Problem 4 - Algebra for College Students

Question

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{r} 3 \\ -9 \\ 7 \end{array}\right], \quad B=\left[\begin{array}{r} -6 \\ 12 \\ 9 \end{array}\right] $$

Step-by-Step Solution

Answer

The solutions are: \(A+B = \begin{bmatrix} -3 \\ 3 \\ 16 \end{bmatrix}\), \(A-B = \begin{bmatrix} 9 \\ -21 \\ -2 \end{bmatrix}\), \(2A + 3B = \begin{bmatrix} -12 \\ 18 \\ 41 \end{bmatrix}\), \(4A - 2B = \begin{bmatrix} 24 \\ -60 \\ 10 \end{bmatrix}\).

Step 1: Find A + B

To add two matrices, add the corresponding elements. For matrix \(A\) and \(B\), we perform the following additions: - First element: \(3 + (-6) = -3\) - Second element: \(-9 + 12 = 3\)- Third element: \(7 + 9 = 16\)The resulting matrix \(A + B\) is:\[A+B = \begin{bmatrix} -3 \ 3 \ 16 \end{bmatrix}\]

Step 2: Find A - B

To subtract two matrices, subtract the corresponding elements of the second matrix from the first. For matrix \(A\) and \(B\), we have:- First element: \(3 - (-6) = 9\)- Second element: \(-9 - 12 = -21\)- Third element: \(7 - 9 = -2\)The resulting matrix \(A - B\) is:\[A-B = \begin{bmatrix} 9 \ -21 \ -2 \end{bmatrix}\]

Step 3: Find 2A + 3B

First, multiply matrix \(A\) by 2 and matrix \(B\) by 3:\[2A = 2 \times \begin{bmatrix} 3 \ -9 \ 7 \end{bmatrix} = \begin{bmatrix} 6 \ -18 \ 14 \end{bmatrix}\]\[3B = 3 \times \begin{bmatrix} -6 \ 12 \ 9 \end{bmatrix} = \begin{bmatrix} -18 \ 36 \ 27 \end{bmatrix}\]Now, add the matrices \(2A\) and \(3B\):- First element: \(6 + (-18) = -12\)- Second element: \(-18 + 36 = 18\)- Third element: \(14 + 27 = 41\)The resulting matrix \(2A + 3B\) is:\[2A + 3B = \begin{bmatrix} -12 \ 18 \ 41 \end{bmatrix}\]

Step 4: Find 4A - 2B

First, multiply matrix \(A\) by 4 and matrix \(B\) by 2:\[4A = 4 \times \begin{bmatrix} 3 \ -9 \ 7 \end{bmatrix} = \begin{bmatrix} 12 \ -36 \ 28 \end{bmatrix}\]\[2B = 2 \times \begin{bmatrix} -6 \ 12 \ 9 \end{bmatrix} = \begin{bmatrix} -12 \ 24 \ 18 \end{bmatrix}\]Now, subtract matrix \(2B\) from \(4A\):- First element: \(12 - (-12) = 24\)- Second element: \(-36 - 24 = -60\)- Third element: \(28 - 18 = 10\)The resulting matrix \(4A - 2B\) is:\[4A - 2B = \begin{bmatrix} 24 \ -60 \ 10 \end{bmatrix}\]

Key Concepts

Matrix Addition

Matrix addition is a straightforward operation where corresponding entries of two matrices are added together. This is analogous to regular addition, but it is applied element-wise over matrices. To perform matrix addition, ensure both matrices have the same dimensions. In the given problem, matrices \(A\) and \(B\) are both column matrices with three elements each. Here's how we add them:

• Start by adding each corresponding element: for instance, add the first element of matrix \(A\) and matrix \(B\), which are \(3\) and \(-6\), respectively.
• Continue this process for each element. This leads us to compute: \(3 + (-6) = -3\) for the first element, \(-9 + 12 = 3\) for the second, and \(7 + 9 = 16\) for the third.

Hence, the resulting matrix after addition, \(A + B\), becomes \[A + B = \begin{bmatrix} -3 \ 3 \ 16 \end{bmatrix}.\]Remember, matrix addition is only defined if both matrices are of the same size. This ensures each element has a counterpart in the other matrix.

Matrix Subtraction

Just like matrix addition, matrix subtraction also involves the element-wise operation but with subtraction. Again, both matrices must have identical dimensions to perform subtraction. The process is simple:

• Subtract each element of matrix \(B\) from the corresponding element of matrix \(A\).
• For the given problem: For the first elements: \(3 - (-6) = 9\), for the second elements: \(-9 - 12 = -21\), and for the third elements: \(7 - 9 = -2\).

The result of the subtraction \(A - B\) is a new matrix: \[A - B = \begin{bmatrix} 9 \ -21 \ -2 \end{bmatrix}.\]It's crucial to pay attention to the signs during subtraction, especially since matrix elements can be negative. Keep in mind, this operation "subtracts" each corresponding element in your matrices.

Scalar Multiplication

In scalar multiplication, each element of a matrix is multiplied by a scalar (a constant number). This operation is applied uniformly to every element of the matrix. Let's delve into the problem:

• When finding \(2A\), you multiply each element of matrix \(A\) by \(2\). So, \(3\) becomes \(6\), \(-9\) becomes \(-18\), and \(7\) becomes \(14\).
• Similarly, for \(3B\), each element of matrix \(B\) is multiplied by \(3\): \(-6\) becomes \(-18\), \(12\) becomes \(36\), and \(9\) becomes \(27\).

The resulting matrices here are: \[2A = \begin{bmatrix} 6 \ -18 \ 14 \end{bmatrix}, \ 3B = \begin{bmatrix} -18 \ 36 \ 27 \end{bmatrix}. \]Scalar multiplication often sets the stage for more complex operations like linear combinations of matrices, as seen when combining with addition and subtraction.

Matrix Arithmetic Problems

Matrix arithmetic problems can often involve a combination of addition, subtraction, and scalar multiplication. These operations can be pieced together to solve more complex problems. Let's break down an example from the problem:To find \(2A + 3B\):

• First, calculate \(2A\) and \(3B\) using scalar multiplication.
• Then, simply add these matrices: resulting in the matrix \(2A + 3B = \begin{bmatrix} -12 \ 18 \ 41 \end{bmatrix}\).

Similarly, consider \(4A - 2B\):

• Compute \(4A\) and \(2B\) using scalar multiplication.
• Subtract the two: yielding \(4A - 2B = \begin{bmatrix} 24 \ -60 \ 10 \end{bmatrix}\).

Understanding these operations in isolation helps tackle matrix arithmetic problems as a whole, providing answers by systematically applying step-by-step matrix operations.