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.