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Solve the system of linear equation, using matrix method $$2x + y + z = 1; x - 2y - z = \frac{3}{2};\,\,3y - 5z = 9$$
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## QuestionMathsClass 12

Solve the system of linear equation, using matrix method $$2x + y + z = 1; x - 2y - z = \frac{3}{2};\,\,3y - 5z = 9$$

$$x = 1,y = \frac{1}{2}$$ and $$z = -\frac{3}{2}$$
4.6
4.6

## 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 $$z =- \frac{3}{2}$$