Problem 7 - College Algebra and Calculus: An Applied Approach

Question

Find (a) \(A+B\), (b) \(A-B\), (c) \(3 A\), and (d) \(3 A-2 B\). $$ A=\left|\begin{array}{rr} 6 & -1 \\ 2 & 4 \\ -3 & 5 \end{array}\right|, B=\left|\begin{array}{rr} 1 & 4 \\ -1 & 5 \\ 1 & 10 \end{array}\right| $$

Step-by-Step Solution

Answer

(a) \(A + B = \begin{array}{cc} 7 & 3 \ 1 & 9 \ -2 & 15 \end{array} \), (b) \(A - B = \begin{array}{cc} 5 & -5 \ 3 & -1 \ -4 & -5 \ \end{array} \), (c) \(3A = \begin{array}{cc} 18 & -3 \ 6 & 12 \ -9 & 15 \ \end{array} \), (d) \(3A - 2B = \begin{array}{cc} 16 & -11 \ 8 & 2 \ -11 & -5 \ \end{array} \).

Step 1: Find \(A + B\)

The addition of two matrices, say A and B, can be performed by adding the corresponding elements. Hence, \(A + B\) is:\[\begin{array}{rr}6 & -1 \2 & 4 \-3 & 5\end{array}+\begin{array}{rr}1 & 4 \-1 & 5 \1 & 10\end{array}=\begin{array}{cc}(6+1) & (-1+4) \(2-1) & (4+5) \(-3+1) & (5+10)\end{array}=\begin{array}{cc}7 & 3 \1 & 9 \-2 & 15\end{array}\]

Step 2: Find \(A - B\)

Subtraction of two matrices involves subtracting corresponding elements of the two matrices. Hence \(A - B\) is:\[\begin{array}{rr}6 & -1 \2 & 4 \-3 & 5\end{array}-\begin{array}{rr}1 & 4 \-1 & 5 \1 & 10\end{array}=\begin{array}{cc}(6-1) & (-1-4) \(2-(-1)) & (4-5) \(-3-1) & (5-10)\end{array}=\begin{array}{cc}5 & -5 \3 & -1 \-4 & -5\end{array}\]

Step 3: Find \(3A\)

Multiplication of a matrix by a scalar involves multiplying each element of the matrix by the scalar value. Hence \(3A\) is:\[3 \begin{array}{rr}6 & -1 \2 & 4 \-3 & 5\end{array}=\begin{array}{cc}3*6 & 3*(-1) \3*2 & 3*4 \3*(-3) & 3*5\end{array}=\begin{array}{cc}18 & -3 \6 & 12 \-9 & 15\end{array}\]

Step 4: Find \(3 A - 2 B\)

This involves first multiplying the scalar values with their respective matrices and then performing a subtractions of the two resulting matrices. So \(3A - 2B\) is:\[3 \begin{array}{rr}6 & -1 \2 & 4 \-3 & 5\end{array}-2\begin{array}{cc}1 & 4 \-1 & 5 \1 & 10\end{array}=\begin{array}{rr}18 & -3 \6 & 12 \-9 & 15\end{array}-\begin{array}{cc}2 & 8 \-2 & 10 \2 & 20\end{array}=\begin{array}{cc}(18-2) & (-3-8) \(6-(-2)) & (12-10) \(-9-2) & (15-20)\end{array}=\begin{array}{cc}16 & -11 \8 & 2 \-11 & -5\end{array}\]

Key Concepts

Matrix Addition

Adding matrices is like adding a bunch of numbers, but in a structured way. We're working with two matrices here, matrix \(A\) and matrix \(B\). To add these, we look at their corresponding elements: basically, just pair the numbers that are in the same spot in each matrix.

For example:

• In the top left corner of both matrices, \(6\) (from \(A\)) and \(1\) (from \(B\)) simply become \((6 + 1) = 7\).
• Do this for all corresponding numbers: top right, bottom left, and so on.

When you’re done, you should have a new matrix with all new numbers at each spot thanks to this element-wise addition.

The result from adding matrices \(A\) and \(B\) appears as:\[A + B = \left[\begin{array}{rr}7 & 3 \1 & 9 \-2 & 15\end{array}\right]\]

Matrix Subtraction

Subtracting matrices is very similar to matrix addition, but with a little twist. Instead of adding the numbers in the same position, we subtract them. Take the number from one matrix and subtract the number in the same position in the other matrix.

Here's how it works:

• Take the top left value of \(A\) which is \(6\), and subtract the top left value of \(B\) which is \(1\): \(6 - 1 = 5\).
• Do this for every pair of corresponding elements in both matrices.

This operation takes care of each position one by one, giving you a clear matrix that's the result of these subtractions.

The matrix \(A - B\) becomes:\[A - B = \left[\begin{array}{rr}5 & -5 \3 & -1 \-4 & -5\end{array}\right]\]

Scalar Multiplication

Scalar multiplication is kind of like making a cake bigger by multiplying its ingredients. When we multiply a matrix by a scalar, every element in the matrix gets multiplied by that number.

For instance, multiplying matrix \(A\) by the scalar \(3\) is simply:

• Multiply the top left element (\(6\)) by \(3\): \(3 \times 6 = 18\).
• Repeat for every element in the matrix.
• This means every single number changes, but stays in the same position.

The whole matrix scales up by the scalar value we use.

When we do this for the matrix \(A\), we find:\[3A = \left[\begin{array}{rr}18 & -3 \6 & 12 \-9 & 15\end{array}\right]\]

Linear Algebra

Linear algebra is the area of mathematics that helps us with things like matrix operations. It's really about understanding vectors and spaces and how they interact. Matrices, as we saw with addition, subtraction, and scalar multiplication, are a fundamental building block in this field.

Here's why it’s useful:

• Matrices are like maps for transforming data, like scaling a 2D shape or rotating it in space.
• Operations like the ones we performed can model real-world systems such as computer graphics, physics simulations, and even solving systems of equations.

Understanding these basics—how to add, subtract, and scale matrices—gives us a strong foundation to explore more complex concepts in linear algebra.

This knowledge is crucial for diving deeper into the stunning world of mathematics called linear algebra, where matrices become more than just numbers in a grid.