Problem 32 - Algebra 2

Question

Use matrices \(A, B, C,\) and \(D\) to find the following. $$A=\left[\begin{array}{rr}{5} & {7} \\ {-1} & {6} \\ {3} & {-9}\end{array}\right] \quad B=\left[\begin{array}{ll}{8} & {3} \\ {5} & {1} \\\ {4} & {4}\end{array}\right] \quad C=\left[\begin{array}{rr}{0} & {4} \\\ {-2} & {5} \\ {7} & {-1}\end{array}\right] \quad D=\left[\begin{array}{ll}{6} & {2} \\ {9} & {0} \\ {-3} & {0}\end{array}\right]$$ \(6 B-2 A\)

Step-by-Step Solution

Answer

The result is \( \begin{bmatrix} 38 & 4 \\ 32 & -6 \\ 18 & 42 \end{bmatrix} \).

Step 1: Multiply matrix B by 6

To begin, let's find the scalar multiplication of matrix B by 6. Each entry in matrix B should be multiplied by 6.\[ 6 imes B = 6 \times \begin{bmatrix} 8 & 3 \ 5 & 1 \ 4 & 4 \end{bmatrix} = \begin{bmatrix} 48 & 18 \ 30 & 6 \ 24 & 24 \end{bmatrix} \]

Step 2: Multiply matrix A by 2

Now, let's perform the scalar multiplication of matrix A by 2. Each entry in matrix A should be multiplied by 2.\[ 2 imes A = 2 \times \begin{bmatrix} 5 & 7 \ -1 & 6 \ 3 & -9 \end{bmatrix} = \begin{bmatrix} 10 & 14 \ -2 & 12 \ 6 & -18 \end{bmatrix} \]

Step 3: Subtract matrices

With both scalar multiplications completed, subtract the resulting matrix from Step 2 from the resulting matrix in Step 1.\[ \begin{bmatrix} 48 & 18 \ 30 & 6 \ 24 & 24 \end{bmatrix} - \begin{bmatrix} 10 & 14 \ -2 & 12 \ 6 & -18 \end{bmatrix} = \begin{bmatrix} 48-10 & 18-14 \ 30 - (-2) & 6 - 12 \ 24 - 6 & 24 - (-18) \end{bmatrix} \]

Step 4: Simplify the result

Finally, simplify the matrix by calculating the entries obtained from the subtraction:\[ \begin{bmatrix} 38 & 4 \ 32 & -6 \ 18 & 42 \end{bmatrix} \]

Key Concepts

Scalar Multiplication

Scalar multiplication is an essential operation when dealing with matrices. It involves multiplying every element of a matrix by a specific scalar value. Here, the entire matrix scales uniformly, much like stretching or compressing the matrix by the scalar factor.
Imagine you have a matrix, such as matrix \( B \) from our example, and a scalar, like 6. To perform the scalar multiplication \( 6 \times B \), simply multiply each entry of \( B \) by 6. For instance, if an entry is 8, it becomes \( 8 \times 6 = 48 \) after scaling. This operation repeats for each element of the matrix, leading to a new rescaled matrix.
Scalar multiplication:

• Is straightforward and involves basic arithmetic operations.
• Results in a matrix of the same size as the original.

Thus, scalar multiplication is a fundamental technique in matrix algebra, allowing matrices to adjust in "size" throughout calculations.

Matrix Subtraction

Matrix subtraction is akin to subtracting numerical values but performed between matrices. However, it's crucial that both matrices involved in the subtraction have the same dimensions (number of rows and columns).
Think of this as element-wise subtraction, where corresponding elements from one matrix are subtracted from the other. For example, if you have matrix \( X \) and matrix \( Y \), and both have dimensions 3x2, the subtraction \( X - Y \) computes as the difference between every corresponding pair of elements.
Here's the pattern of matrix subtraction:

• Align the matrices side by side.
• Subtract each element in the first matrix from the corresponding element in the second.

In our exercise, subtracting \( 2A \) from \( 6B \) involves this component-wise method:

• For the top-left element: the result is \( 48 - 10 = 38 \).
• For other elements, similar calculations apply, yielding the final subtraction matrix.

Following these steps will always guide you to the correct outcome.

Algebraic Expressions

Algebraic expressions with matrices incorporate basic operations like addition, subtraction, and scalar multiplication. Each operation must adhere to the rules of matrix mathematics to derive accurate results. Algebra here acts as a helpful tool to simplify and combine matrix expressions efficiently.
Consider the expression \( 6B - 2A \). It's a mathematical sentence where matrices and scalars are combined using operators. The goal is to systematically execute each operation based on the correct order of operations, which involves:

• Performing scalar multiplications first, e.g., \( 6 \times B \) and \( 2 \times A \).
• Subsequently, conducting matrix subtraction, in this case, \( 6B - 2A \).

By understanding each part's behaviour in an expression, you can seamlessly simplify or evaluate it. This expression handling becomes vital in more complex problems and larger systems of equations, where multiple matrices interact.

Matrices

Matrices are fascinating structures in mathematics, composed of rows and columns that form rectangular arrays. Each position in this array holds an element, and these matrices represent data compactly, enabling operations that manipulate and analyze batches of information altogether.
Matrices have various applications across different fields, from solving linear equations in engineering to graphics imaging in computer science. Understanding their structure and classification is key. Generally, matrices can be identified by:

• Size: Defined by its number-of-rows-by-number-of-columns, such as 3x2 matrices seen in our exercise matrices \( A, B, C, \) and \( D \).
• Shape: Determined by whether it is a square matrix, rectangular, or forms another identifiable structure.
• Type: May include identity matrices, zero matrices, or diagonal matrices, each serving unique purposes.

In No time will you be able to navigate between matrices and employ operations like multiplication, addition, and subtraction proficiently. Familiarity with this versatile tool adds a powerful arsenal of techniques at your disposal.