1. **Stating the problem:** We are demonstrating Theorem (14) which involves expanding a 3×3 determinant using minors and cofactors. 2. **Formula used:** The determinant of a 3×3 matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $ can be expanded along the first row as: $$\det(A) = \sum_{j=1}^3 (-1)^{1+j} a_{1j} M_{1j}$$ where $M_{1j}$ is the minor determinant obtained by deleting the first row and $j^{th}$ column. 3. **Applying the formula to the given matrix:** $$A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$$ The expansion along the first row is: $$\det(A) = (-1)^{1+1} \cdot 1 \cdot \begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} + (-1)^{1+2} \cdot 2 \cdot \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix} + (-1)^{1+3} \cdot 3 \cdot \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix}$$ 4. **Calculating each minor:** - $\begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} = 2 \times 6 - 3 \times 5 = 12 - 15 = -3$ - $\begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix} = 1 \times 6 - 3 \times 4 = 6 - 12 = -6$ - $\begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} = 1 \times 5 - 2 \times 4 = 5 - 8 = -3$ 5. **Substituting back:** $$\det(A) = 1 \times (-3) - 2 \times (-6) + 3 \times (-3) = -3 + 12 - 9$$ 6. **Simplifying:** $$-3 + 12 - 9 = 0$$ 7. **Conclusion:** The determinant of the matrix is 0, demonstrating the expansion and calculation of minors and cofactors as stated in Theorem (14). 8. **Theorem (15) statement:** If any row or column of a determinant is multiplied by a nonzero number $k$, the determinant is also multiplied by $k$.