Problem 12 - Algebra and Trigonometry

Question

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{rrr} {3} & {1} & {1} \\ {-1} & {2} & {5} \end{array}\right], \quad B=\left[\begin{array}{rrr} {2} & {-3} & {6} \\ {-3} & {1} & {-4} \end{array}\right] $$

Step-by-Step Solution

Answer

\[ A+B= \left[\begin{array}{rrr} {5} & {-2} & {7} \ {-4} & {3} & {1} \end{array}\right], A-B= \left[\begin{array}{rrr} {1} & {4} & {-5} \ {2} & {1} & {9} \end{array}\right], -4A= \left[\begin{array}{rrr} {-12} & {-4} & {-4} \ {4} & {-8} & {-20} \end{array}\right], 3A+2B= \left[\begin{array}{rrr} {13} & {-3} & {15} \ {-9} & {8} & {7} \end{array}\right] \]

Step 1: Compute A + B

To compute \( A+B \), simply add corresponding elements in matrices \( A \) and \( B \). This gives the matrix: \[ \left[\begin{array}{rrr} {3+2} & {1+(-3)} & {1+6} \ {-1+(-3)} & {2+1} & {5+(-4)} \end{array}\right] \] which simplifies to \[ \left[\begin{array}{rrr} {5} & {-2} & {7} \ {-4} & {3} & {1} \end{array}\right] \].

Step 2: Compute A - B

The computation for \( A - B \) is similar to that of \( A + B \). Simply subtract the corresponding values in matrices \( A \) and \( B \), yielding \[ \left[\begin{array}{rrr} {3-2} & {1-(-3)} & {1-6} \ {-1-(-3)} & {2-1} & {5-(-4)} \end{array}\right] \] which simplifies to \[ \left[\begin{array}{rrr} {1} & {4} & {-5} \ {2} & {1} & {9} \end{array}\right] \].

Step 3: Compute -4A

For scalar multiplication, multiply each element in matrix \( A \) by -4, resulting in \[ \left[\begin{array}{rrr} {-4(3)} & {-4(1)} & {-4(1)} \ {-4(-1)} & {-4(2)} & {-4(5)} \end{array}\right] \] which simplifies to \[ \left[\begin{array}{rrr} {-12} & {-4} & {-4} \ {4} & {-8} & {-20} \end{array}\right] \].

Step 4: Compute 3A + 2B

Here, perform scalar multiplication first, then add the resulting matrices. We get the matrix \[ \left[\begin{array}{rrr} 3(3)+2(2) & 3(1)+2(-3) & 3(1)+2(6) \ 3(-1)+2(-3) & 3(2)+2(1) & 3(5)+2(-4) \end{array}\right] \] which simplifies to \[ \left[\begin{array}{rrr} {13} & {-3} & {15} \ {-9} & {8} & {7} \end{array}\right] \].

Key Concepts

Matrix Addition

Matrix addition is a straightforward process where you add corresponding elements of two matrices to produce a new matrix. It's simple and intuitive, much like adding numbers on paper. Imagine you have two matrices, say matrix \( A \) and matrix \( B \). These matrices must be of the same dimension, meaning they should have the same number of rows and columns.

If matrix \( A \) is \( 2 \times 3 \), matrix \( B \) also needs to be \( 2 \times 3 \). For every element in \( A \), you add the corresponding element in \( B \). For example, sum the first row, first column of \( A \) with the first row, first column of \( B \).

By following this method across all elements, you will get your resulting matrix \( C \). It looks like this: for \( A+B \) of the given matrices:

• The top left corner is \( 3+2 \).
• The element next to it is \( 1+(-3) \).
• Continue this pattern for all elements.

This gives the final matrix:
\[ \begin{bmatrix} 5 & -2 & 7 \ -4 & 3 & 1 \end{bmatrix} \] which results from calculating element-wise sums.

Matrix Subtraction

Matrix subtraction is quite similar to matrix addition, the main difference being that you subtract the elements of the second matrix from the first. It requires both matrices to be of the same size. Let’s take matrices \( A \) and \( B \), both being \( 2 \times 3 \) matrices, as an example.

When performing subtraction, begin by taking the element in the first row and first column of \( A \) and subtract the corresponding element in \( B \). This process is repeated for every element of the matrices.

• For instance, your first calculation will be \( 3-2 \) for the \( A-B \).
• The subsequent calculation for the same row would be \( 1-(-3) \).

Subtraction follows the formula:
\[ \begin{bmatrix} a-b & c-d \ e-f & g-h \end{bmatrix} \] where each letter represents the corresponding elements being subtracted.

For the example in question, the final matrix is:
\[ \begin{bmatrix} 1 & 4 & -5 \ 2 & 1 & 9 \end{bmatrix} \]

Scalar Multiplication

Scalar multiplication involves multiplying every element of a matrix by a single number, known as the scalar. This process transforms the matrix uniformly. Imagine matrix \( A \) and a scalar \(-4\). In this context, each element of \( A \) is multiplied by \(-4\).

The process is simple and looks like this:
\[ \text{To compute } -4A, \begin{bmatrix} -4(3) & -4(1) & -4(1) \ -4(-1) & -4(2) & -4(5) \end{bmatrix} \]

This operation changes each element by scaling it with the scalar, resulting in a new matrix:

• The top left element becomes \(-12\).
• The first element of the second row becomes \(4\) due to the negative sign involved in multiplying two negatives yielding a positive.

Thus, after completing this multiplication, matrix \( -4A \) is represented as:
\[ \begin{bmatrix} -12 & -4 & -4 \ 4 & -8 & -20 \end{bmatrix} \].
It’s important to notice how each element maintains its position while being scaled.

Algebraic Computations in Matrices

Algebraic computations combine different matrix operations to achieve a final expression or form. Such computations can involve both scalar multiplications and matrix additions or subtractions. Let's take an example: \( 3A + 2B \).

To perform this calculation, first multiply each element of matrix \( A \) by \( 3 \) and each element of matrix \( B \) by \( 2 \). This is scalar multiplication executed separately for both matrices. Here is how it looks:

• For \( 3A \), multiply each element by \( 3 \).
• For \( 2B \), multiply each element of \( B \) by \( 2 \).

Once the above calculations are done, add the resulting matrices using matrix addition.

After applying these operations to the problem, you get:

• Matrix after \( 3A \): \( \begin{bmatrix} 9 & 3 & 3 \ -3 & 6 & 15 \end{bmatrix} \)
• Matrix after \( 2B \): \( \begin{bmatrix} 4 & -6 & 12 \ -6 & 2 & -8 \end{bmatrix} \)
• Adding these matrices, we arrive at:
  \[ \begin{bmatrix} 13 & -3 & 15 \ -9 & 8 & 7 \end{bmatrix} \]

Through algebraic computations, we seamlessly integrate scalar multiplication and matrix operations for diverse and complex results.