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Miscellaneous Exercise 2(B) II · Q92

Q.Express the following equation in matrix form and solve them by the method of reduction. x+y=1, y+z=53, z+x=43x+y=1,\ y+z=\dfrac{5}{3},\ z+x=\dfrac{4}{3}

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Step 1: x+y=1, y+z=53, z+x=43x+y=1,\ y+z=\tfrac53,\ z+x=\tfrac43 becomes [110011101][xyz]=[15343]\begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\\tfrac53\\\tfrac43\end{bmatrix}.

Step 2: Use R3→R3−R1R_3\to R_3-R_1: row 3 becomes (0,−1,1)(0,-1,1), constant becomes 43−1=13\tfrac43-1=\tfrac13.

Step 3: [1100110−11][xyz]=[15313]\begin{bmatrix}1&1&0\\0&1&1\\0&-1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\\tfrac53\\\tfrac13\end{bmatrix}. Use R3→R3+R2R_3\to R_3+R_2: row 3 becomes (0,0,2)(0,0,2), constant becomes 13+53=2\tfrac13+\tfrac53=2. …

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