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Miscellaneous Exercise 2(B) II · Q96

Q.Express the following equation in matrix form and solve them by the method of reduction. x+3y+2z=6, 3x−2y+5z=5x+3y+2z=6,\ 3x-2y+5z=5 and 2x−3y+6z=72x-3y+6z=7

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Step 1: x+3y+2z=6, 3x−2y+5z=5, 2x−3y+6z=7x+3y+2z=6,\ 3x-2y+5z=5,\ 2x-3y+6z=7 becomes [1323−252−36][xyz]=[657]\begin{bmatrix}1&3&2\\3&-2&5\\2&-3&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\5\\7\end{bmatrix}.

Step 2: Use R2→R2−3R1R_2\to R_2-3R_1: row 2 becomes (0,−11,−1)(0,-11,-1), constant 5−18=−135-18=-13. Use R3→R3−2R1R_3\to R_3-2R_1: row 3 becomes (0,−9,2)(0,-9,2), constant 7−12=−57-12=-5. …

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