Problem 5 - Algebra for College Students

Question

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 4 A-3 B $$

Step-by-Step Solution

Answer

The matrix is \( \begin{bmatrix} -2 & 1 \\ -3 & 19 \end{bmatrix} \).

Step 1: Multiply Matrix A by 4

Multiply each element of matrix \( A \) by 4:\[ 4A = 4 \times \begin{bmatrix} 1 & -2 \ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 \times 1 & 4 \times (-2) \ 4 \times 3 & 4 \times 4 \end{bmatrix} = \begin{bmatrix} 4 & -8 \ 12 & 16 \end{bmatrix} \]

Step 2: Multiply Matrix B by 3

Multiply each element of matrix \( B \) by 3:\[ 3B = 3 \times \begin{bmatrix} 2 & -3 \ 5 & -1 \end{bmatrix} = \begin{bmatrix} 3 \times 2 & 3 \times (-3) \ 3 \times 5 & 3 \times (-1) \end{bmatrix} = \begin{bmatrix} 6 & -9 \ 15 & -3 \end{bmatrix} \]

Step 3: Subtract 3B from 4A

Subtract the corresponding elements of the matrix \( 3B \) from \( 4A \):\[ 4A - 3B = \begin{bmatrix} 4 & -8 \ 12 & 16 \end{bmatrix} - \begin{bmatrix} 6 & -9 \ 15 & -3 \end{bmatrix} = \begin{bmatrix} 4 - 6 & -8 - (-9) \ 12 - 15 & 16 - (-3) \end{bmatrix} \]This gives:\[ \begin{bmatrix} -2 & 1 \ -3 & 19 \end{bmatrix} \]

Key Concepts

Matrix Multiplication

Matrix multiplication is not as straightforward as multiplying individual numbers. Instead, it involves applying a specific rule: the elements of the rows in one matrix are multiplied by the elements of the columns in another matrix, and the resulting products are summed up. The process is repeated for each element in the result matrix.
For two matrices to be multiplied, the number of columns in the first matrix must equal the number of rows in the second matrix. If matrix \(A\) has dimensions \( m \times n \) and matrix \(B\) has dimensions \( n \times p \), the resultant matrix will have dimensions \( m \times p \). This rule ensures the multiplication is defined.

• Start with the first row of the first matrix and the first column of the second matrix.
• Multiply each element of the row by the corresponding element of the column.
• Add the results to get the element in the product matrix.
• Repeat the process for all rows and columns.

Remember, the order of multiplication matters in matrices, unlike numbers. So \(AB\) is not always equal to \(BA\).

Matrix Subtraction

Matrix subtraction involves the subtraction of corresponding elements from two matrices of the same dimensions. This step requires aligning the matrices perfectly to ensure each element has a matching counterpart to subtract it from.
Matrix subtraction is fairly straightforward as long as both matrices involved have identical rows and columns.

• Align both matrices one above the other.
• Subtract each corresponding element in the two matrices.
• The dimensions of the resulting matrix will be the same as the original matrices.

Understanding this helps simplify operations with matrices, as seen in the solution where \(3B\) is subtracted from \(4A\) by directly subtracting corresponding elements.

Matrix Scalar Multiplication

Matrix scalar multiplication is one of the simplest matrix operations. It involves multiplying every element of a matrix by a single scalar number. This operation scales the entire matrix by the scalar value.
In the original exercise, the matrices \(A\) and \(B\) are multiplied by scalars 4 and 3, respectively.

• Take the scalar value and multiply it with each element of the matrix.
• This operation maintains the dimensions of the matrix, as only values change, not structure.
• The operation is distributive, meaning scalar multiplication is performed before any addition or subtraction, as seen in the solution steps with \(4A\) and \(3B\).

Matrix scalar multiplication is helpful for scaling and balancing matrix equations effectively.