation is straightforward and merely requires ensuring each matrix entry is handled precisely.

Element-wise Operations

Element-wise operations refer to conducting arithmetic operations on corresponding elements of matrices. This is seen often in matrices with the same dimensions.

In element-wise operations:

• Each element in the first matrix directly interacts with the corresponding element of the second matrix.
• Operations conducted include addition, subtraction, multiplication, and division.

In the exercise problem, element-wise subtraction was used to find matrix \( C \). Then, element-wise division (by 2) transformed matrix \( C \) into matrix \( X \). It’s vital to always ensure matrices being used for element-wise operations are compatible dimension-wise for these operations to be valid.