1. **Stating the problem:** We have two vertical matrices (or column vectors) and their operations: $$\begin{pmatrix} A \\ c \\ 11 \\ 12 \end{pmatrix} + \begin{pmatrix} B \\ n \\ 11 \\ 2 \end{pmatrix} = 14$$ and $$\begin{pmatrix} A \\ c \\ 11 \\ 12 \end{pmatrix} - \begin{pmatrix} B \\ n \\ 11 \\ 2 \end{pmatrix} = 9$$ 2. **Understanding the problem:** The sum and difference of these two vectors equal the scalars 14 and 9 respectively. This implies the sum and difference of corresponding components equal these values. 3. **Using vector addition and subtraction:** For vectors $$\mathbf{X} = \begin{pmatrix} A \\ c \\ 11 \\ 12 \end{pmatrix}$$ and $$\mathbf{Y} = \begin{pmatrix} B \\ n \\ 11 \\ 2 \end{pmatrix}$$, $$\mathbf{X} + \mathbf{Y} = \begin{pmatrix} A+B \\ c+n \\ 11+11 \\ 12+2 \end{pmatrix} = \begin{pmatrix} A+B \\ c+n \\ 22 \\ 14 \end{pmatrix}$$ $$\mathbf{X} - \mathbf{Y} = \begin{pmatrix} A-B \\ c-n \\ 11-11 \\ 12-2 \end{pmatrix} = \begin{pmatrix} A-B \\ c-n \\ 0 \\ 10 \end{pmatrix}$$ 4. **Equating to given scalars:** The problem states the sum equals 14 and the difference equals 9. Since the vectors have 4 components, this likely means the sums and differences of the first components equal these scalars, or the problem is about the sums of the entire vectors. 5. **Sum of all components:** Sum of $$\mathbf{X} + \mathbf{Y}$$ components: $$ (A+B) + (c+n) + 22 + 14 = 14 $$ This is impossible since 22 + 14 = 36 already exceeds 14. 6. **Sum of first components only:** Assuming the problem means: $$ A + B = 14 $$ $$ A - B = 9 $$ 7. **Solving for A and B:** Add the two equations: $$ (A+B) + (A-B) = 14 + 9 $$ $$ 2A = 23 $$ $$ A = \frac{23}{2} = 11.5 $$ Substitute back to find B: $$ 11.5 + B = 14 $$ $$ B = 14 - 11.5 = 2.5 $$ 8. **No information about c and n is given for sums or differences, so they remain unknown.** **Final answer:** $$ A = 11.5, \quad B = 2.5 $$