Problem 9 - College Algebra and Calculus: An Applied Approach

Question

Find (a) \(A+B\), (b) \(A-B\), (c) \(3 A\), and (d) \(3 A-2 B\). $$ \mathbf{A}=\left[\begin{array}{rrr} 2 & 2 & -1 \\ 1 & 1 & -2 \\ 1 & -1 & 3 \end{array}\right], B=\left[\begin{array}{rrr} 1 & 1 & -1 \\ -3 & 4 & 9 \\ 0 & -7 & 8 \end{array}\right] $$

Step-by-Step Solution

Answer

(a) \( \mathbf{C}=\left[\begin{array}{rrr} 3 & 3 & -2 \ -2 & 5 & 7 \ 1 & -8 & 11 \end{array}\right] \n \), (b) \( \mathbf{D}=\left[\begin{array}{rrr} 1 & 1 & 0 \ 4 & -3 & -11 \ 1 & 6 & -5 \end{array}\right] \n \), (c) \( \mathbf{E}=\left[\begin{array}{rrr} 6 & 6 & -3 \ 3 & 3 & -6 \ 3 & -3 & 9 \end{array}\right] \n \),(d) \[\mathbf{F}=\left[\begin{array}{rrr} 4 & 4 & -1 \ 9 & -5 & -12 \ 3 & 11 & -1 \end{array}\right] \n \]

Step 1: Matrix Addition

Matrix addition is performed element by element across the matrices. When adding matrix A to matrix B we get a new matrix: \n \( \mathbf{C}=\left[\begin{array}{rrr} 2+1 & 2+1 & -1+(-1) \ 1+(-3) & 1+4 & -2+9 \ 1+0 & -1+(-7) & 3+8\end{array}\right] \n\) The resulting matrix is: \( \mathbf{C}=\left[\begin{array}{rrr} 3 & 3 & -2 \ -2 & 5 & 7 \ 1 & -8 & 11\end{array}\right] \n\)

Step 2: Matrix Subtraction

Matrix subtraction is also performed element by element across the matrices. When subtracting matrix B from matrix A we get a new matrix: \n \( \mathbf{D}=\left[\begin{array}{rrr} 2-1 & 2-1 & -1-(-1) \ 1-(-3) & 1-4 & -2-9 \ 1-0 & -1-(-7) & 3-8 \end{array}\right] \n\) The resulting matrix is: \( \mathbf{D}=\left[\begin{array}{rrr} 1 & 1 & 0 \ 4 & -3 & -11 \ 1 & 6 & -5 \end{array}\right] \n\)

Step 3: Scalar Multiplication

Scalar multiplication involves multiplying each element of the matrix by the scalar. Hence, the matrix \(3A\) becomes: \n \( \mathbf{E}=\left[\begin{array}{rrr} 3×2 & 3×2 & 3×(-1) \ 3×1 & 3×1 & 3×(-2) \ 3×1 & 3×(-1) & 3×3 \end{array}\right] \n\) The resulting matrix is: \( \mathbf{E}=\left[\begin{array}{rrr} 6 & 6 & -3 \ 3 & 3 & -6 \ 3 & -3 & 9 \end{array}\right] \n\)

Step 4: Combined operation

The combined operation involves scalar multiplication and matrix subtraction. \(3A-2B\) becomes: \n \[\mathbf{F}=3\mathbf{A}-2\mathbf{B}=\left[\begin{array}{cc} 3×2-2×1 & 3×2-2×1 & 3×(-1)-2×(-1) \ 3×1-2×(-3) & 3×1-2×4 & 3×(-2)-2×9 \ 3×1-2×0 & 3×(-1)-2×(-7) & 3×3-2×8 \end{array}\right] \] The resulting matrix is: \[ \mathbf{F}=\left[\begin{array}{rrr} 4 & 4 & -1 \ 9 & -5 & -12 \ 3 & 11 & -1 \end{array}\right] \n \]

Key Concepts

Matrix Addition

Matrix addition is a fundamental technique in matrix operations. It involves combining two matrices by adding their corresponding elements. To perform matrix addition, both matrices must be of the same dimension. For example, if Matrix A and Matrix B both have dimensions 3x3, they can be added together.

• Element-wise Addition: Each element in the resulting matrix is the sum of the elements from the two original matrices that are in the same position.
• Zero-based Indexing: In programming and higher-level mathematics, matrices are often indexed starting from zero, so remember to add elements based on their position, starting from 0,0 up to n,n.

Here is how it's done using the matrices from our example:The first element of the resulting matrix, located at the top-left corner, will be the sum of the first elements from both Matrix A and Matrix B \((2 + 1) = 3\). This process is repeated for each element. In the end, the result of adding Matrix A and Matrix B gives us a new matrix: \(\mathbf{C}=\begin{bmatrix} 3 & 3 & -2 \ -2 & 5 & 7 \ 1 & -8 & 11\end{bmatrix}\).

Matrix Subtraction

Matrix subtraction, much like matrix addition, involves performing element-wise operations, but instead of summing the elements, they are subtracted. This process also requires the matrices to be of the same size, ensuring that each element can be matched and subtracted accurately from its counterpart.

• Element-by-element Subtraction: Each element in the resulting matrix is obtained by subtracting an element from Matrix B from the corresponding element in Matrix A.
• Negative Values Understanding: When subtracting, it's important to be cautious of negative numbers as they behave differently from positive ones in subtraction.

For example, with the given matrices, we subtract each element of Matrix B from the corresponding element of Matrix A. The calculation begins with the first element: \((2 - 1) = 1\). This operation is repeated for all corresponding elements, resulting in the matrix \(\mathbf{D}=\begin{bmatrix} 1 & 1 & 0 \ 4 & -3 & -11 \ 1 & 6 & -5 \end{bmatrix}\).

Scalar Multiplication

Scalar multiplication is a process where each element of a matrix is multiplied by a constant scalar value. This operation allows you to scale the matrix up or down, effectively altering the values while maintaining their relative proportions within the matrix.

• Consistent Scaling: The scalar multiplies every individual entry in the matrix, ensuring that the transformation applies uniformly to all elements.
• Preserve Matrix Size: While the numerical values of the matrix elements change, the dimensions of the matrix remain the same.

Applying scalar multiplication to a matrix involves multiplying each component by the scalar value. For the given matrix A and a scalar value of 3, multiply every entry in Matrix A by 3. For instance, the top-left element \(2\) is multiplied by 3 to yield \(6\). Repeating this for all elements of Matrix A results in the matrix \(\mathbf{E}=\begin{bmatrix} 6 & 6 & -3 \ 3 & 3 & -6 \ 3 & -3 & 9 \end{bmatrix}\). This demonstrates the straightforward nature of scalar multiplication.