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Miscellaneous Exercise 2(B) II · Q95

Q.Express the following equation in matrix form and solve them by the method of reduction. x+2y+z=8, 2x+3y−z=1x+2y+z=8,\ 2x+3y-z=1 and 3x−y−2z=53x-y-2z=5

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Step 1: x+2y+z=8, 2x+3y−z=1, 3x−y−2z=5x+2y+z=8,\ 2x+3y-z=1,\ 3x-y-2z=5 becomes [12123−13−1−2][xyz]=[815]\begin{bmatrix}1&2&1\\2&3&-1\\3&-1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8\\1\\5\end{bmatrix}.

Step 2: Use R2→R2−2R1R_2\to R_2-2R_1: row 2 becomes (0,−1,−3)(0,-1,-3), constant 1−16=−151-16=-15. Use R3→R3−3R1R_3\to R_3-3R_1: row 3 becomes (0,−7,−5)(0,-7,-5), constant 5−24=−195-24=-19.

Step 3: [1210−1−30−7−5][xyz]=[8−15−19]\begin{bmatrix}1&2&1\\0&-1&-3\\0&-7&-5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8\\-15\\-19\end{bmatrix}. Use R3→R3−7R2R_3\to R_3-7R_2: row 3 becomes (0,0,−5+21)=(0,0,16)(0,0,-5+21)=(0,0,16), constant −19−7(−15)=−19+105=86-19-7(-15)=-19+105=86.

Step 4: 16z=86⇒z=8616=43816z=86\Rightarrow z=\dfrac{86}{16}=\dfrac{43}{8}. From row 2: −y−3z=−15⇒y=15−3z=15−1298=120−1298=−98-y-3z=-15\Rightarrow y=15-3z=15-\dfrac{129}{8}=\dfrac{120-129}{8}=-\dfrac{9}{8}. …

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