Problem 15 - Algebra and Trigonometry

Question

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{rrr} {2} & {-10} & {-2} \\ {14} & {12} & {10} \\ {4} & {-2} & {2} \end{array}\right], \quad B=\left[\begin{array}{rrr} {6} & {10} & {-2} \\ {0} & {-12} & {-4} \\ {-5} & {2} & {-2} \end{array}\right] $$

Step-by-Step Solution

Answer

a. \( A + B = \left[\begin{array}{ccc}{8} & {0} & {-4} \ {14} & {0} & {6} \ {-1} & {0} & {0}\end{array}\right] \) \n b. \( A - B = \left[\begin{array}{ccc}{-4} & {-20} & {0} \ {14} & {24} & {14} \ {9} & {-4} & {4}\end{array}\right] \) \n c. \( -4A = \left[\begin{array}{ccc}{-8} & {40} & {8} \ {-56} & {-48} & {-40} \ {-16} & {8} & {-8}\end{array}\right] \) \n d. \( 3A + 2B = \left[\begin{array}{ccc}{18} & {0} & {-10} \ {42} & {0} & {26} \ {2} & {0} & {2}\end{array}\right] \)

Step 1: Adding the Matrices A and B

To add two matrices, they must be of the same dimension, which indeed matrix A and B are. Addition is performed element-wise, i.e., each corresponding element of the two matrices are added together. Hence \( A + B = \left[\begin{array}{ccc}{2+6} & {-10+10} & {-2+(-2)} \ {14+0} & {12+(-12)} & {10+(-4)} \ {4+(-5)} & {-2+2} & {2+(-2)}\end{array}\right] = \left[\begin{array}{ccc}{8} & {0} & {-4} \ {14} & {0} & {6} \ {-1} & {0} & {0}\end{array}\right] \)

Step 2: Subtracting Matrix B from A

Subtraction is also performed element-wise. Hence \( A - B = \left[\begin{array}{ccc}{2-6} & {-10-10} & {-2+2} \ {14-0} & {12+12} & {10+4} \ {4+5} & {-2-2} & {2+2}\end{array}\right] = \left[\begin{array}{ccc}{-4} & {-20} & {0} \ {14} & {24} & {14} \ {9} & {-4} & {4}\end{array}\right] \)

Step 3: Multiply Matrix A by -4

Multiplication of a matrix by a scalar includes multiplying each element of the matrix by the scalar. So, \( -4A = \left[\begin{array}{ccc}{-4*2} & {-4*-10} & {-4*-2} \ {-4*14} & {-4*12} & {-4*10} \ {-4*4} & {-4*-2} & {-4*2}\end{array}\right] = \left[\begin{array}{ccc}{-8} & {40} & {8} \ {-56} & {-48} & {-40} \ {-16} & {8} & {-8}\end{array}\right] \)

Step 4: The linear combination of matrices A and B

The operation is \( 3A + 2B = \left[\begin{array}{ccc}{3*2+2*6} & {3*-10+2*10} & {3*-2+2*-2} \ {3*14+2*0} & {3*12+2*-12} & {3*10+2*-4} \ {3*4+2*(-5)} & {3*-2+2*2} & {3*2+2*-2}\end{array}\right] = \left[\begin{array}{ccc}{18} & {0} & {-10} \ {42} & {0} & {26} \ {2} & {0} & {2}\end{array}\right] \)

Key Concepts

Matrix Addition

Matrix addition is a simple yet crucial operation in linear algebra. When adding matrices, it is essential that the matrices have the same dimensions. This means they must have the same number of rows and columns. The addition process involves adding corresponding elements from each matrix. For example, if we have two matrices, A and B, each with the same dimensions, you simply add each element from matrix A with its matching element in matrix B.

Here's a quick breakdown of performing matrix addition:

• Ensure matrices are the same size.
• Add each element in matrix A to its corresponding element in matrix B: \( C_{ij} = A_{ij} + B_{ij} \).

For example, with matrices:\[A = \begin{bmatrix}2 & -10 & -2 \14 & 12 & 10 \4 & -2 & 2\end{bmatrix}, \quadB = \begin{bmatrix}6 & 10 & -2 \0 & -12 & -4 \-5 & 2 & -2\end{bmatrix}\] The added matrix \( C = A + B \) will be:\[C = \begin{bmatrix}8 & 0 & -4 \14 & 0 & 6 \-1 & 0 & 0\end{bmatrix}\]This method is straightforward and ensures a new matrix, where each entry is the sum of the corresponding entries of the original matrices.

Matrix Subtraction

Matrix subtraction is quite similar to matrix addition. It requires the same condition that the matrices involved have the same dimensions, meaning the same number of rows and columns. The subtraction process involves taking corresponding elements from both matrices and finding their difference.

To subtract matrix B from matrix A, follow these steps:

• Make sure matrices A and B are of identical dimensions.
• Subtract each element of matrix B from its corresponding element in matrix A: \( C_{ij} = A_{ij} - B_{ij} \).

For the example matrices:\[A = \begin{bmatrix}2 & -10 & -2 \14 & 12 & 10 \4 & -2 & 2\end{bmatrix}, \quadB = \begin{bmatrix}6 & 10 & -2 \0 & -12 & -4 \-5 & 2 & -2\end{bmatrix}\]The resulting difference, \( C = A - B \), is:\[C = \begin{bmatrix}-4 & -20 & 0 \14 & 24 & 14 \9 & -4 & 4\end{bmatrix}\]Matrix subtraction is just as easy as addition, as long as you carefully perform the element-wise operations.

Scalar Multiplication

Scalar multiplication involves multiplying every element of a matrix by a scalar value, which is just a single real number. This operation changes the magnitude of the matrix but not its direction. Scalar multiplication is useful in scaling matrices when solving equations and linear transformations.

Let's say you want to multiply matrix A by the scalar \(-4\). Here’s how you do it:

• Multiply each element of matrix A by -4: \( C_{ij} = -4 imes A_{ij} \).

For example, with matrix:\[A = \begin{bmatrix}2 & -10 & -2 \14 & 12 & 10 \4 & -2 & 2\end{bmatrix}\]The result of \(-4A\) will be:\[C = \begin{bmatrix}-8 & 40 & 8 \-56 & -48 & -40 \-16 & 8 & -8\end{bmatrix}\]Each entry in the original matrix \( A \) is multiplied by the scalar \(-4\), giving a matrix that is simply scaled by -4. This is a key operation that is frequently used to manipulate matrices efficiently.

Linear Combination of Matrices

A linear combination of matrices is an operation where matrices are combined using both addition (or subtraction) and scalar multiplication. Each matrix is multiplied by a scalar, and the results are then added together. This is frequently used in various algebra and calculus problems, including finding the vector space.

To form a linear combination like \( 3A + 2B \):

• Multiply each element of matrix A by its scalar, \( 3 \).
• Multiply each element of matrix B by its scalar, \( 2 \).
• Add the resulting matrices from the previous two steps.

For example, with matrices:\[A = \begin{bmatrix}2 & -10 & -2 \14 & 12 & 10 \4 & -2 & 2\end{bmatrix}, \quadB = \begin{bmatrix}6 & 10 & -2 \0 & -12 & -4 \-5 & 2 & -2\end{bmatrix}\]The combination \( 3A + 2B \) yields:\[C = \begin{bmatrix}18 & 0 & -10 \42 & 0 & 26 \2 & 0 & 2\end{bmatrix}\]This process involves both scalar multiplication and matrix addition, demonstrating the flexibility and power of linear algebra operations.