91Ó°ÊÓ

Note that matrix \(\mathbf{A}\) has dimensions 3x3, and matrix \(\mathbf{B}\) also has dimensions 3x3. Since the number of columns in \(\mathbf{A}\) matches the number of rows in \(\mathbf{B}\), the multiplication \(\mathbf{AB}\) is well-defined.

To find the product \(\mathbf{AB}\), we need to compute the dot product of each row in matrix \(\mathbf{A}\) with each column in matrix \(\mathbf{B}\). That is: $$ \mathbf{AB}=\left(\begin{array}{c} a_{1,1} \cdot b_{1,1} + a_{1,2} \cdot b_{2,1} + a_{1,3} \cdot b_{3,1} \\\ a_{2,1} \cdot b_{1,1} + a_{2,2} \cdot b_{2,1} + a_{2,3} \cdot b_{3,1} \\\ a_{3,1} \cdot b_{1,1} + a_{3,2} \cdot b_{2,1} + a_{3,3} \cdot b_{3,1} \end{array}\right. \, \left.\begin{array}{c} a_{1,1} \cdot b_{1,2} + a_{1,2} \cdot b_{2,2} + a_{1,3} \cdot b_{3,2} \\\ a_{2,1} \cdot b_{1,2} + a_{2,2} \cdot b_{2,2} + a_{2,3} \cdot b_{3,2} \\\ a_{3,1} \cdot b_{1,2} + a_{3,2} \cdot b_{2,2} + a_{3,3} \cdot b_{3,2} \end{array}\right. \, \left.\begin{array}{c} a_{1,1} \cdot b_{1,3} + a_{1,2} \cdot b_{2,3} + a_{1,3} \cdot b_{3,3} \\\ a_{2,1} \cdot b_{1,3} + a_{2,2} \cdot b_{2,3} + a_{2,3} \cdot b_{3,3} \\\ a_{3,1} \cdot b_{1,3} + a_{3,2} \cdot b_{2,3} + a_{3,3} \cdot b_{3,3} \end{array}\right) $$ Substituting the given matrix values and simplify, $$ \mathbf{AB}= \begin{pmatrix} (1\cdot6)+(2\cdot3)+(3\cdot0) & (1\cdot4)+(2\cdot-1)+(3\cdot2) & (1\cdot2)+(2\cdot-3)+(3\cdot-4) \\\ (4\cdot6)+(-1\cdot3)+(-2\cdot0) & (4\cdot4)+(-1\cdot-1)+(-2\cdot2) & (4\cdot2)+(-1\cdot-3)+(-2\cdot-4) \\\ (-1\cdot6)+(0\cdot3)+(5\cdot0) & (-1\cdot4)+(0\cdot-1)+(5\cdot2) & (-1\cdot2)+(0\cdot-3)+(5\cdot-4) \end{pmatrix} $$ Calculating the elements of the product \(\mathbf{AB}\) gives: $$ \mathbf{AB}= \begin{pmatrix} 12 & 4 & -10 \\\ 21 & 19 & 2 \\\ -6 & 10 & -22 \end{pmatrix} $$

Since the matrix \(\mathbf{B}\) has dimensions 3x3 and matrix \(\mathbf{A}\) also has dimensions 3x3, the number of columns in \(\mathbf{B}\) matches the number of rows in \(\mathbf{A}\), making the multiplication \(\mathbf{BA}\) well-defined.

Now, multiply matrix \(\mathbf{B}\) by matrix \(\mathbf{A}\), calculating the dot product of each row in matrix \(\mathbf{B}\) with each column in matrix \(\mathbf{A}\). After substituting the given matrix values, we get: $$ \mathbf{BA} = \begin{pmatrix} 6\cdot1+4\cdot4+2\cdot-1 & 6\cdot2+4\cdot-1+2\cdot0 & 6\cdot3+4\cdot-2+2\cdot5 \\\ 3\cdot1-1\cdot4+(-3)\cdot-1 & 3\cdot2-1\cdot-1+(-3)\cdot0 & 3\cdot3-1\cdot-2+(-3)\cdot5 \\\ 0\cdot1+2\cdot4+(-4)\cdot-1 & 0\cdot2+2\cdot-1+(-4)\cdot0 & 0\cdot3+2\cdot-2+(-4)\cdot5 \end{pmatrix} $$ Calculating the elements of the product \(\mathbf{BA}\) gives: $$ \mathbf{BA}= \begin{pmatrix} 25 & 3 & 10 \\\ 2 & 12 & -21 \\\ -8 & 3 & -34 \end{pmatrix} $$ Thus, the product matrices are: $$ \mathbf{AB}= \begin{pmatrix} 12 & 4 & -10 \\\ 21 & 19 & 2 \\\ -6 & 10 & -22 \end{pmatrix} $$ and $$ \mathbf{BA}= \begin{pmatrix} 25 & 3 & 10 \\\ 2 & 12 & -21 \\\ -8 & 3 & -34 \end{pmatrix} $$