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Exercise 2.3 · Q30

Q.Solve the following equations by reduction method. 3x−y=1, 4x+y=63x - y = 1,\ 4x + y = 6

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Step 1: 3x−y=1, 4x+y=63x-y=1,\ 4x+y=6 becomes [3−141][xy]=[16]\begin{bmatrix}3&-1\\4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\6\end{bmatrix}.

Step 2: Use R2→3R2−4R1R_2\to 3R_2-4R_1 (clears (2,1): 3(4)−4(3)=03(4)-4(3)=0): new row 2 of AA is (0, 3(1)−4(−1))=(0,7)(0,\ 3(1)-4(-1))=(0,7); new entry of BB is 3(6)−4(1)=143(6)-4(1)=14. …

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