Solve the system of linear equation, using matrix method \(2x + y + z = 1; x - 2y - z = \frac{3}{2};\,\,3y - 5z = 9\)
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Solve the system of linear equation, using matrix method \(2x + y + z = 1; x - 2y - z = \frac{3}{2};\,\,3y - 5z = 9\)
Answer
\(x = 1,y = \frac{1}{2}\) and \(z = -\frac{3}{2}\)
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Solution
Matrix form of given equations is AX = B
\(\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]\)
Here \(A = \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]\)and \(B = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]\)
Here,\(A_{11}=13,A_{12}=5,A_{13}=3,\\ A_{21}=8,A_{22}=-10,A_{23}=-6, \\ A_{31}=1,A_{32}=3,A_{33}=-5\)
adj A=\(\left[ {\begin{array}{*{20}{c}} 13&8&1 \\ 5&{ - 10}&{ 3} \\ 3&-6&{ - 5} \end{array}} \right]\)
\(\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right| \)
\(= 2(10 + 3) - 1(-5)+1(3) = 26 + 5 + 3 \)
\( = 34 \ne 0\)
Therefore, solution is unique and \(X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B\)
\(\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13}&8&1 \\ 5&{ - 10}&3 \\ 3&{ - 6}&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]\)
\(= \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13 + 12 + 9} \\ {5 - 15 + 27} \\ {3 - 9 - 45} \end{array}} \right]\)
\( = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {34} \\ {17} \\ { - 51} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{1}{2}} \\ {\frac{{ - 3}}{2}} \end{array}} \right]\)
Therefore, \(x = 1,y = \frac{1}{2}\) and \(z =- \frac{3}{2}\)
\(\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]\)
Here \(A = \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]\)and \(B = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]\)
Here,\(A_{11}=13,A_{12}=5,A_{13}=3,\\ A_{21}=8,A_{22}=-10,A_{23}=-6, \\ A_{31}=1,A_{32}=3,A_{33}=-5\)
adj A=\(\left[ {\begin{array}{*{20}{c}} 13&8&1 \\ 5&{ - 10}&{ 3} \\ 3&-6&{ - 5} \end{array}} \right]\)
\(\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right| \)
\(= 2(10 + 3) - 1(-5)+1(3) = 26 + 5 + 3 \)
\( = 34 \ne 0\)
Therefore, solution is unique and \(X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B\)
\(\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13}&8&1 \\ 5&{ - 10}&3 \\ 3&{ - 6}&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]\)
\(= \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13 + 12 + 9} \\ {5 - 15 + 27} \\ {3 - 9 - 45} \end{array}} \right]\)
\( = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {34} \\ {17} \\ { - 51} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{1}{2}} \\ {\frac{{ - 3}}{2}} \end{array}} \right]\)
Therefore, \(x = 1,y = \frac{1}{2}\) and \(z =- \frac{3}{2}\)
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