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2x + Y − 2z = 4 X − 2y + Z = − 2 5x − 5y + Z = − 2 - Mathematics

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Question

2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2

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Solution

Using the equations we get
\[D = \begin{vmatrix}2 & 1 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 5 & 1\end{vmatrix}\] 
\[ \Rightarrow 2\left( - 2 + 5 \right) - 1\left( 1 - 5 \right) - 2\left( - 5 + 10 \right) = 0\] 
\[ D_1 = \begin{vmatrix}4 & 1 & - 2 \\ - 2 & - 2 & 1 \\ - 2 & - 5 & 1\end{vmatrix}\] 
\[ \Rightarrow 4\left( - 2 + 5 \right) - 1\left( - 2 + 2 \right) - 2\left( 10 - 4 \right) = 0\] 
\[ D_2 = \begin{vmatrix}2 & 4 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 2 & 1\end{vmatrix}\] 
\[ \Rightarrow 2\left( - 2 + 2 \right) - 4\left( 1 - 5 \right) - 2\left( - 2 + 10 \right) = 0\] 
\[ D_3 = \begin{vmatrix}2 & 1 & 4 \\ 1 & - 2 & - 2 \\ 5 & - 5 & - 2\end{vmatrix}\] 
\[ \Rightarrow 2\left( 4 - 10 \right) - 1\left( - 2 + 10 \right) + 4\left( - 5 + 10 \right) = 0\] 
\[\therefore D = D_1 = D_2 = 0\]
Hence, the system of linear equations has infinitely many solutions.

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Chapter 6: Determinants - Exercise 6.4 [Page 85]

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RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.4 | Q 29 | Page 85

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