Problem 24 - Precalculus: Mathematics for Calculus

Question

The matrices \(A, B, C, D, E, F, G\) and \(H\) are defined as follows. $$A=\left[\begin{array}{rr}2 & -5 \\\0 & 7\end{array}\right] \quad B=\left[\begin{array}{rrr}3 &\frac{1}{2} & 5 \\\1 & -1 & 3\end{array}\right] \quad C=\left[\begin{array}{rrr}2 & -\frac{5}{2} &0 \\\0 & 2 & -3\end{array}\right]$$ $$D=\left[\begin{array}{lll}7 & 3\end{array}\right] \quad E=\left[\begin{array}{l}1 \\\2 \\\0\end{array}\right] \quad F=\left[\begin{array}{lll}1 & 0 & 0 \\\0 & 1 & 0 \\\0 & 0 & 1\end{array}\right]$$ $$G=\left[\begin{array}{rrr}5 & -3 & 10 \\\6 & 1 & 0 \\\\-5 & 2 & 2\end{array}\right] \quadH=\left[\begin{array}{rr}3 & 1 \\\2 & -1\end{array}\right]$$ Carry out the indicated algebraic operation, or explain why it cannot be performed. (a) \(C-B\) (b) \(2 C-6 B\)

Step-by-Step Solution

Answer

(a) \([-1, -3, -5; -1, 3, -6]\) (b) \([-14, -8, -30; -6, 10, -24]\)

Step 1: Determining Matrix Dimensions for (a)

Matrices \(C\) and \(B\) are both 2x3 matrices. To perform subtraction \(C-B\), the matrices must have the same dimensions. Since both matrices have dimensions of 2x3, they can be subtracted.

Step 2: Matrix Subtraction for (a)

For each element in the resulting matrix, subtract the corresponding elements in \(B\) from \(C\):- First row first column: \(2 - 3 = -1\)- First row second column: \(-\frac{5}{2} - \frac{1}{2} = -3\)- First row third column: \(0 - 5 = -5\)- Second row first column: \(0 - 1 = -1\)- Second row second column: \(2 - (-1) = 3\)- Second row third column: \(-3 - 3 = -6\).

Step 3: Writing the Result for (a)

The result of \(C-B\) is: \[\left[\begin{array}{rrr} -1 & -3 & -5 \ -1 & 3 & -6 \end{array}\right]\]

Step 4: Determine Operation Feasibility for (b)

For the expression \(2C - 6B\), scalar multiplication involves multiplying each element of the matrices by their respective scalars.

Step 5: Scalar Multiplication for C in (b)

Multiply each element of matrix \(C\) by 2:- First row: \([2 * 2, 2 * -\frac{5}{2}, 2 * 0] = [4, -5, 0]\)- Second row: \([2 * 0, 2 * 2, 2 * -3] = [0, 4, -6]\).This results in the matrix: \[\left[\begin{array}{rrr} 4 & -5 & 0 \ 0 & 4 & -6 \end{array}\right]\].

Step 6: Scalar Multiplication for B in (b)

Multiply each element of matrix \(B\) by 6:- First row: \([6 * 3, 6 * \frac{1}{2}, 6 * 5] = [18, 3, 30]\)- Second row: \([6 * 1, 6 * -1, 6 * 3] = [6, -6, 18]\).This results in the matrix: \[\left[\begin{array}{rrr} 18 & 3 & 30 \ 6 & -6 & 18 \end{array}\right]\].

Step 7: Matrix Subtraction in (b)

Perform the matrix subtraction \((2C) - (6B)\):- First row: \((4 - 18, -5 - 3, 0 - 30) = (-14, -8, -30)\)- Second row: \((0 - 6, 4 - (-6), -6 - 18) = (-6, 10, -24)\).The result is: \[\left[\begin{array}{rrr} -14 & -8 & -30 \ -6 & 10 & -24 \end{array}\right]\].

Key Concepts

Matrix Subtraction

Matrix subtraction is an important operation in linear algebra, where two matrices of the exact same dimensions are subtracted from each other. To perform matrix subtraction, each corresponding element in one matrix is subtracted from the element in the same position in the other matrix.
For example, if you have two matrices, say matrix \(A\) and matrix \(B\), with sizes \(m \times n\), their subtraction \(A - B\) results in a new matrix where each element is calculated as \((a_{ij} - b_{ij})\).
This means if you are subtracting matrix \(B\) from matrix \(C\), you simply take each element from matrix \(C\) and subtract the corresponding element from matrix \(B\).
For instance, in our exercise where we subtract matrix \(B\) from matrix \(C\):
- For the first element, located in the first row and first column, if the element in \(C\) is 2 and in \(B\) is 3, then the result is \(2 - 3 = -1\).
Such step-by-step processing occurs for all elements at the corresponding positions, resulting in a new matrix with the combined effects of the subtraction.

Matrix Dimensions

Understanding matrix dimensions is crucial before executing any matrix operations, as operations like addition and subtraction require matrices to share the same dimensions. A matrix is defined by its dimensions, which are specified by the number of rows \(m\) and columns \(n\) it contains, commonly written as\(m \times n\).
For matrix subtraction, both matrices should have identical dimensions. This means each must have the same number of rows and columns.
In our exercise, both matrices \(C\) and \(B\) are described as being 2x3 matrices.

• 2 represents the number of rows.
• 3 represents the number of columns.

Maintaining these dimensions ensures that every element has a corresponding counterpart for operations such as subtraction. If dimensions don’t match, subtraction cannot be performed, because there would be elements in one matrix without a partner in the other to operate with. Hence, recognizing dimensions is the first step in safely carrying out matrix operations.

Scalar Multiplication

Scalar multiplication is a straightforward operation where each element of a matrix is multiplied by a constant value, known as a scalar. This operation is useful for scaling matrices before further operations, such as addition or subtraction, are performed.
In algebraic terms, if a matrix \(M\) is multiplied by a scalar \(k\), then each element \(m_{ij}\) in \(M\) is multiplied by \(k\), giving us a new matrix \(kM\).
In our example, when calculating \(2C - 6B\), each element of matrix \(C\) is first multiplied by 2. This scaling transforms the original values, modifying the matrix table but preserving its dimensions. Similarly, each element of matrix \(B\) is multiplied by 6.
Take matrix \(C\) as an illustration: if the element in the first row and the first column is 2, by multiplying this element by the scalar 2, the new value becomes \(2 \times 2 = 4\).
Such processes repeat for every element in the matrix. The resulting scaled matrices can then be used in subtraction operations, as demonstrated in the exercise.