1. **State the problem:** We are given six matrices A, B, C, D, E, and F and asked to find: (a) $A + B$ (b) $5D - 4C$ (c) $FE$ (d) $F^{-1}$ 2. **Recall matrix operations and rules:** - Matrix addition is done element-wise and requires matrices to be the same size. - Scalar multiplication multiplies every element by the scalar. - Matrix multiplication requires the number of columns in the first matrix to equal the number of rows in the second. - The inverse of a matrix $F$ (if it exists) satisfies $FF^{-1} = I$, where $I$ is the identity matrix. --- ### (a) Calculate $A + B$ Given: $$A = \begin{bmatrix}-1 & 2 & 0 \\ 4 & 5 & 3\end{bmatrix}, \quad B = \begin{bmatrix}7 & 1 & -3 \\ 2 & 0 & 6\end{bmatrix}$$ Add element-wise: $$A + B = \begin{bmatrix}-1+7 & 2+1 & 0+(-3) \\ 4+2 & 5+0 & 3+6\end{bmatrix} = \begin{bmatrix}6 & 3 & -3 \\ 6 & 5 & 9\end{bmatrix}$$ --- ### (b) Calculate $5D - 4C$ Given: $$C = \begin{bmatrix}1 & 2 \\ -4 & 9\end{bmatrix}, \quad D = \begin{bmatrix}11 & 5 \\ 0 & -2\end{bmatrix}$$ Calculate scalar multiples: $$5D = 5 \times \begin{bmatrix}11 & 5 \\ 0 & -2\end{bmatrix} = \begin{bmatrix}55 & 25 \\ 0 & -10\end{bmatrix}$$ $$4C = 4 \times \begin{bmatrix}1 & 2 \\ -4 & 9\end{bmatrix} = \begin{bmatrix}4 & 8 \\ -16 & 36\end{bmatrix}$$ Subtract: $$5D - 4C = \begin{bmatrix}55-4 & 25-8 \\ 0-(-16) & -10-36\end{bmatrix} = \begin{bmatrix}51 & 17 \\ 16 & -46\end{bmatrix}$$ --- ### (c) Calculate $FE$ Given: $$F = \begin{bmatrix}1 & 2 & 0 \\ 0 & 1 & 2 \\ -1 & 3 & 4\end{bmatrix}, \quad E = \begin{bmatrix}0 & 1 \\ 1 & 0 \\ 0 & 3\end{bmatrix}$$ Multiply $F$ (3x3) by $E$ (3x2): Element $(i,j)$ of $FE$ is the dot product of row $i$ of $F$ and column $j$ of $E$. Calculate each element: - $(1,1) = 1\times0 + 2\times1 + 0\times0 = 2$ - $(1,2) = 1\times1 + 2\times0 + 0\times3 = 1$ - $(2,1) = 0\times0 + 1\times1 + 2\times0 = 1$ - $(2,2) = 0\times1 + 1\times0 + 2\times3 = 6$ - $(3,1) = -1\times0 + 3\times1 + 4\times0 = 3$ - $(3,2) = -1\times1 + 3\times0 + 4\times3 = -1 + 12 = 11$ So, $$FE = \begin{bmatrix}2 & 1 \\ 1 & 6 \\ 3 & 11\end{bmatrix}$$ --- ### (d) Calculate $F^{-1}$ Given: $$F = \begin{bmatrix}1 & 2 & 0 \\ 0 & 1 & 2 \\ -1 & 3 & 4\end{bmatrix}$$ We find $F^{-1}$ by using the formula for the inverse of a 3x3 matrix: $$F^{-1} = \frac{1}{\det(F)} \mathrm{adj}(F)$$ **Step 1: Calculate $\det(F)$** $$\det(F) = 1 \times \begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix} - 2 \times \begin{vmatrix}0 & 2 \\ -1 & 4\end{vmatrix} + 0 \times \begin{vmatrix}0 & 1 \\ -1 & 3\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix} = 1 \times 4 - 2 \times 3 = 4 - 6 = -2$$ $$\begin{vmatrix}0 & 2 \\ -1 & 4\end{vmatrix} = 0 \times 4 - 2 \times (-1) = 0 + 2 = 2$$ So, $$\det(F) = 1 \times (-2) - 2 \times 2 + 0 = -2 - 4 = -6$$ **Step 2: Calculate the matrix of cofactors** Cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor of element $(i,j)$. Calculate each minor: - $C_{11}$: Minor of element (1,1): $$\begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix} = -2$$ Cofactor: $$C_{11} = (+1) \times (-2) = -2$$ - $C_{12}$: Minor of element (1,2): $$\begin{vmatrix}0 & 2 \\ -1 & 4\end{vmatrix} = 2$$ Cofactor: $$C_{12} = (-1) \times 2 = -2$$ - $C_{13}$: Minor of element (1,3): $$\begin{vmatrix}0 & 1 \\ -1 & 3\end{vmatrix} = 0 \times 3 - 1 \times (-1) = 1$$ Cofactor: $$C_{13} = (+1) \times 1 = 1$$ - $C_{21}$: Minor of element (2,1): $$\begin{vmatrix}2 & 0 \\ 3 & 4\end{vmatrix} = 2 \times 4 - 0 \times 3 = 8$$ Cofactor: $$C_{21} = (-1) \times 8 = -8$$ - $C_{22}$: Minor of element (2,2): $$\begin{vmatrix}1 & 0 \\ -1 & 4\end{vmatrix} = 1 \times 4 - 0 \times (-1) = 4$$ Cofactor: $$C_{22} = (+1) \times 4 = 4$$ - $C_{23}$: Minor of element (2,3): $$\begin{vmatrix}1 & 2 \\ -1 & 3\end{vmatrix} = 1 \times 3 - 2 \times (-1) = 3 + 2 = 5$$ Cofactor: $$C_{23} = (-1) \times 5 = -5$$ - $C_{31}$: Minor of element (3,1): $$\begin{vmatrix}2 & 0 \\ 1 & 2\end{vmatrix} = 2 \times 2 - 0 \times 1 = 4$$ Cofactor: $$C_{31} = (+1) \times 4 = 4$$ - $C_{32}$: Minor of element (3,2): $$\begin{vmatrix}1 & 0 \\ 0 & 2\end{vmatrix} = 1 \times 2 - 0 \times 0 = 2$$ Cofactor: $$C_{32} = (-1) \times 2 = -2$$ - $C_{33}$: Minor of element (3,3): $$\begin{vmatrix}1 & 2 \\ 0 & 1\end{vmatrix} = 1 \times 1 - 2 \times 0 = 1$$ Cofactor: $$C_{33} = (+1) \times 1 = 1$$ **Step 3: Form the cofactor matrix:** $$\mathrm{Cof}(F) = \begin{bmatrix}-2 & -2 & 1 \\ -8 & 4 & -5 \\ 4 & -2 & 1\end{bmatrix}$$ **Step 4: Transpose the cofactor matrix to get adjugate:** $$\mathrm{adj}(F) = \begin{bmatrix}-2 & -8 & 4 \\ -2 & 4 & -2 \\ 1 & -5 & 1\end{bmatrix}$$ **Step 5: Calculate inverse:** $$F^{-1} = \frac{1}{-6} \begin{bmatrix}-2 & -8 & 4 \\ -2 & 4 & -2 \\ 1 & -5 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & \frac{4}{3} & -\frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ -\frac{1}{6} & \frac{5}{6} & -\frac{1}{6}\end{bmatrix}$$ --- **Final answers:** (a) $A + B = \begin{bmatrix}6 & 3 & -3 \\ 6 & 5 & 9\end{bmatrix}$ (b) $5D - 4C = \begin{bmatrix}51 & 17 \\ 16 & -46\end{bmatrix}$ (c) $FE = \begin{bmatrix}2 & 1 \\ 1 & 6 \\ 3 & 11\end{bmatrix}$ (d) $$F^{-1} = \begin{bmatrix}\frac{1}{3} & \frac{4}{3} & -\frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ -\frac{1}{6} & \frac{5}{6} & -\frac{1}{6}\end{bmatrix}$$