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Solve the system of linear equation, using matrix method x - y + z = 4; 2x + y - 3z = 0; x + y + z = 2
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Solve the system of linear equation, using matrix method x - y + z = 4; 2x + y - 3z = 0; x + y + z = 2

Answer

Matrix form of given equations is AX = B
\(\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\ 2&1&{ - 3} \\ 1&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ 2 \end{array}} \right]\)
Here \(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\ 2&1&{ - 3} \\ 1&1&1 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ 2 \end{array}} \right]\)
\(\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\ 2&1&{ - 3} \\ 1&1&1 \end{array}} \right|\)
 = 1(1 + 3) - ( - 1)(2 + 3) + 1(2 - 1)
\( = 4 + 5 + 1 = 10 \ne 0\)
Therefore, solution is unique and \(X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B\)
\(\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2 \\ { - 5}&0&5 \\ 1&2&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ 2 \end{array}} \right]\)
\(= \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {16 + 0 + 4} \\ { - 20 + 0 + 10} \\ {4 +0 + 6} \end{array}} \right]\)
\(= \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {20} \\ { - 10} \\ {10} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right]\)
Therefore, x = 2, y = - 1 and z = 1
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