Problem 1 - Precalculus

Question

For each pair of matrices \(A\) and \(B\) in Exercises \(1-7,\) find the following, if defined \- \(3 A\) \- \(-B\) \- \(A^{2}\) \- \(A-2 B\) \- \(A B\) \- \(B A\) $$ A=\left[\begin{array}{rr} 2 & -3 \\ 1 & 4 \end{array}\right], B=\left[\begin{array}{rr} 5 & -2 \\ 4 & 8 \end{array}\right] $$

Step-by-Step Solution

Answer

The results are \(3A = \begin{bmatrix} 6 & -9 \\ 3 & 12 \end{bmatrix}\), \(-B = \begin{bmatrix} -5 & 2 \\ -4 & -8 \end{bmatrix}\), \(A^2 = \begin{bmatrix} 1 & -18 \\ 6 & 13 \end{bmatrix}\), \(A - 2B = \begin{bmatrix} -8 & 1 \\ -7 & -12 \end{bmatrix}\), \(AB = \begin{bmatrix} -2 & -28 \\ 21 & 30 \end{bmatrix}\), and \(BA = \begin{bmatrix} 8 & -23 \\ 16 & 20 \end{bmatrix}\).

Step 1: Compute 3A

To find \(3A\), multiply each entry of matrix \(A\) by 3.\[3A = 3 \times \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} = \begin{bmatrix} 3 \times 2 & 3 \times (-3) \ 3 \times 1 & 3 \times 4 \end{bmatrix} = \begin{bmatrix} 6 & -9 \ 3 & 12 \end{bmatrix}\]

Step 2: Compute -B

To find \(-B\), multiply each entry of matrix \(B\) by -1 to negate the matrix.\[-B = -1 \times \begin{bmatrix} 5 & -2 \ 4 & 8 \end{bmatrix} = \begin{bmatrix} -5 & 2 \ -4 & -8 \end{bmatrix}\]

Step 3: Compute A^2

To find \(A^{2}\), multiply matrix \(A\) by itself.\[A^{2} = \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} = \begin{bmatrix} (2 \times 2 + (-3) \times 1) & (2 \times -3 + (-3) \times 4) \ (1 \times 2 + 4 \times 1) & (1 \times -3 + 4 \times 4) \end{bmatrix} = \begin{bmatrix} 1 & -18 \ 6 & 13 \end{bmatrix}\]

Step 4: Compute A - 2B

To find \(A - 2B\), first find \(2B\) and then subtract it from \(A\).\[2B = 2 \times \begin{bmatrix} 5 & -2 \ 4 & 8 \end{bmatrix} = \begin{bmatrix} 10 & -4 \ 8 & 16 \end{bmatrix}\]Now subtract \(2B\) from \(A\):\[A - 2B = \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} - \begin{bmatrix} 10 & -4 \ 8 & 16 \end{bmatrix} = \begin{bmatrix} 2 - 10 & -3 + 4 \ 1 - 8 & 4 - 16 \end{bmatrix} = \begin{bmatrix} -8 & 1 \ -7 & -12 \end{bmatrix}\]

Step 5: Compute AB

To find the product \(AB\), multiply matrix \(A\) by matrix \(B\).\[AB = \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} \times \begin{bmatrix} 5 & -2 \ 4 & 8 \end{bmatrix} = \begin{bmatrix} (2 \times 5 + (-3) \times 4) & (2 \times -2 + (-3) \times 8) \ (1 \times 5 + 4 \times 4) & (1 \times -2 + 4 \times 8) \end{bmatrix} = \begin{bmatrix} -2 & -28 \ 21 & 30 \end{bmatrix}\]

Step 6: Compute BA

To find the product \(BA\), multiply matrix \(B\) by matrix \(A\).\[BA = \begin{bmatrix} 5 & -2 \ 4 & 8 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} = \begin{bmatrix} (5 \times 2 + (-2) \times 1) & (5 \times -3 + (-2) \times 4) \ (4 \times 2 + 8 \times 1) & (4 \times -3 + 8 \times 4) \end{bmatrix} = \begin{bmatrix} 8 & -23 \ 16 & 20 \end{bmatrix}\]

Key Concepts

Scalar Multiplication

Scalar multiplication of matrices makes calculations more efficient and intuitive by stretching or shrinking matrix entries by a single number, called a 'scalar'. You perform this operation by taking each element of the matrix and multiplying it by the scalar. For instance, when you have a matrix \( A \) and a scalar 3, multiplying \( A \) by 3, written as \( 3A \), involves taking every element of \( A \) and multiplying it by 3.

Let's take the matrix \( A = \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} \). To compute \( 3A \), you multiply each element:

• \( 3 \times 2 = 6 \)
• \( 3 \times (-3) = -9 \)
• \( 3 \times 1 = 3 \)
• \( 3 \times 4 = 12 \)

So, the resulting matrix is \( \begin{bmatrix} 6 & -9 \ 3 & 12 \end{bmatrix} \). This process is straightforward but powerful, especially when used in more complex calculations.

Matrix Addition

When you add matrices, you must ensure they have the same dimensions. This is because matrix addition is done element-wise, pairing each entry in one matrix with its counterpart in the other matrix.

For example, if you have two matrices \( A \) and \( B \), both of size 2x2, say:\( A = \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 5 & -2 \ 4 & 8 \end{bmatrix} \), you add them like this:

• Add the top-left corners: \( 2 + 5 = 7 \).
• Add the top-right corners: \( -3 + (-2) = -5 \).
• Add the bottom-left corners: \( 1 + 4 = 5 \).
• Add the bottom-right corners: \( 4 + 8 = 12 \).

Therefore, the resulting matrix is \( \begin{bmatrix} 7 & -5 \ 5 & 12 \end{bmatrix} \). This simple process makes matrix addition an efficient and essential operation in linear algebra.

Matrix Multiplication

Matrix multiplication is more complex than addition but provides powerful results, especially when no commutativity is involved, which means \( AB eq BA \) for matrices \( A \) and \( B \).

To perform matrix multiplication, you examine rows and columns. For matrix \( A \) to multiply matrix \( B \), the number of columns in \( A \) must equal the number of rows in \( B \). So, if \( A = \begin{bmatrix} 2 & -3 \ 1 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 5 & -2 \ 4 & 8 \end{bmatrix} \), both 2x2 matrices, you can compute the product \( AB \):

• Compute the top-left entry by multiplying the first row of \( A \) by the first column of \( B \): \( 2 \times 5 + (-3) \times 4 = -2 \).
• Compute the top-right entry by multiplying the first row of \( A \) by the second column of \( B \): \( 2 \times -2 + (-3) \times 8 = -28 \).
• Compute the bottom-left entry by multiplying the second row of \( A \) by the first column of \( B \): \( 1 \times 5 + 4 \times 4 = 21 \).
• Compute the bottom-right entry by multiplying the second row of \( A \) by the second column of \( B \): \( 1 \times -2 + 4 \times 8 = 30 \).

The result is \( \begin{bmatrix} -2 & -28 \ 21 & 30 \end{bmatrix} \). The distinctiveness of each multiplication highlights each matrix's contribution to the overall product.

Matrix Negation

Negating a matrix is a simple yet useful operation, particularly when solving systems of linear equations or performing other calculations involving inversions.

To negate a matrix, you multiply every entry by -1. For the matrix \( B = \begin{bmatrix} 5 & -2 \ 4 & 8 \end{bmatrix} \), the negation \( -B \) is calculated by:

• \( -1 \times 5 = -5 \)
• \( -1 \times (-2) = 2 \)
• \( -1 \times 4 = -4 \)
• \( -1 \times 8 = -8 \)

As a result, the negated matrix is \( \begin{bmatrix} -5 & 2 \ -4 & -8 \end{bmatrix} \). Each element simply changes sign, making it an easy operation to visualize and perform. Although it seems straightforward, negation is crucial in many larger, more complex computations.