## College Algebra (6th Edition)

To evaluate an nth-order determinant, where $n > 2$, 1. Select a row or column about which to expand. 2. Multiply each element $a_{ij}$ in the row or column by $(-1)^{i+j}$ times the determinant obtained by deleting the ith row and the $j\mathrm{t}\mathrm{h}$ column in the given array of numbers. 3. The value of the determinant is the sum of the products found in step 2. ----- We select row 1, ($1$s are easy to work with). The first position, (i,j)=(1,1), 1+1=2, defines the sequence of alternating signs starting with a plus. $\left|\begin{array}{lll} 1 & 1 & 1\\ 2 & 2 & 2\\ -3 & 4 & -5 \end{array}\right|=$ $=+1\cdot\left|\begin{array}{ll} 2 & 2\\ 4 & -5 \end{array}\right|-1\cdot\left|\begin{array}{ll} 2 & 2\\ -3 & -5 \end{array}\right|+1\cdot\left|\begin{array}{ll} 2 & 2\\ -3 & 4 \end{array}\right|$ $=(-10-8)-(-10+6)+(8+6)$ $=-18+4+14$ $=0$