Problem 37 - Algebra 2

Question

Perform the indicated matrix operation. If the matrix does not exist, write impossible. $$ \frac{1}{2}\left[\begin{array}{ll}{4} & {6} \\ {3} & {0}\end{array}\right]-\frac{2}{3}\left[\begin{array}{cc}{9} & {27} \\ {0} & {3}\end{array}\right] $$

Step-by-Step Solution

Answer

The resulting matrix is \(\begin{bmatrix} 8 & 21 \\ \frac{3}{2} & 2 \end{bmatrix}\).

Step 1: Scalar Multiplication of the First Matrix

First, multiply the matrix \(\begin{bmatrix} 4 & 6 \ 3 & 0 \end{bmatrix}\) by \(\frac{1}{2}\). This involves multiplying each element of the matrix by \(\frac{1}{2}\):\[\frac{1}{2} \times \begin{bmatrix} 4 & 6 \ 3 & 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \times 4 & \frac{1}{2} \times 6 \ \frac{1}{2} \times 3 & \frac{1}{2} \times 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 \ \frac{3}{2} & 0 \end{bmatrix}\]

Step 2: Scalar Multiplication of the Second Matrix

Now, multiply the matrix \(\begin{bmatrix} 9 & 27 \ 0 & 3 \end{bmatrix}\) by \(-\frac{2}{3}\). The negative sign is important because we are subtracting:\[-\frac{2}{3} \times \begin{bmatrix} 9 & 27 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} \times 9 & -\frac{2}{3} \times 27 \ -\frac{2}{3} \times 0 & -\frac{2}{3} \times 3 \end{bmatrix} = \begin{bmatrix} -6 & -18 \ 0 & -2 \end{bmatrix}\]

Step 3: Matrix Subtraction

Subtract the resulting matrix from Step 2 from the result of Step 1. Ensure that subtraction is element-wise:\[\begin{bmatrix} 2 & 3 \ \frac{3}{2} & 0 \end{bmatrix} - \begin{bmatrix} -6 & -18 \ 0 & -2 \end{bmatrix} = \begin{bmatrix} 2 - (-6) & 3 - (-18) \ \frac{3}{2} - 0 & 0 - (-2) \end{bmatrix}\]This simplifies to:\[\begin{bmatrix} 8 & 21 \ \frac{3}{2} & 2 \end{bmatrix}\]

Step 4: Final Solution

The resulting matrix after performing all operations is:\[\begin{bmatrix} 8 & 21 \ \frac{3}{2} & 2 \end{bmatrix}\]

Key Concepts

Scalar Multiplication

Scalar multiplication in matrices is a straightforward yet powerful operation where each element of the matrix is multiplied by a scalar value. Think of a scalar as a single number that stretches or shrinks the entire matrix. In the example above, we scaled two matrices—a process also known as scaling by scalars.

• For the matrix \( \begin{bmatrix} 4 & 6 \ 3 & 0 \end{bmatrix} \), each element is multiplied by the scalar \( \frac{1}{2} \).
• This means each element is halved: \( 4 \to 2 \), \( 6 \to 3 \), \( 3 \to \frac{3}{2} \), and \( 0 \to 0 \).
• The result is \( \begin{bmatrix} 2 & 3 \ \frac{3}{2} & 0 \end{bmatrix} \).

Scalar multiplication modifies the matrix's size by the scalar, which affects each element equally across the matrix. When dealing with subtraction, as seen later, this scalar factor must be accurately applied to ensure correct computation.

Matrix Subtraction

Matrix subtraction is another fundamental operation that involves the element-wise subtraction of corresponding elements from two matrices. For subtraction to be possible, the matrices must have identical dimensions, meaning they need to have the same number of rows and columns.

• In our original problem, we subtract the matrix \( \begin{bmatrix} -6 & -18 \ 0 & -2 \end{bmatrix} \) from \( \begin{bmatrix} 2 & 3 \ \frac{3}{2} & 0 \end{bmatrix} \).
• Perform subtraction for each corresponding element: \( 2 - (-6) \) simplifies to 8, and \( 3 - (-18) \) simplifies to 21.
• For the other elements, we have \( \frac{3}{2} - 0 = \frac{3}{2} \) and \( 0 - (-2) = 2 \).

Thus, matrix subtraction is a simple yet precise operation requiring careful attention to each element’s arithmetic operations. The accurate application leads to the new matrix \( \begin{bmatrix} 8 & 21 \ \frac{3}{2} & 2 \end{bmatrix} \).

Element-wise Subtraction

Element-wise subtraction is a specific type of matrix operation where we subtract corresponding elements from two matrices. This method ensures each element’s operation is handled individually and consistently.

• In the exercise, each element from the two matrices is independently subtracted: \( a_{ij} - b_{ij} \), where \( i \) and \( j \) denote the row and column.
• Example: For the element at the first row and first column, we performed \( 2 - (-6) \), resulting in 8.
• Similarly, for the first row, second column, the operation was \( 3 - (-18) \), simplified to 21.

This type of subtraction ensures that the matrix's original structure is preserved and that each element is attentively and accurately processed. It’s a cornerstone of matrix arithmetic, vital for deriving correct results in matrix operations.