Problem 23 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The result is \( \left[\begin{array}{rrr} -14 & -8 & -30 \\ -6 & 10 & -24 \end{array}\right] \).

Step 1: Multiply Matrix C by 2

The first task is to multiply each element of the matrix \( C \) by 2.Given:\[ C=\left[\begin{array}{rrr} 2 & -\frac{5}{2} & 0 \ 0 & 2 & -3 \end{array}\right] \]Multiply each element by 2:\[ 2C=\left[\begin{array}{rrr} 4 & -5 & 0 \ 0 & 4 & -6 \end{array}\right] \]

Step 2: Multiply Matrix B by 6

Next, we multiply each element of the matrix \( B \) by 6.Given:\[ B=\left[\begin{array}{rrr} 3 & \frac{1}{2} & 5 \ 1 & -1 & 3 \end{array}\right] \]Multiply each element by 6:\[ 6B=\left[\begin{array}{rrr} 18 & 3 & 30 \ 6 & -6 & 18 \end{array}\right] \]

Step 3: Subtract 6B from 2C

Finally, subtract the resulting matrix \( 6B \) from \( 2C \).Given:\[ 2C = \left[\begin{array}{rrr} 4 & -5 & 0 \ 0 & 4 & -6 \end{array}\right] \] and \[ 6B = \left[\begin{array}{rrr} 18 & 3 & 30 \ 6 & -6 & 18 \end{array}\right] \]The subtraction is performed element by element:\[2C - 6B = \left[\begin{array}{rrr}4-18 & -5-3 & 0-30 \0-6 & 4+6 & -6-18\end{array}\right] = \left[\begin{array}{rrr}-14 & -8 & -30 \-6 & 10 & -24\end{array}\right]\]

Key Concepts

Matrix Multiplication

Matrix multiplication involves taking two matrices and combining them into a new matrix by performing a specified calculation on their corresponding elements. The key rule in matrix multiplication is that the number of columns in the first matrix must equal the number of rows in the second matrix. This ensures that the operation can be performed.

Here's how it works:

• Take a row from the first matrix and a column from the second matrix.
• Multiply the corresponding elements and sum these products to get a single entry in the resulting matrix.

Matrix multiplication is not commutative, meaning that even if one can multiply matrix A by B, it might not be possible to multiply B by A. For instance, when considering matrices that we have, one can't simply multiply every matrix because the dimensions often do not align for multiplication.
Understanding this principle ensures clarity when dealing with complex matrix expressions.

Matrix Subtraction

Matrix subtraction is perhaps one of the more straightforward matrix operations, where you subtract the corresponding elements of two matrices. To perform matrix subtraction, it's important that both matrices are of the same size. This means they must have the same number of rows and columns. If this condition is met, the subtraction can be carried out seamlessly.

For example:

• With matrices, like in our exercise (\[2Cdot - 6B\]), each element in the resulting matrix is the difference of the corresponding elements in the two matrices.
• It is done element by element: \((a - b)_{ij} = a_{ij} - b_{ij}\).

The operation essentially reflects the same characteristics as regular subtraction but applied to every element respectively. This fundamental logic is what was used in the original exercise you worked through.

Scalar Multiplication

Scalar multiplication occurs when you multiply every element of a matrix by a single numerical value, or scalar. This is a simple yet powerful operation. Taking an example from our exercise:

• To perform the operation, each entry of the matrix is multiplied by the scalar.
• If the scalar is 'k', each element \(a_{ij}\) in matrix \(A\) becomes \(k \times a_{ij}\).

In our problem, this operation was key to obtaining matrices \(2C\) and \(6B\). It effectively stretches or shrinks the matrix values in proportion to the scalar. In this exercise, multiplying by scalars adjusted matrices to the context necessary for subtraction.

Element-wise Operations

Element-wise operations involve performing an operation on corresponding elements of the matrices involved. This is evident in both the subtraction and scalar multiplication described in our example. The subtraction of \((2C - 6B)\) involved an element-wise operation where each respective element from \(6B\) was subtracted from \(2C\).

Some key traits of element-wise operations include:

• Consistency of dimensions: The matrices involved must be the same size to perform element-wise operations.
• Direct correspondence: Each element in one matrix corresponds directly with the element in the same position of the other matrix.

These operations can extend beyond subtraction to include addition and other mathematical operations, provided that matrices are of consistent dimensions. Remembering these concepts is essential for understanding the original complex algebraic exercise you tackled.