silemus.

15Inverse of lower triangular AA

Let A=[100210341]\displaystyle A = \begin{bmatrix}1 & 0 & 0\\2 & 1 & 0\\3 & 4 & 1\end{bmatrix}. Find the entry in row 3, column 1 of A−1\displaystyle A^{-1}.

The (3,1) entry of A⁻¹ is the third coordinate of the solution to Ax = (1,0,0). Solving forward using A's rows gives x1=1, x2=−2, x3=5, so the entry you want is 5. Check: 3(1)+4(−2)+5 = 0 ✓.

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