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Exercise 2.1 · Q6

Q.A=[1−13210331]A = \begin{bmatrix} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}, apply C3+2C2C_3 + 2C_2 and then 3R33R_3. What do you conclude from ex. 5 and ex. 6?

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Step 1: A=[1−13210331]A=\begin{bmatrix} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}. Apply C3→C3+2C2C_3\to C_3+2C_2 first: new column 3, row by row, is 3+2(−1)=13+2(-1)=1, 0+2(1)=20+2(1)=2, 1+2(3)=71+2(3)=7, giving A∼[1−11212337]A\sim\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 3 & 3 & 7 \end{bmatrix}.

Step 2: Now apply 3R33R_3 to THIS matrix: row 3 becomes 3×(3,3,7)=(9,9,21)3\times(3,3,7)=(9,9,21), giving A∼[1−112129921]A\sim\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 9 & 9 & 21 \end{bmatrix}.

Step 3: Compare with Question 5's result (same starting matrix, same two operations, opposite order): both routes land on [1−112129921]\begin{bmatrix}1&-1&1\\2&1&2\\9&9&21\end{bmatrix}. …

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