Problem 15

Question

If possible, find each of the following. (a) \(A+B\) (b) \(3 A\) (c) \(2 A-3 B\) $$A=\left[\begin{array}{rr}2 & -6 \\\3& 1\end{array}\right]$$ $$B=\left[\begin{array}{ll}-1 & 0 \\\\-2 & 3\end{array}\right]$$

Step-by-Step Solution

Verified
Answer
(a) \(\begin{pmatrix} 1 & -6 \\ 1 & 4 \end{pmatrix}\), (b) \(\begin{pmatrix} 6 & -18 \\ 9 & 3 \end{pmatrix}\), (c) \(\begin{pmatrix} 7 & -12 \\ 12 & -7 \end{pmatrix}\)."}
1Step 1: Add Matrices A and B
To add two matrices, simply add corresponding elements. For matrices A and B, \[A = \begin{pmatrix} 2 & -6 \ 3 & 1 \end{pmatrix}\] \[B = \begin{pmatrix} -1 & 0 \ -2 & 3 \end{pmatrix}\].Calculate each element: \[A + B = \begin{pmatrix} 2 + (-1) & -6 + 0 \ 3 + (-2) & 1 + 3 \end{pmatrix} = \begin{pmatrix} 1 & -6 \ 1 & 4 \end{pmatrix}\].
2Step 2: Multiply Matrix A by Scalar 3
To find \(3A\), multiply each element of matrix A by 3:\[A = \begin{pmatrix} 2 & -6 \ 3 & 1 \end{pmatrix}\]Then, \[3A = \begin{pmatrix} 3 \times 2 & 3 \times (-6) \ 3 \times 3 & 3 \times 1 \end{pmatrix} = \begin{pmatrix} 6 & -18 \ 9 & 3 \end{pmatrix}\].
3Step 3: Find 2A - 3B
First, calculate \(2A\) by multiplying each element of matrix A by 2:\[2A = \begin{pmatrix} 2 \times 2 & 2 \times (-6) \ 2 \times 3 & 2 \times 1 \end{pmatrix} = \begin{pmatrix} 4 & -12 \ 6 & 2 \end{pmatrix}\].Then, calculate \(3B\) by multiplying each element of matrix B by 3:\[3B = \begin{pmatrix} 3 \times (-1) & 3 \times 0 \ 3 \times (-2) & 3 \times 3 \end{pmatrix} = \begin{pmatrix} -3 & 0 \ -6 & 9 \end{pmatrix}\].Subtract \(3B\) from \(2A\):\[2A - 3B = \begin{pmatrix} 4 - (-3) & -12 - 0 \ 6 - (-6) & 2 - 9 \end{pmatrix} = \begin{pmatrix} 7 & -12 \ 12 & -7 \end{pmatrix}\].

Key Concepts

Matrix AdditionScalar MultiplicationMatrix Subtraction
Matrix Addition
Matrix addition is a fundamental operation where two matrices of the same dimensions are combined. Each element from one matrix is added to the corresponding element of the other matrix. For this to work, both matrices must have identical dimensions, meaning the same number of rows and columns.

Let's look at an example with matrices \( A \) and \( B \):
  • \( A = \begin{pmatrix} 2 & -6 \ 3 & 1 \end{pmatrix} \)
  • \( B = \begin{pmatrix} -1 & 0 \ -2 & 3 \end{pmatrix} \)
For each element in the resulting matrix, simply add the elements from \( A \) and \( B \):
  • Top left: \( 2 + (-1) = 1 \)
  • Top right: \( -6 + 0 = -6 \)
  • Bottom left: \( 3 + (-2) = 1 \)
  • Bottom right: \( 1 + 3 = 4 \)
The final matrix after addition is \( \begin{pmatrix} 1 & -6 \ 1 & 4 \end{pmatrix} \). Notice how each element is determined independently from others, making the process straightforward.
Scalar Multiplication
Scalar multiplication involves multiplying every element of a matrix by a single number, known as a scalar. This operation is crucial when dealing with systems of linear equations or modifications in vector spaces.

Consider matrix \( A = \begin{pmatrix} 2 & -6 \ 3 & 1 \end{pmatrix} \) and a scalar 3. To find \( 3A \), multiply each element in \( A \) by 3:
  • Top left: \( 3 \times 2 = 6 \)
  • Top right: \( 3 \times (-6) = -18 \)
  • Bottom left: \( 3 \times 3 = 9 \)
  • Bottom right: \( 3 \times 1 = 3 \)
The result of the scalar multiplication is \( \begin{pmatrix} 6 & -18 \ 9 & 3 \end{pmatrix} \). Notice how every element scales by the same factor, making loud transformations visually apparent when comparing matrices.
Matrix Subtraction
Matrix subtraction works similarly to addition, but involves subtracting elements from one matrix by those of another. As with addition, both matrices must have the same dimensions to perform subtraction.

Take matrices \( 2A \) and \( 3B \), which are results from earlier calculations:
  • \( 2A = \begin{pmatrix} 4 & -12 \ 6 & 2 \end{pmatrix} \)
  • \( 3B = \begin{pmatrix} -3 & 0 \ -6 & 9 \end{pmatrix} \)
Subtract corresponding elements from \( 3B \) from \( 2A \):
  • Top left: \( 4 - (-3) = 7 \)
  • Top right: \( -12 - 0 = -12 \)
  • Bottom left: \( 6 - (-6) = 12 \)
  • Bottom right: \( 2 - 9 = -7 \)
The resulting matrix from subtraction is \( \begin{pmatrix} 7 & -12 \ 12 & -7 \end{pmatrix} \). Remember, subtraction requires careful attention to signs and can visually indicate changes or differences between the datasets represented by the matrices.