Question

Given \(A=\left[\begin{array}{rr}7 & -1 \\ 6 & 9\end{array}\right], B=\left[\begin{array}{rr}0 & 4 \\ 3 & -2\end{array}\right],\) and \(C=\left[\begin{array}{ll}8 & 3 \\ 6 & 1\end{array}\right],\) find (a) \(A+B\) (b) \(C-A\) \((c) 3 A\) \((d) 4 B+2 C\)

Short answer

(a) \( \begin{bmatrix} 7 & 3 \\ 9 & 7 \end{bmatrix} \); (b) \( \begin{bmatrix} 1 & 4 \\ 0 & -8 \end{bmatrix} \); (c) \( \begin{bmatrix} 21 & -3 \\ 18 & 27 \end{bmatrix} \); (d) \( \begin{bmatrix} 16 & 22 \\ 24 & -6 \end{bmatrix} \)."

Step-by-step solution

Add Matrices A and B

To find \( A + B \), add the corresponding elements of matrices \( A \) and \( B \).\[A+B = \begin{bmatrix} 7 & -1 \ 6 & 9 \end{bmatrix} + \begin{bmatrix} 0 & 4 \ 3 & -2 \end{bmatrix} = \begin{bmatrix} 7+0 & -1+4 \ 6+3 & 9-2 \end{bmatrix} = \begin{bmatrix} 7 & 3 \ 9 & 7 \end{bmatrix}\]

Subtract Matrix A from C

To evaluate \( C - A \), subtract the corresponding elements of matrix \( A \) from \( C \).\[C-A = \begin{bmatrix} 8 & 3 \ 6 & 1 \end{bmatrix} - \begin{bmatrix} 7 & -1 \ 6 & 9 \end{bmatrix} = \begin{bmatrix} 8-7 & 3+1 \ 6-6 & 1-9 \end{bmatrix} = \begin{bmatrix} 1 & 4 \ 0 & -8 \end{bmatrix}\]

Multiply Matrix A by 3

Multiply each element of matrix \( A \) by 3 for \( 3A \).\[3A = 3 \times \begin{bmatrix} 7 & -1 \ 6 & 9 \end{bmatrix} = \begin{bmatrix} 3 \times 7 & 3 \times (-1) \ 3 \times 6 & 3 \times 9 \end{bmatrix} = \begin{bmatrix} 21 & -3 \ 18 & 27 \end{bmatrix}\]

Calculate 4B

Multiply each element of matrix \( B \) by 4.\[4B = 4 \times \begin{bmatrix} 0 & 4 \ 3 & -2 \end{bmatrix} = \begin{bmatrix} 4 \times 0 & 4 \times 4 \ 4 \times 3 & 4 \times (-2) \end{bmatrix} = \begin{bmatrix} 0 & 16 \ 12 & -8 \end{bmatrix}\]

Calculate 2C

Multiply each element of matrix \( C \) by 2.\[2C = 2 \times \begin{bmatrix} 8 & 3 \ 6 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 8 & 2 \times 3 \ 2 \times 6 & 2 \times 1 \end{bmatrix} = \begin{bmatrix} 16 & 6 \ 12 & 2 \end{bmatrix}\]

Add 4B and 2C

Add the matrices \( 4B \) and \( 2C \) from previous steps.\[4B + 2C = \begin{bmatrix} 0 & 16 \ 12 & -8 \end{bmatrix} + \begin{bmatrix} 16 & 6 \ 12 & 2 \end{bmatrix} = \begin{bmatrix} 0+16 & 16+6 \ 12+12 & -8+2 \end{bmatrix} = \begin{bmatrix} 16 & 22 \ 24 & -6 \end{bmatrix}\]

Key concepts

Matrix Addition

To add two matrices, you simply combine their corresponding elements. It is crucial that the matrices have the same dimensions, meaning the same number of rows and columns. Let’s take matrices \(A\) and \(B\) as an example: \(A = \left[ \begin{array}{rr} 7 & -1 \ 6 & 9 \end{array} \right]\) and \(B = \left[ \begin{array}{rr} 0 & 4 \ 3 & -2 \end{array} \right]\).Adding these involves the following steps:

• Add the elements in the first row, first column: \(7 + 0 = 7\)
• Add the elements in the first row, second column: \(-1 + 4 = 3\)
• Add the elements in the second row, first column: \(6 + 3 = 9\)
• Add the elements in the second row, second column: \(9 + (-2) = 7\)

So, the resulting matrix is \(\begin{bmatrix} 7 & 3 \ 9 & 7 \end{bmatrix}\). Matrix addition can be thought of as combining the matrices to create a new matrix that represents the sum of their corresponding components.

Matrix Subtraction

Matrix subtraction is similar to matrix addition, but instead of adding the corresponding elements, you subtract them. For example, to find \(C - A\) using matrices \(C = \left[ \begin{array}{rr} 8 & 3 \ 6 & 1 \end{array} \right]\) and \(A = \left[ \begin{array}{rr} 7 & -1 \ 6 & 9 \end{array} \right]\), you subtract the elements of \(A\) from \(C\):

• Subtract the elements in the first row, first column: \(8 - 7 = 1\)
• Subtract the elements in the first row, second column: \(3 - (-1) = 4\)
• Subtract the elements in the second row, first column: \(6 - 6 = 0\)
• Subtract the elements in the second row, second column: \(1 - 9 = -8\)

The resulting matrix becomes \(\begin{bmatrix} 1 & 4 \ 0 & -8 \end{bmatrix}\). Subtracting matrices can be viewed as finding the difference in value for each corresponding component, creating a new matrix that reflects these differences.

Scalar Multiplication

Scalar multiplication involves multiplying every entry in a matrix by a given scalar value. It is an easy yet powerful operation that scales the entire matrix by a uniform factor. Consider matrix \(A = \left[ \begin{array}{rr} 7 & -1 \ 6 & 9 \end{array} \right]\) and a scalar of 3.To find \(3A\):

• Multiply every element of \(A\) by 3:
  • \(3 \times 7 = 21\)
  • \(3 \times (-1) = -3\)
  • \(3 \times 6 = 18\)
  • \(3 \times 9 = 27\)

The result is \(\begin{bmatrix} 21 & -3 \ 18 & 27 \end{bmatrix}\). Scalar multiplication effectively scales every component of the matrix, which can be used in various applications, such as stretching or shrinking vectors in geometric transformations.

Matrix Operations

Matrix operations encompass several basic actions you can perform on matrices, including addition, subtraction, and scalar multiplication. These operations are foundational for more complex tasks in linear algebra, such as matrix multiplication, finding determinants, and solving systems of linear equations. One key aspect of performing these operations is understanding how they interact:

• Addition and Subtraction: Both require matrices to have the same size, ensuring element-wise operations are meaningful.
• Scalar Multiplication: Scales every element of the matrix uniformly, independent of its size.

These basic operations allow us to manipulate data efficiently and are essential in areas such as computer graphics, engineering, physics, and data science. By mastering these operations, you can simplify or solve complex matrices, facilitating a deeper understanding of problems where matrices are applied.