1. **Problem statement:** (a) Find the product of matrices $A$ and $B$ where $$A = \begin{pmatrix}1 & 2 \\ 3 & -1 \\ 2 & 5\end{pmatrix}, \quad B = \begin{pmatrix}5 & 4 \\ 3 & 1 \\ -2 & 2\end{pmatrix}$$ (b) Solve the system of equations using matrix methods: $$\begin{cases} 2i_1 + i_2 - i_3 = 8 \\ i_1 - i_2 + i_3 = -5 \\ 3i_1 + 2i_2 = 9 \end{cases}$$ --- 2. **Matrix multiplication rules:** To multiply two matrices $A$ (of size $m \times n$) and $B$ (of size $n \times p$), the number of columns in $A$ must equal the number of rows in $B$. The resulting matrix $AB$ will be of size $m \times p$. Each element $(AB)_{ij}$ is computed as the dot product of the $i$th row of $A$ and the $j$th column of $B$: $$ (AB)_{ij} = \sum_{k=1}^n A_{ik} B_{kj} $$ 3. **Step (a): Find $AB$** Matrix $A$ is $3 \times 2$, matrix $B$ is $3 \times 2$. Since the number of columns in $A$ (2) does not equal the number of rows in $B$ (3), the product $AB$ is **not defined**. However, if the question intended $B$ to be $2 \times 3$ (transposed), then $AB$ would be defined. Let's check $B^T$: $$ B^T = \begin{pmatrix}5 & 3 & -2 \\ 4 & 1 & 2\end{pmatrix} $$ $A$ is $3 \times 2$, $B^T$ is $2 \times 3$, so $AB^T$ is $3 \times 3$. Calculate $AB^T$: Row 1 of $A$: $(1, 2)$ - $(AB^T)_{11} = 1 \times 5 + 2 \times 4 = 5 + 8 = 13$ - $(AB^T)_{12} = 1 \times 3 + 2 \times 1 = 3 + 2 = 5$ - $(AB^T)_{13} = 1 \times (-2) + 2 \times 2 = -2 + 4 = 2$ Row 2 of $A$: $(3, -1)$ - $(AB^T)_{21} = 3 \times 5 + (-1) \times 4 = 15 - 4 = 11$ - $(AB^T)_{22} = 3 \times 3 + (-1) \times 1 = 9 - 1 = 8$ - $(AB^T)_{23} = 3 \times (-2) + (-1) \times 2 = -6 - 2 = -8$ Row 3 of $A$: $(2, 5)$ - $(AB^T)_{31} = 2 \times 5 + 5 \times 4 = 10 + 20 = 30$ - $(AB^T)_{32} = 2 \times 3 + 5 \times 1 = 6 + 5 = 11$ - $(AB^T)_{33} = 2 \times (-2) + 5 \times 2 = -4 + 10 = 6$ So, $$ AB^T = \begin{pmatrix}13 & 5 & 2 \\ 11 & 8 & -8 \\ 30 & 11 & 6\end{pmatrix} $$ 4. **Step (b): Solve the system using matrix method** Write the system as $AX = B$ where $$ A = \begin{pmatrix}2 & 1 & -1 \\ 1 & -1 & 1 \\ 3 & 2 & 0\end{pmatrix}, \quad X = \begin{pmatrix}i_1 \\ i_2 \\ i_3\end{pmatrix}, \quad B = \begin{pmatrix}8 \\ -5 \\ 9\end{pmatrix} $$ We solve for $X$ by computing $$ X = A^{-1} B $$ Calculate determinant of $A$: $$ \det(A) = 2 \begin{vmatrix} -1 & 1 \\ 2 & 0 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} + (-1) \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} $$ Calculate minors: - $\begin{vmatrix} -1 & 1 \\ 2 & 0 \end{vmatrix} = (-1)(0) - (1)(2) = -2$ - $\begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = 1 \times 0 - 1 \times 3 = -3$ - $\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = 1 \times 2 - (-1) \times 3 = 2 + 3 = 5$ So, $$ \det(A) = 2(-2) - 1(-3) + (-1)(5) = -4 + 3 - 5 = -6 $$ Since $\det(A) \neq 0$, $A$ is invertible. Calculate $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$ where adjoint is transpose of cofactor matrix. Cofactors: - $C_{11} = (-1)^{2} \times (-2) = -2$ - $C_{12} = (-1)^{3} \times (-3) = 3$ - $C_{13} = (-1)^{4} \times 5 = 5$ - $C_{21} = (-1)^{3} \times \begin{vmatrix}1 & -1 \\ 2 & 0\end{vmatrix} = -1 \times (1 \times 0 - (-1) \times 2) = -1 \times 2 = -2$ - $C_{22} = (-1)^{4} \times \begin{vmatrix}2 & -1 \\ 3 & 0\end{vmatrix} = 1 \times (2 \times 0 - (-1) \times 3) = 3$ - $C_{23} = (-1)^{5} \times \begin{vmatrix}2 & 1 \\ 3 & 2\end{vmatrix} = -1 \times (2 \times 2 - 1 \times 3) = -1 \times (4 - 3) = -1$ - $C_{31} = (-1)^{4} \times \begin{vmatrix}1 & -1 \\ -1 & 1\end{vmatrix} = 1 \times (1 \times 1 - (-1) \times -1) = 1 \times (1 - 1) = 0$ - $C_{32} = (-1)^{5} \times \begin{vmatrix}2 & -1 \\ 1 & 1\end{vmatrix} = -1 \times (2 \times 1 - (-1) \times 1) = -1 \times (2 + 1) = -3$ - $C_{33} = (-1)^{6} \times \begin{vmatrix}2 & 1 \\ 1 & -1\end{vmatrix} = 1 \times (2 \times -1 - 1 \times 1) = 1 \times (-2 - 1) = -3$ Cofactor matrix: $$ \begin{pmatrix} -2 & 3 & 5 \\ -2 & 3 & -1 \\ 0 & -3 & -3 \end{pmatrix} $$ Adjoint (transpose): $$ \text{adj}(A) = \begin{pmatrix} -2 & -2 & 0 \\ 3 & 3 & -3 \\ 5 & -1 & -3 \end{pmatrix} $$ Calculate $X$: $$ X = \frac{1}{-6} \begin{pmatrix} -2 & -2 & 0 \\ 3 & 3 & -3 \\ 5 & -1 & -3 \end{pmatrix} \begin{pmatrix}8 \\ -5 \\ 9\end{pmatrix} $$ Multiply: - First row: $(-2)(8) + (-2)(-5) + 0(9) = -16 + 10 + 0 = -6$ - Second row: $3(8) + 3(-5) + (-3)(9) = 24 - 15 - 27 = -18$ - Third row: $5(8) + (-1)(-5) + (-3)(9) = 40 + 5 - 27 = 18$ Divide by $-6$: $$ i_1 = \frac{-6}{-6} = 1, \quad i_2 = \frac{-18}{-6} = 3, \quad i_3 = \frac{18}{-6} = -3 $$ --- **Final answers:** (a) The product $AB$ is not defined as given, but $AB^T$ is: $$ AB^T = \begin{pmatrix}13 & 5 & 2 \\ 11 & 8 & -8 \\ 30 & 11 & 6\end{pmatrix} $$ (b) The currents are: $$ i_1 = 1, \quad i_2 = 3, \quad i_3 = -3 $$