Problem 21 - College Algebra with Modeling and Visualization

Question

Evaluate the matrix expression. $$2\left[\begin{array}{rr}2 & -1 \\\5 & 1 \\\0 & 3\end{array}\right]+\left[\begin{array}{rr} 5 & 0 \\\7 & -3 \\\1 & 1\end{array}\right]-\left[\begin{array}{rr}9 & -4 \\\4 & 4 \\\1 & 6 \end{array}\right]$$

Step-by-Step Solution

Answer

The result of the matrix expression is \( \begin{bmatrix} 0 & 2 \\ 13 & -5 \\ 0 & 1 \end{bmatrix} \).

Step 1: Scalar Multiplication of the First Matrix

Multiply each element of the first matrix \( \begin{bmatrix} 2 & -1 \ 5 & 1 \ 0 & 3 \end{bmatrix} \) by 2. This yields the matrix \( \begin{bmatrix} 4 & -2 \ 10 & 2 \ 0 & 6 \end{bmatrix} \).

Step 2: Addition of Two Matrices

Add the matrices obtained from Step 1 and the second given matrix directly: \( \begin{bmatrix} 4 & -2 \ 10 & 2 \ 0 & 6 \end{bmatrix} + \begin{bmatrix} 5 & 0 \ 7 & -3 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 & -2 \ 17 & -1 \ 1 & 7 \end{bmatrix} \). To perform this addition, add corresponding elements from each matrix.

Step 3: Subtraction of Matrices

Subtract the third matrix \( \begin{bmatrix} 9 & -4 \ 4 & 4 \ 1 & 6 \end{bmatrix} \) from the result of Step 2: \( \begin{bmatrix} 9 & -2 \ 17 & -1 \ 1 & 7 \end{bmatrix} - \begin{bmatrix} 9 & -4 \ 4 & 4 \ 1 & 6 \end{bmatrix} = \begin{bmatrix} 0 & 2 \ 13 & -5 \ 0 & 1 \end{bmatrix} \). Subtract each element of the second matrix from the corresponding element of the first matrix.

Key Concepts

Scalar Multiplication

When you perform scalar multiplication on a matrix, you are multiplying every element within the matrix by a single number, known as the scalar. Consider it like stretching or shrinking the entire matrix by the same factor. It is a straightforward yet impactful operation in matrix algebra.
Let's look at our exercise example. We have the matrix \[\begin{bmatrix} 2 & -1 \5 & 1 \0 & 3 \end{bmatrix}\]. To perform scalar multiplication with the scalar 2, we multiply each of its elements by 2.

• Multiply the first row: \(2 \times 2 = 4\) and \(2 \times (-1) = -2\).
• Move to the second row: \(2 \times 5 = 10\) and \(2 \times 1 = 2\).
• Finally, multiply the third row: \(2 \times 0 = 0\) and \(2 \times 3 = 6\).

Completing these calculations gives you the new matrix:\[\begin{bmatrix} 4 & -2 \10 & 2 \0 & 6 \end{bmatrix}\]. By maintaining the same operation throughout, scalar multiplication remains consistent and simple.

Matrix Addition

Matrix addition involves combining two matrices by adding each corresponding element together. It is essential that the matrices have the same dimensions, otherwise the addition cannot be performed. This shared structure allows a seamless addition of corresponding elements.
After finishing the scalar multiplication, we move to adding the resultant matrix \[\begin{bmatrix} 4 & -2 \10 & 2 \0 & 6 \end{bmatrix}\] with the second given matrix \[\begin{bmatrix} 5 & 0 \7 & -3 \1 & 1 \end{bmatrix}\].
Here's how it works:

• For the first row, add \(4 + 5 = 9\) and \(-2 + 0 = -2\).
• In the second row, use \(10 + 7 = 17\) and \(2 + (-3) = -1\).
• For the third row, combine \(0 + 1 = 1\) and \(6 + 1 = 7\).

Following this operation, you get the resulting matrix:\[\begin{bmatrix} 9 & -2 \17 & -1 \1 & 7 \end{bmatrix}\].
Matrix addition is like a simple arithmetic task done on each pair of corresponding numbers within the matrices.

Matrix Subtraction

Matrix subtraction, much like matrix addition, requires matrices to be of the same size. Instead of adding, you subtract each corresponding element of the second matrix from the first. It’s just as straightforward as matrix addition but focuses on differences rather than sums.
From our previous result of matrix addition, we have \[\begin{bmatrix} 9 & -2 \17 & -1 \1 & 7 \end{bmatrix}\]. We need to subtract the third given matrix \[\begin{bmatrix} 9 & -4 \4 & 4 \1 & 6 \end{bmatrix}\].
The steps involved are:

• First row subtraction: \(9 - 9 = 0\) and \(-2 - (-4) = 2\).
• Second row subtraction: \(17 - 4 = 13\) and \(-1 - 4 = -5\).
• Third row subtraction: \(1 - 1 = 0\) and \(7 - 6 = 1\).

Performing these operations, you end up with:\[\begin{bmatrix} 0 & 2 \13 & -5 \0 & 1 \end{bmatrix}\].
Matrix subtraction is crucial for finding differences and manipulating data, closely mimicking what you might do with basic subtraction in arithmetic.