Problem 13 - Discrete Mathematics with Applications

Question

Let \(A=\left[\begin{array}{ccc}{1} & {0} & {-1} \\ {0} & {2} & {3}\end{array}\right], B=\left[\begin{array}{ccc}{0} & {-2} & {5} \\ {0} & {0} & {1}\end{array}\right],\) and \(C=\left[\begin{array}{ccc}{-3} & {0} & {0} \\\ {0} & {1} & {2}\end{array}\right]\) . Find each. $$3 A+(-2) B$$

Step-by-Step Solution

Answer

The short answer is: \(3A + (-2)B = \left[\begin{array}{ccc}{3} & {4} & {-13} \\\ {0} & {6} & {7}\end{array}\right]\)

Step 1: Multiply A by 3

To achieve this, we multiply each element in matrix A by 3: \(3A = \left[\begin{array}{ccc}{3} & {0} & {-3} \\\ {0} & {6} & {9}\end{array}\right]\)

Step 2: Multiply B by -2

Now, we need to multiply each element in matrix B by -2: \((-2)B = \left[\begin{array}{ccc}{0} & {4} & {-10} \\\ {0} & {0} & {-2}\end{array}\right]\)

Step 3: Add the resulting matrices

Finally, we will add the two resulting matrices obtained in Step 1 and 2 element-wise: \(3A + (-2)B = \left[\begin{array}{ccc}{3} & {0} & {-3} \\\ {0} & {6} & {9}\end{array}\right] + \left[\begin{array}{ccc}{0} & {4} & {-10} \\\ {0} & {0} & {-2}\end{array}\right] = \left[\begin{array}{ccc}{3} & {4} & {-13} \\\ {0} & {6} & {7}\end{array}\right]\) So, the resulting matrix after performing the given operation is: \(\left[\begin{array}{ccc}{3} & {4} & {-13} \\\ {0} & {6} & {7}\end{array}\right]\)

Key Concepts

Matrix Addition

Matrix addition is one of the fundamental operations in matrix algebra. It involves adding two matrices of the same dimensions by performing an element-wise addition. This means that each element in the resulting matrix is the sum of the corresponding elements from the two matrices being added.

For example, consider two matrices:

• Matrix M: \[ \begin{bmatrix} m_{11} & m_{12} \ m_{21} & m_{22} \end{bmatrix} \]
• Matrix N: \[ \begin{bmatrix} n_{11} & n_{12} \ n_{21} & n_{22} \end{bmatrix} \]

To find the sum, M + N, we calculate:

• \[\begin{bmatrix} m_{11} + n_{11} & m_{12} + n_{12} \ m_{21} + n_{21} & m_{22} + n_{22} \end{bmatrix}\]

It is important to note that matrix addition is only defined for matrices having the same number of rows and columns. If the dimensions do not match, the addition cannot be performed. Remember, matrix addition is commutative, meaning that \( A + B = B + A \).

Matrix Multiplication

Matrix multiplication is a core operation in linear algebra and has a slightly more complex procedure than addition. The product of two matrices is only defined if the number of columns in the first matrix equals the number of rows in the second matrix. This condition ensures that matrix multiplication is feasible.

Consider this example of multiplying Matrices X and Y:

• Matrix X with elements arranged in \( i \) rows and \( j \) columns.
• Matrix Y with elements arranged in \( j \) rows and \( k \) columns.

The resulting matrix XY will have dimensions of \( i \times k \). Each element \( {c_{ik}} \) in the product matrix is the sum of the products of the corresponding elements from the row of the first matrix and the column of the second matrix:

• \[c_{ik} = x_{i1}y_{1k} + x_{i2}y_{2k} + \dots + x_{ij}y_{jk}\]

Matrix multiplication is associative, meaning \( (AB)C = A(BC) \), and it is distributive over addition, \( A(B + C) = AB + AC \), but it is not commutative. Therefore, \( AB \) does not necessarily equal \( BA \).

Scalar Multiplication

Scalar multiplication involves multiplying each element of a matrix by a scalar, which is a single number. This operation changes the magnitude of the matrix's elements but keeps its dimensions unchanged, essentially scaling the matrix by the given scalar.

Suppose we have a matrix D:

• \[ \begin{bmatrix} d_{11} & d_{12} \ d_{21} & d_{22} \end{bmatrix} \]

If we multiply this matrix by a scalar \( k \), we apply \( k \) to each element in the matrix:

• \[kD = \begin{bmatrix} kd_{11} & kd_{12} \ kd_{21} & kd_{22} \end{bmatrix}\]

This operation is straightforward as it involves performing the same calculation for each element. Scalar multiplication is easy to visualize and understand, making it a concept that students usually grasp quickly. In many mathematical problems, including the one you are studying, scalar multiplication is often the first step in a series of operations.