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X − Y + 3z = 6 X + 3y − 3z = − 4 5x + 3y + 3z = 10

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Question

x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10

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Solution

Using the equations, we get 
\[D = \begin{vmatrix}1 & - 1 & 3 \\ 1 & 3 & - 3 \\ 5 & 3 & 3\end{vmatrix} = 1(9 + 9) + 1(3 + 15) + 3(3 - 15)\] 
\[ = 18 + 18 - 36 = 0\] 
\[ D_1 = \begin{vmatrix}6 & - 1 & 3 \\ - 4 & 3 & - 3 \\ 10 & 3 & 3\end{vmatrix} = 6(9 + 9) + 1( - 12 + 30) + 3( - 12 - 30)\] 
\[ = 108 + 18 - 126 = 0\] 
\[ D_2 = \begin{vmatrix}1 & 6 & 3 \\ 1 & - 4 & - 3 \\ 5 & 10 & 3\end{vmatrix} = 1( - 12 + 30) - 6(3 + 15) + 3(10 + 20)\] 
\[ = 18 - 108 + 90 = 0\] 
\[ D_3 = \begin{vmatrix}1 & - 1 & 6 \\ 1 & 3 & - 4 \\ 5 & 3 & 10\end{vmatrix} = 1(30 + 12) + 1(10 + 20) + 6(3 - 15)\] 
\[ = 42 + 30 - 72 = 0\] 
\[ \therefore D = D_1 = D_2 = D_3 = 0\]
Hence, the system of equations has infinitely many solutions.

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Chapter 5: Determinants - Exercise 6.4 [Page 85]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.4 | Q 30 | Page 85

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