12th NCERT Matrices Exercise 4.6 Questions 16

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Examine the consistency of the system of equations in Exercises 1 to 6

Question (1)

\[x + 2y = 2;\;2x + 3y = 3\]

Solution

\[x + 2y = 2;\;2x + 3y = 3\] \[\left[ {\begin{array}{*{20}{c}}1&2\\2&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\] \[AX = B\] \[A = \left[ {\begin{array}{*{20}{c}}1&2\\2&3\end{array}} \right]\] \[\left| A \right| = 3 - 4 = - 1 \ne 0\] So it is consistant

Question (2)

\[2x - y = 5;\;x + y = 4\]

Solution

\[2x - y = 5;\;x + y = 4\] \[\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\] \[AX = B\] \[A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\1&1\end{array}} \right]\] \[\left| A \right| = 2 + 1 = 3 \ne 0\] So it is consistant

Question (3)

\[x + 3y = 5;\;2x + 6y = 8\]

Solution

\[\left[ {\begin{array}{*{20}{c}}1&3\\2&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\8\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}1&3\\2&6\end{array}} \right]\] \[\left| A \right| = 6 - 6 = 0\] so inconsistant

Question (4)

\[x + y + z = 1;\;2x + 3y + 2z = 2\] \[ax + ay + 2az = 4\]

Solution

\[\left[ {\begin{array}{*{20}{c}}1&1&1\\2&3&2\\a&a&{2a}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\2\\4\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}1&1&1\\2&3&2\\a&a&{2a}\end{array}} \right]\] \[\left| A \right| = 1\left( {6a - 2a} \right) - 1\left( {4a - 2a} \right) + \left( {2a - 3a} \right)\] \[\left| A \right| = 4a - 2a - a\] \[\left| A \right| = a \ne 0\] So consistent

Question (5)

\[3x - y - 2z = 2;\;\;2y - z = - 1\] \[3x - 5y = 3\]

Solution

\[\left[ {\begin{array}{*{20}{c}}3&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\3\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right]\] \[\left| A \right| = 3\left( {0 - 5} \right) + 1\left( {0 + 3} \right) - 2\left( {0 - 6} \right)\] \[\left| A \right| = - 15 + 3 + 12 = 0\] ∴ Inconsistent

Question (6)

\[5x - y + 4z = 5;\;\;2x + 3y + 5z = 2\] \[5x - 2y + 6z = - 1\]

Solution

\[\left[ {\begin{array}{*{20}{c}}5&{ - 1}&4\\2&3&5\\5&{ - 2}&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\2\\{ - 1}\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}5&{ - 1}&4\\2&3&5\\5&{ - 2}&6\end{array}} \right]\] \[\left| A \right| = 5\left( {18 + 10} \right) + 1\left( {12 - 25} \right) + 4\left( { - 4 - 15} \right)\] \[\left| A \right| = 140 - 13 - 76 = 51 \ne 0\] ∴ Consistent

Solve system of linear equations, using matrix method, in Exercise 7 to 14

Question (7)

\[5x + 2y = 4;\;\;7x + 3y = 5\]

Solution

\[\left[ {\begin{array}{*{20}{c}}5&2\\7&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\5\end{array}} \right]\] \[AX = B\] \[\therefore X = {A^{ - 1}}B\] \[A = \left[ {\begin{array}{*{20}{c}}5&2\\7&3\end{array}} \right]\] \[\left| A \right| = 15 - 14 = 1\] \[\text{cofactor of}A = \left[ {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right]\] adjA=(cofactorA)'
\[adjA = \left[ {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 7}&5\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA = \left[ {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 7}&5\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 7}&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}4\\5\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{12 - 10}\\{ - 28 + 25}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right]\] \[x = 2;\quad y = - 3\]

Question (8)

\[2x - y = - 2;\;\;3x + 4y = 3\]

Solution

\[\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\3&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\3&4\end{array}} \right]\] \[\left| A \right| = 8 + 3 = 11\] \[ \text{cofactor} A = \left[ {\begin{array}{*{20}{c}}4&{ - 3}\\1&2\end{array}} \right]\] adjA=(cofactorA)'
\[adjA = \left[ {\begin{array}{*{20}{c}}4&1\\{ - 3}&2\end{array}} \right]\] \[ \Rightarrow {A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}4&1\\{ - 3}&2\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}4&1\\{ - 3}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right]\] \[X = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{ - 8 + 3}\\{6 + 6}\end{array}} \right]\] \[X = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}}{ - 5}\\{12}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{ - 5}}{{11}}}\\{\frac{{12}}{{11}}}\end{array}} \right]\] \[ \Rightarrow x = \frac{{ - 5}}{{11}};\quad y = \frac{{12}}{{11}}\]

Question (9)

\[4x - 3y = 3;\;\;3x - 5y = 7\]

Solution

\[\left[ {\begin{array}{*{20}{c}}4&{ - 3}\\3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\7\end{array}} \right]\] \[AX = B\] \[ \therefore X = {A^{ - 1}}B\] \[A = \left[ {\begin{array}{*{20}{c}}4&{ - 3}\\3&{ - 5}\end{array}} \right]\] \[\left| A \right| = - 20 + 9 = - 11 \ne 0\] \[ \text{cofactor of} A = \left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 3}\\3&4\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}}{ - 5}&3\\{ - 3}&4\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}}{ - 5}&3\\{ - 3}&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\7\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}}6\\{19}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{ - 6}}{{11}}}\\{\frac{{ - 19}}{{11}}}\end{array}} \right]\] \[x = \frac{6}{{11}};\;y = \frac{{ - 19}}{{11}}\]

Question (10)

\[5x + 2y = 3;\;\;3x + 2y = 5\]

Solution

\[\left[ {\begin{array}{*{20}{c}}5&2\\3&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[A = \left[ {\begin{array}{*{20}{c}}5&2\\3&2\end{array}} \right]\] \[\left| A \right| = 10 - 6 = 4\] \[ \text{cofactor of} A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 2}&5\end{array}} \right]\] adjA=(cofactorA)'
\[adjA = \left[ {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 3}&5\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 3}&5\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 3}&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right]\] \[X = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}{ - 4}\\{16}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\] \[x = - 1;\quad y = 4\]

Question (11)

\[2x + y + z = 1;\;\;x - 2y - z = \frac{3}{2}\] \[3y - 5z = 9\]

Solution

\[\left[ {\begin{array}{*{20}{c}}2&1&1\\1&{ - 2}&{ - 1}\\0&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{\frac{3}{2}}\\9\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[\left| A \right| = \left| {\begin{array}{*{20}{c}}2&1&1\\1&{ - 2}&{ - 1}\\0&3&{ - 5}\end{array}} \right|\] \[\left| A \right| = 2\left( {13} \right) - 1\left( { - 5} \right) + 1\left( 3 \right) = 34\] \[ \text{cofactor A} = \left[ {\begin{array}{*{20}{c}}{13}&5&3\\8&{ - 10}&{ - 6}\\1&3&{ - 5}\end{array}} \right]\] adjA = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\{\frac{3}{2}}\\9\end{array}} \right]\] \[X = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}}{34}\\{17}\\{ - 51}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{\frac{1}{2}}\\{ - \frac{3}{2}}\end{array}} \right]\] \[x = 1;y = \frac{1}{2};z = \frac{{ - 3}}{2}\]

Question (12)

\[x - y + z = 4;\;\;2x + y - 2z = 0\] \[x + y + z = 2\]

Solution

\[\left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\0\\2\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[\left| A \right| = \left| {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right|\] \[\left| A \right| = \left| {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right|\] \[\left| A \right| = 1\left( 4 \right) + 1\left( 5 \right) + 1\left( 1 \right) = 10\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}4&{ - 5}&1\\2&0&{ - 2}\\2&5&3\end{array}} \right]\] adjA = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}4\\0\\2\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}{20}\\{ - 10}\\{10}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\1\end{array}} \right]\] \[x = 2;y = - 1;z = 1\]

Question (13)

\[2x + 3y + 3z = 5;\;\;x - 2y + z = - 4\] \[3x - y - 2z = 3\]

Solution

\[\left[ {\begin{array}{*{20}{c}}2&3&3\\1&{ - 2}&1\\3&{ - 1}&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\3\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[\left| A \right| = \left| {\begin{array}{*{20}{c}}2&3&3\\1&{ - 2}&1\\3&{ - 1}&{ - 2}\end{array}} \right|\] \[\left| A \right| = 2\left( 5 \right) - 3\left( { - 5} \right) + 3\left( 5 \right) = 40\] \[ \text{Cofactor}A = \left| {\begin{array}{*{20}{c}}5&5&5\\3&{ - 13}&{11}\\9&1&{ - 7}\end{array}} \right|\] adjA = (cofactor A)'
\[{\rm{adj}}A = \left| {\begin{array}{*{20}{c}}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right|\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\3\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}}{40}\\{80}\\{ - 40}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\end{array}} \right]\] \[x = 1,y = 2,z = - 1\]

Question (14)

\[x - y + 2z = 7;\;\;3x + 4y - 5z = - 5\] \[2x - y + 3z = 12\]

Solution

\[\left| {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&4&{ - 5}\\2&{ - 1}&3\end{array}} \right|\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7\\{ - 5}\\{12}\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[\left| A \right| = \left| {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&4&{ - 5}\\2&{ - 1}&3\end{array}} \right|\] \[\left| A \right| = 1\left( 7 \right) + 1\left( {19} \right) + 2\left( { - 11} \right)\] \[\left| A \right| = 7 + 19 - 22 = 4\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}7&{ - 19}&{ - 11}\\1&{ - 1}&{ - 1}\\{ - 3}&{11}&7\end{array}} \right]\] adjA = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right]\left[ {\begin{array}{*{20}{c}}7\\{ - 5}\\{12}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}8\\4\\{12}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\1\\3\end{array}} \right]\] \[x = 2,y = 1,z = 3\]

Question (15)

If $A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right]$, find A^-1. Using A^-1 solve the system of equations
\[2x - 3y + 5z = 11;\;\;3x + 2y - 4z = - 5\] \[x + y - 2z = - 3\]

Solution

\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right]\] \[\left| A \right| = \left| {\begin{array}{*{20}{c}}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right|\] \[\left| A \right| = 2\left( 0 \right) + 3\left( { - 2} \right) + 5\left( 1 \right) = -1\] \[cofactorA = \left[ {\begin{array}{*{20}{c}}0&2&1\\{ - 1}&{ - 9}&{ - 5}\\2&{23}&{13}\end{array}} \right]\] adjA = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}0&{ - 1}&2\\2&{ - 9}&{23}\\1&{ - 5}&{13}\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}0&{ - 1}&2\\2&{ - 9}&{23}\\1&{ - 5}&{13}\end{array}} \right]\] \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}0&1&{ - 2}\\{ - 2}&9&{ - 23}\\{ - 1}&5&{ - 13}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{11}\\{ - 5}\\{ - 3}\end{array}} \right]\] \[AX = B\] \[X = {A^{ - 1}}B\] \[X = \left[ {\begin{array}{*{20}{c}}0&1&{ - 2}\\{ - 2}&9&{ - 23}\\{ - 1}&5&{ - 13}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{11}\\{ - 5}\\{ - 3}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]\] \[x = 1,y = 2,z = 3\]

Question (16)

The cost of 4kg onion, 3 kg wheat and 2kg rice is Rs60. the cost of 2kg onion, 4kg wheat and 6kg rice is Rs 90. the cost of 6kg onion 2kg wheat and 3kg rice is Rs70. Find cost of each item per kg by matrix method

Solution

Let cost of onion be = Rs x/kg
Let cost of wheat be = Rs y/kg
Let cost of rice be = Rs z/kg
From the informatio
\[4x + 3y + 2z = 60\] \[2x + 4y + 6z = 90\] \[6x + 2y + 3z = 70\] \[\left[ {\begin{array}{*{20}{c}}4&3&2\\2&4&6\\6&2&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{60}\\{90}\\{70}\end{array}} \right]\] \[AX = B\] \[\therefore X = {A^{ - 1}}B\] \[\left| A \right| = 4\left( 0 \right) - 3\left( { - 30} \right) + 2\left( { - 20} \right)\] \[\left| A \right| = 90 - 40 = 50\] \[ \texr{cofactor A} = \left[ {\begin{array}{*{20}{c}}0&{30}&{ - 20}\\{ - 5}&0&{10}\\{10}&{ - 20}&{10}\end{array}} \right]\] adjA = (cofactor A)'
\[adjA = \left[ {\begin{array}{*{20}{c}}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right]\] \[{A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA\] \[{A^{ - 1}} = \frac{1}{{50}}\left[ {\begin{array}{*{20}{c}}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right]\] \[X = {A^{ - 1}}B\] \[X = \frac{1}{{50}}\left[ {\begin{array}{*{20}{c}}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{60}\\{90}\\{70}\end{array}} \right]\] \[X = \frac{1}{{50}}\left[ {\begin{array}{*{20}{c}}{250}\\{400}\\{400}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\8\\8\end{array}} \right]\] \[x = 5;y = 8;z = 8\] cost of onion is 5/kg
cost of wheat is 8/kg
cost of rice is 8/kg