Problem 27 - A Graphical Approach to College Algebra

Question

Perform each operation if possible. $$2\left[\begin{array}{rr} 2 & -1 \\ 5 & 1 \\ 0 & 3 \end{array}\right]+\left[\begin{array}{rr} 5 & 0 \\ 7 & -3 \\ 1 & 1 \end{array}\right]-\left[\begin{array}{rr} 9 & -4 \\ 4 & 4 \\ 1 & 6 \end{array}\right]$$

Step-by-Step Solution

Answer

The final matrix is \( \begin{bmatrix} 0 & 2 \\ 13 & -5 \\ 0 & 1 \end{bmatrix} \).

Step 1: Multiply First Matrix by Scalar

We need to multiply each element of the first matrix by 2:\[2 \times \begin{bmatrix}2 & -1 \5 & 1 \0 & 3\end{bmatrix} = \begin{bmatrix}2 \times 2 & 2 \times (-1) \2 \times 5 & 2 \times 1 \2 \times 0 & 2 \times 3\end{bmatrix} = \begin{bmatrix}4 & -2 \10 & 2 \0 & 6\end{bmatrix}\]

Step 2: Add First and Second Matrix

Add the resulting matrix from Step 1 to the second matrix:\[\begin{bmatrix}4 & -2 \10 & 2 \0 & 6\end{bmatrix} + \begin{bmatrix}5 & 0 \7 & -3 \1 & 1\end{bmatrix} = \begin{bmatrix}4+5 & -2+0 \10+7 & 2-3 \0+1 & 6+1\end{bmatrix} = \begin{bmatrix}9 & -2 \17 & -1 \1 & 7\end{bmatrix}\]

Step 3: Subtract Third Matrix

Subtract the third matrix from the result of Step 2:\[\begin{bmatrix}9 & -2 \17 & -1 \1 & 7\end{bmatrix} - \begin{bmatrix}9 & -4 \4 & 4 \1 & 6\end{bmatrix} = \begin{bmatrix}9-9 & -2-(-4) \17-4 & -1-4 \1-1 & 7-6\end{bmatrix} = \begin{bmatrix}0 & 2 \13 & -5 \0 & 1\end{bmatrix}\]

Key Concepts

Scalar Multiplication

Scalar multiplication is a process where each entry of a matrix is multiplied by a fixed number, called a scalar. It's like applying a simple, uniform change to every value within the matrix.
When you multiply a matrix by a scalar, you stretch or shrink the matrix, depending on the scalar. If the scalar is greater than 1, the matrix is stretched; if it is between 0 and 1, it is shrunk. If the scalar is negative, it not only scales but also mirrors the matrix elements across the origin.
For example, let's say we have a scalar of 2 and a matrix \[\begin{bmatrix} 2 & -1 \ 5 & 1 \ 0 & 3 \end{bmatrix} ,\]we multiply each individual element by 2:

• The element at row 1, column 1: \(2 \times 2 = 4\)
• The element at row 1, column 2: \(2 \times -1 = -2\)
• The element at row 2, column 1: \(2 \times 5 = 10\)
• The element at row 2, column 2: \(2 \times 1 = 2\)
• The element at row 3, column 1: \(2 \times 0 = 0\)
• The element at row 3, column 2: \(2 \times 3 = 6\)

After performing these calculations, we get the new matrix: \[\begin{bmatrix} 4 & -2 \ 10 & 2 \ 0 & 6 \end{bmatrix} .\]

Matrix Addition

Matrix addition involves adding matrices of the same order, meaning they must have the same number of rows and columns. This operation is performed element-wise, much like adding two lists of numbers.
The key rule is that corresponding elements are added together.
In practice, consider two matrices:\[\begin{bmatrix} 4 & -2 \ 10 & 2 \ 0 & 6 \end{bmatrix}\]added to\[\begin{bmatrix} 5 & 0 \ 7 & -3 \ 1 & 1 \end{bmatrix} .\]The resulting matrix is computed by:

• The first row, first column: \(4 + 5 = 9\)
• The first row, second column: \(-2 + 0 = -2\)
• The second row, first column: \(10 + 7 = 17\)
• The second row, second column: \(2 - 3 = -1\)
• The third row, first column: \(0 + 1 = 1\)
• The third row, second column: \(6 + 1 = 7\)

This yields the combined matrix:\[\begin{bmatrix} 9 & -2 \ 17 & -1 \ 1 & 7 \end{bmatrix} .\]

Matrix Subtraction

Matrix subtraction is similar to matrix addition but involves subtracting corresponding elements of the matrices.
Like addition, subtraction requires matrices of the same order.
Each element of the first matrix is reduced by the corresponding element in the second matrix.
For instance, to subtract the matrix \[\begin{bmatrix} 9 & -4 \ 4 & 4 \ 1 & 6 \end{bmatrix} \]from \[\begin{bmatrix} 9 & -2 \ 17 & -1 \ 1 & 7 \end{bmatrix} ,\]we do the following calculations:

• The first row, first column: \(9 - 9 = 0\)
• The first row, second column: \(-2 + 4 = 2\)
• The second row, first column: \(17 - 4 = 13\)
• The second row, second column: \(-1 - 4 = -5\)
• The third row, first column: \(1 - 1 = 0\)
• The third row, second column: \(7 - 6 = 1\)

After performing these operations, the resulting matrix is:\[\begin{bmatrix} 0 & 2 \ 13 & -5 \ 0 & 1 \end{bmatrix} .\]