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3x − Y + 2z = 3 2x + Y + 3z = 5 X − 2y − Z = 1

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Question

3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1

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Solution

Given: 3x − y + 2z = 3
            2x + y + 3z = 5    
            x − 2y − z = 1

\[D = \begin{vmatrix}3 & - 1 & 2 \\ 2 & 1 & 3 \\ 1 & - 2 & - 1\end{vmatrix}\] 
\[ = 3\left( - 1 + 6 \right) + 1\left( - 2 - 3 \right) + 2\left( - 4 - 1 \right)\] 
\[ = 0\] 
\[ D_{1 =} \begin{vmatrix}3 & - 1 & 2 \\ 5 & 1 & 3 \\ 1 & - 2 & - 1\end{vmatrix}\] 
\[ = 3\left( - 1 + 6 \right) + 1\left( - 5 - 3 \right) + 2\left( - 10 - 1 \right)\] 
\[ = - 15\] 
\[ D_2 = \begin{vmatrix}3 & 3 & 2 \\ 2 & 5 & 3 \\ 1 & 1 & - 1\end{vmatrix}\] 
\[ = 3\left( - 5 - 3 \right) - 3\left( - 2 - 3 \right) + 2\left( 2 - 5 \right)\] 
\[ = - 15\]
\[ D_3 = \begin{vmatrix}3 & - 1 & 3 \\ 2 & 1 & 5 \\ 1 & - 2 & 1\end{vmatrix}\] 
\[ = 3\left( 1 + 10 \right) + 1\left( 2 - 5 \right) + 3\left( - 4 - 1 \right)\] 
\[ = - 15\] Here, D is zero, but D1, D2 and D3​ are non-zero. Thus, the system of linear equations is inconsistent.

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Chapter 5: Determinants - Exercise 6.4 [Page 84]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.4 | Q 24 | Page 84

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