Matrix Operations: Multiplication and Subtraction

Task 4: Matrix Operations

This task requires us to perform a series of matrix multiplications, scalar multiplications, and matrix subtraction.

Let the given expression be \(E\). We can write it as:
\(E = 4 \cdot A \cdot B - 3 \cdot C \cdot D\)

Where:
\(A = \begin{bmatrix} 6 & -1 & 4 \\ 2 & 3 & -2 \\ 3 & 0 & -3 \\ 4 & 2 & 7 \end{bmatrix}\)
\(B = \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ -1 & 0 \end{bmatrix}\)
\(C = \begin{bmatrix} 7 & -1 \\ 1 & -5 \\ -2 & 0 \\ 3 & 2 \end{bmatrix}\)
\(D = \begin{bmatrix} -1 & 2 \\ -4 & 1 \end{bmatrix}\)

Let's break down the calculation into steps.

✨ Step 1: Calculate the product \(A \cdot B\)
Matrix \(A\) is a \(4 \times 3\) matrix and matrix \(B\) is a \(3 \times 2\) matrix. The resulting matrix will be a \(4 \times 2\) matrix.

\(A \cdot B = \begin{bmatrix} (6)(1)+(-1)(0)+(4)(-1) & (6)(2)+(-1)(1)+(4)(0) \\ (2)(1)+(3)(0)+(-2)(-1) & (2)(2)+(3)(1)+(-2)(0) \\ (3)(1)+(0)(0)+(-3)(-1) & (3)(2)+(0)(1)+(-3)(0) \\ (4)(1)+(2)(0)+(7)(-1) & (4)(2)+(2)(1)+(7)(0) \end{bmatrix}\)

\(A \cdot B = \begin{bmatrix} 6+0-4 & 12-1+0 \\ 2+0+2 & 4+3+0 \\ 3+0+3 & 6+0+0 \\ 4+0-7 & 8+2+0 \end{bmatrix}\)

\(A \cdot B = \begin{bmatrix} 2 & 11 \\ 4 & 7 \\ 6 & 6 \\ -3 & 10 \end{bmatrix}\)

✨ Step 2: Calculate \(4 \cdot (A \cdot B)\)
Now, we multiply the result from Step 1 by the scalar 4.

\(4 \cdot (A \cdot B) = 4 \cdot \begin{bmatrix} 2 & 11 \\ 4 & 7 \\ 6 & 6 \\ -3 & 10 \end{bmatrix} = \begin{bmatrix} 4 \cdot 2 & 4 \cdot 11 \\ 4 \cdot 4 & 4 \cdot 7 \\ 4 \cdot 6 & 4 \cdot 6 \\ 4 \cdot (-3) & 4 \cdot 10 \end{bmatrix}\)

\(4 \cdot (A \cdot B) = \begin{bmatrix} 8 & 44 \\ 16 & 28 \\ 24 & 24 \\ -12 & 40 \end{bmatrix}\)

Let's call this matrix \(M_1\).

✨ Step 3: Calculate the product \(C \cdot D\)
Matrix \(C\) is a \(4 \times 2\) matrix and matrix \(D\) is a \(2 \times 2\) matrix. The resulting matrix will be a \(4 \times 2\) matrix.

\(C \cdot D = \begin{bmatrix} (7)(-1)+(-1)(-4) & (7)(2)+(-1)(1) \\ (1)(-1)+(-5)(-4) & (1)(2)+(-5)(1) \\ (-2)(-1)+(0)(-4) & (-2)(2)+(0)(1) \\ (3)(-1)+(2)(-4) & (3)(2)+(2)(1) \end{bmatrix}\)

\(C \cdot D = \begin{bmatrix} -7+4 & 14-1 \\ -1+20 & 2-5 \\ 2+0 & -4+0 \\ -3-8 & 6+2 \end{bmatrix}\)

\(C \cdot D = \begin{bmatrix} -3 & 13 \\ 19 & -3 \\ 2 & -4 \\ -11 & 8 \end{bmatrix}\)

✨ Step 4: Calculate \(3 \cdot (C \cdot D)\)
Now, we multiply the result from Step 3 by the scalar 3.

\(3 \cdot (C \cdot D) = 3 \cdot \begin{bmatrix} -3 & 13 \\ 19 & -3 \\ 2 & -4 \\ -11 & 8 \end{bmatrix} = \begin{bmatrix} 3 \cdot (-3) & 3 \cdot 13 \\ 3 \cdot 19 & 3 \cdot (-3) \\ 3 \cdot 2 & 3 \cdot (-4) \\ 3 \cdot (-11) & 3 \cdot 8 \end{bmatrix}\)

\(3 \cdot (C \cdot D) = \begin{bmatrix} -9 & 39 \\ 57 & -9 \\ 6 & -12 \\ -33 & 24 \end{bmatrix}\)

Let's call this matrix \(M_2\).

✨ Step 5: Calculate \(M_1 - M_2\)
Finally, we subtract \(M_2\) from \(M_1\).

\(M_1 - M_2 = \begin{bmatrix} 8 & 44 \\ 16 & 28 \\ 24 & 24 \\ -12 & 40 \end{bmatrix} - \begin{bmatrix} -9 & 39 \\ 57 & -9 \\ 6 & -12 \\ -33 & 24 \end{bmatrix}\)

\(M_1 - M_2 = \begin{bmatrix} 8 - (-9) & 44 - 39 \\ 16 - 57 & 28 - (-9) \\ 24 - 6 & 24 - (-12) \\ -12 - (-33) & 40 - 24 \end{bmatrix}\)

\(M_1 - M_2 = \begin{bmatrix} 8 + 9 & 5 \\ 16 - 57 & 28 + 9 \\ 18 & 24 + 12 \\ -12 + 33 & 16 \end{bmatrix}\)

\(M_1 - M_2 = \begin{bmatrix} 17 & 5 \\ -41 & 37 \\ 18 & 36 \\ 21 & 16 \end{bmatrix}\)

This is the final result of the expression.