Problem 19 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Perform the indicated operations. \(\frac{1}{2}\left[\begin{array}{rrrr}1 & 0 & 0 & -4 \\ 3 & 0 & -1 & 6 \\ -2 & 1 & -4 & 2\end{array}\right]+\frac{4}{3}\left[\begin{array}{rrrr}3 & 0 & -1 & 4 \\ -2 & 1 & -6 & 2 \\ 8 & 2 & 0 & -2\end{array}\right]\) \(-\frac{1}{3}\left[\begin{array}{rrrr}3 & -9 & -1 & 0 \\ 6 & 2 & 0 & -6 \\ 0 & 1 & -3 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

The final result is: \(\left[\begin{array}{rrrr}5.5 & -3 & -\frac{5}{3} & \frac{10}{3} \\\ \frac{5}{3} & 2 & -8.5 & \frac{11}{3} \\\ \frac{29}{3} & \frac{11}{3} & -3 & -\frac{4}{3}\end{array}\right]\)

Step 1: Perform scalar multiplication on each matrix

First, we need to multiply the first matrix by the scalar \(\frac{1}{2}\) and the second matrix by the scalar \(\frac{4}{3}\). \(\frac{1}{2}\left[\begin{array}{rrrr}1 & 0 & 0 & -4 \\\ 3 & 0 & -1 & 6 \\\ -2 & 1 & -4 & 2\end{array}\right]=\left[\begin{array}{rrrr}0.5 & 0 & 0 & -2 \\\ 1.5 & 0 & -0.5 & 3 \\\ -1 & 0.5 & -2 & 1\end{array}\right]\) \(\frac{4}{3}\left[\begin{array}{rrrr}3 & 0 & -1 & 4 \\\ -2 & 1 & -6 & 2 \\\ 8 & 2 & 0 & -2\end{array}\right]=\left[\begin{array}{rrrr}4 & 0 & -\frac{4}{3} & \frac{16}{3} \\\ -\frac{8}{3} & \frac{4}{3} & -8 & \frac{8}{3} \\\ \frac{32}{3} & \frac{8}{3} & 0 & -\frac{8}{3}\end{array}\right]\)

Step 2: Add the two resulting matrices together

Now, we add the two matrices obtained in step 1. \(\left[\begin{array}{rrrr}0.5 & 0 & 0 & -2 \\\ 1.5 & 0 & -0.5 & 3 \\\ -1 & 0.5 & -2 & 1\end{array}\right]+\left[\begin{array}{rrrr}4 & 0 & -\frac{4}{3} & \frac{16}{3} \\\ -\frac{8}{3} & \frac{4}{3} & -8 & \frac{8}{3} \\\ \frac{32}{3} & \frac{8}{3} & 0 & -\frac{8}{3}\end{array}\right]=\left[\begin{array}{rrrr}4.5 & 0 & -\frac{4}{3} & \frac{10}{3} \\\ \frac{1}{3} & \frac{4}{3} & -8.5 & \frac{17}{3} \\\ \frac{29}{3} & \frac{10}{3} & -2 & -\frac{5}{3}\end{array}\right]\)

Step 3: Perform scalar multiplication on the last matrix and subtract from the result obtained in step 2

Now, we need to multiply the last matrix by the scalar \(-\frac{1}{3}\) and subtract this result from the matrix obtained in step 2. \(-\frac{1}{3}\left[\begin{array}{rrrr}3 & -9 & -1 & 0 \\\ 6 & 2 & 0 & -6 \\\ 0 & 1 & -3 & 1\end{array}\right]=\left[\begin{array}{rrrr}-1 & 3 & \frac{1}{3} & 0 \\\ -2 & -\frac{2}{3} & 0 & 2 \\\ 0 & -\frac{1}{3} & 1 & -\frac{1}{3}\end{array}\right]\) \(\left[\begin{array}{rrrr}4.5 & 0 & -\frac{4}{3} & \frac{10}{3} \\\ \frac{1}{3} & \frac{4}{3} & -8.5 & \frac{17}{3} \\\ \frac{29}{3} & \frac{10}{3} & -2 & -\frac{5}{3}\end{array}\right]-\left[\begin{array}{rrrr}-1 & 3 & \frac{1}{3} & 0 \\\ -2 & -\frac{2}{3} & 0 & 2 \\\ 0 & -\frac{1}{3} & 1 & -\frac{1}{3}\end{array}\right]=\left[\begin{array}{rrrr}5.5 & -3 & -\frac{5}{3} & \frac{10}{3} \\\ \frac{5}{3} & 2 & -8.5 & \frac{11}{3} \\\ \frac{29}{3} & \frac{11}{3} & -3 & -\frac{4}{3}\end{array}\right]\) So, the final result is: \(\left[\begin{array}{rrrr}5.5 & -3 & -\frac{5}{3} & \frac{10}{3} \\\ \frac{5}{3} & 2 & -8.5 & \frac{11}{3} \\\ \frac{29}{3} & \frac{11}{3} & -3 & -\frac{4}{3}\end{array}\right]\)

Key Concepts

Scalar Multiplication

Scalar multiplication of matrices involves multiplying each element of a matrix by a given scalar. It's like distributing the scalar across all the elements inside the matrix. Let's say you have a scalar value, such as \( \frac{1}{2} \), and a matrix \( \left[\begin{array}{cc} 2 & 3 \ 4 & 5 \end{array}\right] \). To perform scalar multiplication, multiply every element in the matrix by \( \frac{1}{2} \).
For example:
- \( \text{New element at (1,1)} = 2 \times \frac{1}{2} = 1 \) - \( \text{New element at (1,2)} = 3 \times \frac{1}{2} = 1.5 \) - Continue this with all elements
The resultant matrix is \( \left[\begin{array}{cc} 1 & 1.5 \ 2 & 2.5 \end{array}\right] \).
This operation is simple as it involves basic arithmetic applied to each element individually and can significantly change the size of numbers within the matrix depending on the scalar used.

Matrix Addition

Matrix addition requires you to sum two matrices of the same size. This operation combines elements from each position in the matrices. Consider two 2x2 matrices, A and B:
A = \( \left[\begin{array}{cc} 2 & 3 \ 4 & 5 \end{array}\right] \), B = \( \left[\begin{array}{cc} 1 & 0 \ 2 & 3 \end{array}\right] \).
To add them:
- Compute \( C(1,1) = A(1,1) + B(1,1) = 2 + 1 = 3 \) - Compute \( C(1,2) = A(1,2) + B(1,2) = 3 + 0 = 3 \) - Do this for all elements
The result is matrix C = \( \left[\begin{array}{cc} 3 & 3 \ 6 & 8 \end{array}\right] \).
Matrix addition is straightforward: add corresponding elements of two matrices. You cannot add matrices of different sizes because each position in the matrices must be added with its matching position.
This operation maintains the original structure or size of the matrices involved.

Matrix Subtraction

In matrix subtraction, you subtract corresponding elements of one matrix from another, as long as both matrices are of the same size. For instance, if Matrix D = \( \left[\begin{array}{cc} 6 & 5 \ 4 & 3 \end{array}\right] \) and Matrix E = \( \left[\begin{array}{cc} 1 & 2 \ 3 & 1 \end{array}\right] \), you perform the following steps:
- \( F(1,1) = D(1,1) - E(1,1) = 6 - 1 = 5 \) - \( F(1,2) = D(1,2) - E(1,2) = 5 - 2 = 3 \)
Continue this subtraction for each corresponding element.
The resulting matrix F is \( \left[\begin{array}{cc} 5 & 3 \ 1 & 2 \end{array}\right] \).
Subtraction of matrices, much like addition, allows modifications to the original values in each matrix position. This operation doesn't alter the dimensional structure but changes the elements’ values directly. Both matrices need to maintain identical dimensions for the subtraction process.