Q.Transform $\begin{bmatrix} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}$ into an upper triangular matrix by suitable column transformations.

Step 1: $A=\begin{bmatrix}1&-1&2\\2&1&3\\3&2&4\end{bmatrix}$. To make this upper triangular via COLUMN operations, entries (2,1), (3,1) and (3,2) must become $0$.

Step 2: A column operation that mixes column 1 with column 2 changes row 2's column-1 entry using row 2's own column-2 entry -- so clearing (2,1) using $C_2$ (pivot $a_{22}=1$) needs $C_1\to C_1-\tfrac12C_2$; this alone gives column 1 $=(1-\tfrac12(-1),\ 2-\tfrac12(1),\ 3-\tfrac12(2))=(\tfrac32,\tfrac32,2)$ -- (2,1) is not yet exactly $0$, and a second pass is still needed against column 3.

Step 3: Apply $C_1\to C_1-\tfrac12C_3$ next, using the ORIGINAL column 3 $=(2,3,4)$ (unchanged since only column 1 has been touched so far): column 1 becomes $(\tfrac32-\tfrac12(2),\ \tfrac32-\tfrac12(3),\ 2-\tfrac12(4))=(\tfrac12,0,0)$ -- now BOTH (2,1) and (3,1) are $0$ together, because the two multipliers $-\tfrac12,-\tfrac12$ were chosen to solve $2+a(1)+b(2)=0$ and $3+a(2)+b(4)=0$ simultaneously. …

Zero the (2,1) and (3,1) entries together by combining column 1 with BOTH other original columns (solved simultaneously so row 2's zero is no …

• Clearing (2,1) using column 2 alone and then re-touching column 1 with column 2 again for (3,1), which silently undoes the (2,1) zero just created …