Problem 19 - Advanced Engineering Mathematics

Question

If \(\mathbf{A}=\left(\begin{array}{rr}1 & -2 \\ -2 & 4\end{array}\right), \mathbf{B}=\left(\begin{array}{ll}6 & 3 \\ 2 & 1\end{array}\right)\), and \(\mathbf{C}=\left(\begin{array}{ll}0 & 2 \\ 3 & 4\end{array}\right)\) find (a) \(\mathbf{B C}\), (b) \(\mathbf{A}(\mathbf{B C})\), (c) \(\mathbf{C}(\mathbf{B} \mathbf{A})\), (d) \(\mathbf{A}(\mathbf{B}+\mathbf{C})\).

Step-by-Step Solution

Answer

(a) \(\mathbf{B}\mathbf{C} = \left(\begin{array}{cc} 9 & 24 \\ 3 & 8 \end{array}\right)\), (b) \(\mathbf{A}(\mathbf{B}\mathbf{C}) = \left(\begin{array}{cc} 3 & 8 \\ -6 & -16 \end{array}\right)\), (c) \(\mathbf{C}(\mathbf{B}\mathbf{A}) = \left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right)\), (d) \(\mathbf{A}(\mathbf{B} + \mathbf{C}) = \left(\begin{array}{cc} -4 & -5 \\ 8 & 10 \end{array}\right)\)."

Step 1: Calculate \(\mathbf{B} \mathbf{C}\)

To find the product \(\mathbf{B}\mathbf{C}\), we multiply the elements row by column. Recall that the element in row \(i\) and column \(j\) of the product is found by multiplying each element of row \(i\) of \(\mathbf{B}\) with each corresponding element of column \(j\) of \(\mathbf{C}\), then summing the results.**Calculations:**- Entry (1,1): \(6\times 0 + 3\times 3 = 0 + 9 = 9\)- Entry (1,2): \(6\times 2 + 3\times 4 = 12 + 12 = 24\)- Entry (2,1): \(2\times 0 + 1\times 3 = 0 + 3 = 3\)- Entry (2,2): \(2\times 2 + 1\times 4 = 4 + 4 = 8\)So, \(\mathbf{B}\mathbf{C} = \left(\begin{array}{cc} 9 & 24 \ 3 & 8 \end{array}\right)\).

Step 2: Calculate \(\mathbf{A}(\mathbf{B} \mathbf{C})\)

Multiply matrix \(\mathbf{A}\) by the product \(\mathbf{B}\mathbf{C}\) obtained in Step 1.**Calculations:**- Entry (1,1): \(1\times 9 + (-2)\times 3 = 9 - 6 = 3\)- Entry (1,2): \(1\times 24 + (-2)\times 8 = 24 - 16 = 8\)- Entry (2,1): \((-2)\times 9 + 4\times 3 = -18 + 12 = -6\)- Entry (2,2): \((-2)\times 24 + 4\times 8 = -48 + 32 = -16\)So, \(\mathbf{A}(\mathbf{B}\mathbf{C}) = \left(\begin{array}{cc} 3 & 8 \ -6 & -16 \end{array}\right)\).

Step 3: Calculate \(\mathbf{B} \mathbf{A}\)

Calculate the product of matrices \(\mathbf{B}\) and \(\mathbf{A}\), which is necessary for part (c).**Calculations:**- Entry (1,1): \(6\times 1 + 3\times (-2) = 6 - 6 = 0\)- Entry (1,2): \(6\times (-2) + 3\times 4 = -12 + 12 = 0\)- Entry (2,1): \(2\times 1 + 1\times (-2) = 2 - 2 = 0\)- Entry (2,2): \(2\times (-2) + 1\times 4 = -4 + 4 = 0\)Therefore, \(\mathbf{B} \mathbf{A} = \left(\begin{array}{cc} 0 & 0 \ 0 & 0 \end{array}\right)\).

Step 4: Calculate \(\mathbf{C}(\mathbf{B} \mathbf{A})\)

Since the product \(\mathbf{B} \mathbf{A}\) is the zero matrix, multiplying it by any matrix \(\mathbf{C}\) gives the zero matrix again.Thus, \(\mathbf{C}(\mathbf{B} \mathbf{A}) = \left(\begin{array}{cc} 0 & 0 \ 0 & 0 \end{array}\right)\).

Step 5: Calculate \(\mathbf{B} + \mathbf{C}\)

Add matrices \(\mathbf{B}\) and \(\mathbf{C}\) by adding corresponding elements:**Calculations:**- Entry (1,1): \(6 + 0 = 6\)- Entry (1,2): \(3 + 2 = 5\)- Entry (2,1): \(2 + 3 = 5\)- Entry (2,2): \(1 + 4 = 5\)So, \(\mathbf{B} + \mathbf{C} = \left(\begin{array}{cc} 6 & 5 \ 5 & 5 \end{array}\right)\).

Step 6: Calculate \(\mathbf{A}(\mathbf{B} + \mathbf{C})\)

Now that we have \(\mathbf{B} + \mathbf{C}\), multiply this matrix by \(\mathbf{A}\).**Calculations:**- Entry (1,1): \(1\times 6 + (-2)\times 5 = 6 - 10 = -4\)- Entry (1,2): \(1\times 5 + (-2)\times 5 = 5 - 10 = -5\)- Entry (2,1): \((-2)\times 6 + 4\times 5 = -12 + 20 = 8\)- Entry (2,2): \((-2)\times 5 + 4\times 5 = -10 + 20 = 10\)So, \(\mathbf{A}(\mathbf{B} + \mathbf{C}) = \left(\begin{array}{cc} -4 & -5 \ 8 & 10 \end{array}\right)\).

Key Concepts

Matrix Addition

Matrix addition is a basic operation where you add two matrices by adding their corresponding elements. It requires that both matrices have the same dimensions; that is, they must have the same number of rows and columns.
For example, consider matrices \( \mathbf{B} \) and \( \mathbf{C} \) with dimensions \( 2 \times 2 \). To find \( \mathbf{B} + \mathbf{C} \), you simply add each pair of corresponding elements.

• Entry (1,1) results from \(6 + 0 = 6\).
• Entry (1,2) results from \(3 + 2 = 5\).
• Entry (2,1) results from \(2 + 3 = 5\).
• Entry (2,2) results from \(1 + 4 = 5\).

As a result, the sum of matrices \( \mathbf{B} \) and \( \mathbf{C} \) is a new matrix:\[ \mathbf{B} + \mathbf{C} = \begin{pmatrix} 6 & 5 \ 5 & 5 \end{pmatrix} \] Matrix addition is straightforward but crucial for further matrix operations.

Matrix Product

Matrix multiplication, or the matrix product, is a bit more involved than addition. It combines two matrices via a systematic multiplication and addition of rows and columns. For matrices to be compatible for multiplication, the number of columns in the first matrix must equal the number of rows in the second matrix.
Let's look at finding the product \( \mathbf{B} \times \mathbf{C} \). Here, each entry in the resulting matrix is computed by multiplying respective elements of the row from \( \mathbf{B} \) by the relevant column from \( \mathbf{C} \), then summing these values.

• Entry (1,1): \(6 \times 0 + 3 \times 3 = 9\)
• Entry (1,2): \(6 \times 2 + 3 \times 4 = 24\)
• Entry (2,1): \(2 \times 0 + 1 \times 3 = 3\)
• Entry (2,2): \(2 \times 2 + 1 \times 4 = 8\)

This gives us:\[ \mathbf{B} \cdot \mathbf{C} = \begin{pmatrix} 9 & 24 \ 3 & 8 \end{pmatrix} \] Matrix multiplication is not commutative, meaning \( \mathbf{A} \cdot \mathbf{B} eq \mathbf{B} \cdot \mathbf{A} \). When performing these calculations, one must pay careful attention to the order of multiplication, as altering it usually changes the result.

Matrix Operations

Matrix operations encompass the various ways matrices can be combined or transformed using either addition, multiplication, or other mathematical functions like transposition and inversion.
In the given exercise, after finding individual products and sums, more complex operations involve multiplying matrices with the results of previous operations, like \( \mathbf{A} \cdot (\mathbf{B} \cdot \mathbf{C}) \). Beyond addition and basic multiplication, these operations sometimes include combining several processes.
For example, a major operation in the exercise involves calculating \((\mathbf{B} + \mathbf{C})\) and then multiplying the result by \( \mathbf{A} \), which signifies executing an addition followed by a multiplication. Applying multiple matrix operations successfully requires a strong understanding of the flexibility and constraints inherent in matrix algebra, such as the need for specific dimension matching for multiplication and the associative properties that allow regrouping matrices in products without changing the result. These operations are foundational in disciplines such as computer graphics, robotics, and statistical data analysis.