Problem 10 - College Algebra

Question

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$A=\left[\begin{array}{cc}-2 & 3 \\\0 & 1\end{array}\right], \quad B=\left[\begin{array}{ll} 8 & 1 \\\5 & 4\end{array}\right]$$

Step-by-Step Solution

Answer

The results of the matrix operations are \n (A+B) = \left[ \begin{array}{cc} 6 & 4 \ 5 & 5 \end{array} \right] \n (A-B) = \left[ \begin{array}{cc} -10 & 2 \ -5 & -3 \end{array} \right] \n -4*A = \left[ \begin{array}{cc} -8 & -12 \ 0 & -4 \end{array} \right] \n and finally, 3*A+2*B = \left[ \begin{array}{cc} 10 & 11 \ 10 & 11 \end{array} \right]

Step 1: Computing the Sum Matrix (A+B)

The sum of two matrices is obtained by adding corresponding elements. Therefore, the elements of the sum matrix (A+B) become: \n \[(A+B) = \left[ \begin{array}{cc} (-2+8) & (3+1) \ (0+5) & (1+4) \end{array} \right] = \left[ \begin{array}{cc} 6 & 4 \ 5 & 5 \end{array} \right]\]

Step 2: Computing the Difference Matrix (A-B)

The difference between two matrices is obtained by subtracting corresponding elements. Therefore, the elements of the difference matrix (A-B) become: \n \[(A-B) = \left[ \begin{array}{cc} (-2-8) & (3-1) \ (0-5) & (1-4) \end{array} \right] = \left[ \begin{array}{cc} -10 & 2 \ -5 & -3 \end{array} \right]\]

Step 3: Computing the Scalar Multiplication -4*A

To perform scalar multiplication with a matrix, each element of the matrix should be multiplied by the scalar. Therefore, elements of the matrix after scalar multiplication will be: \n \[(-4*A) = \left[ \begin{array}{cc} 4*(-2) & -4*3 \ -4*0 & -4*1 \end{array} \right] = \left[ \begin{array}{cc} -8 & -12 \ 0 & -4 \end{array} \right]\]

Step 4: Computing the Combined Operation 3*A +2*B

To solve this, we first need to perform scalar multiplication for matrices A and B, and then add the resultant matrices. Therefore, elements of the resultant matrix will be: \n \[3*A = \left[ \begin{array}{cc} 3*(-2) & 3*3 \ 3*0 & 3*1 \end{array} \right] = \left[ \begin{array}{cc} -6 & 9 \ 0 & 3 \end{array} \right] \]\[2*B = \left[ \begin{array}{cc} 2*8 & 2*1 \ 2*5 & 2*4 \end{array} \right] = \left[ \begin{array}{cc} 16 & 2 \ 10 & 8 \end{array} \right] \]\[(3*A + 2*B) = \left[ \begin{array}{cc} (-6+16) & (9+2) \ (0+10) & (3+8) \end{array} \right] = \left[ \begin{array}{cc} 10 & 11 \ 10 & 11 \end{array} \right]\]

Key Concepts

Matrix Addition

Matrix addition is a simple and fundamental operation in matrix algebra. It involves adding two matrices by summing their corresponding elements. Consider matrices \(A\) and \(B\) of the same dimension. To compute \(A + B\), add each element in matrix \(A\) to the corresponding element in matrix \(B\).

Let's take the matrices from the exercise:

• Matrix \(A\) is \(\begin{bmatrix} -2 & 3 \ 0 & 1 \end{bmatrix}\)
• Matrix \(B\) is \(\begin{bmatrix} 8 & 1 \ 5 & 4 \end{bmatrix}\)

To find the sum \(A + B\), we compute:

• \((-2) + 8 = 6\)
• \(3 + 1 = 4\)
• \(0 + 5 = 5\)
• \(1 + 4 = 5\)

Thus, the resulting matrix is \(\begin{bmatrix} 6 & 4 \ 5 & 5 \end{bmatrix}\).

It's essential that both matrices have the same dimensions; otherwise, matrix addition is not defined.

Matrix Subtraction

Just like matrix addition, matrix subtraction involves dealing with corresponding elements of two matrices. The operation is defined when both matrices are of the same size. For matrices \(A\) and \(B\), the difference \(A - B\) requires subtracting each element of \(B\) from the corresponding element in \(A\).

Using our given matrices in the exercise:

• Matrix \(A\) is \(\begin{bmatrix} -2 & 3 \ 0 & 1 \end{bmatrix}\)
• Matrix \(B\) is \(\begin{bmatrix} 8 & 1 \ 5 & 4 \end{bmatrix}\)

The subtraction \(A - B\) is computed as follows:

• \((-2) - 8 = -10\)
• \(3 - 1 = 2\)
• \(0 - 5 = -5\)
• \(1 - 4 = -3\)

The resulting matrix is \(\begin{bmatrix} -10 & 2 \ -5 & -3 \end{bmatrix}\).

This process illustrates that subtraction is as straightforward as addition but involves finding the difference of elements.

Scalar Multiplication

Scalar multiplication involves multiplying every element of a matrix by a constant value or scalar. This operation is useful in scaling matrices while keeping the proportions between the elements the same.

For instance, consider multiplying matrix \(A\) by a scalar \(-4\):

• Start with matrix \(A = \begin{bmatrix} -2 & 3 \ 0 & 1 \end{bmatrix}\)
• Multiply each element by \(-4\):
• \((-4)\times(-2) = 8\)
• \((-4)\times3 = -12\)
• \((-4)\times0 = 0\)
• \((-4)\times1 = -4\)

The resulting matrix after scalar multiplication is \(\begin{bmatrix} -8 & -12 \ 0 & -4 \end{bmatrix}\).

Scalar multiplication is straightforward and is applicable to any size matrix. Just ensure each element is scaled uniformly.

Elementary Algebra in Matrices

Elementary algebra concepts extend naturally into matrix operations. These include combining scalar multiplication and matrix addition or subtraction into a single procedure. For example, suppose we need to find \(3A + 2B\), a common operation in solving linear equations with matrices.

Consider the matrices in the exercise:

• Matrix \(A = \begin{bmatrix} -2 & 3 \ 0 & 1 \end{bmatrix}\)
• Matrix \(B = \begin{bmatrix} 8 & 1 \ 5 & 4 \end{bmatrix}\)

Perform the individual scalar multiplications:

• \(3 \times A = \begin{bmatrix} -6 & 9 \ 0 & 3 \end{bmatrix}\)
• \(2 \times B = \begin{bmatrix} 16 & 2 \ 10 & 8 \end{bmatrix}\)

Add these scaled matrices:

• \(-6 + 16 = 10\)
• \(9 + 2 = 11\)
• \(0 + 10 = 10\)
• \(3 + 8 = 11\)

The resulting matrix is \(\begin{bmatrix} 10 & 11 \ 10 & 11 \end{bmatrix}\).

Understanding these operations helps in solving complex problems by breaking them into simpler, more manageable parts.