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2x − 3y − 4z = 29 − 2x + 5y − Z = − 15 3x − Y + 5z = − 11

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Question

2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11

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Solution

Given: 2x − 3y − 4z = 29
          − 2x + 5y − z = − 15
           3x − y + 5z = − 11


\[D = \begin{vmatrix}2 & - 3 & - 4 \\ - 2 & 5 & - 1 \\ 3 & - 1 & 5\end{vmatrix}\] 
\[ = 2(25 - 1) + 3( - 10 + 3) - 4(2 - 15)\] 
\[ = 2(24) + 3( - 7) - 4( - 13)\] 
\[ = 79\] 
\[ D_1 = \begin{vmatrix}29 & - 3 & - 4 \\ - 15 & 5 & - 1 \\ - 11 & - 1 & 5\end{vmatrix}\] 
\[ = 29(25 - 1) + 3( - 75 - 11) - 4(15 + 55)\] 
\[ = 29(24) + 3( - 86) - 4(70)\] 
\[ = 158\] 
\[ D_2 = \begin{vmatrix}2 & 29 & - 4 \\ - 2 & - 15 & - 1 \\ 3 & - 11 & 5\end{vmatrix}\] 
\[ = 2( - 75 - 11) - 29( - 10 + 3) - 4(22 + 45)\]
\[ = 2( - 86) - 29( - 7) - 4(67)\] 
\[ = - 237\] 
\[ D_3 = \begin{vmatrix}2 & - 3 & 29 \\ - 2 & 5 & - 15 \\ 3 & - 1 & - 11\end{vmatrix}\] 
\[ = 2( - 55 - 15) + 3(22 + 45) + 29(2 - 15)\] 
\[ = 2( - 70) + 3(67) + 29( - 13)\] 
\[ = - 316\] 
Now, 
\[x = \frac{D_1}{D} = \frac{158}{79} = 2\] 
\[y = \frac{D_2}{D} = \frac{- 237}{79} = - 3\] 
\[z = \frac{D_3}{D} = \frac{- 316}{79} = - 4\] 
\[ \therefore x = 2, y = - 3\text{ and }z = - 4\]

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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 17 | Page 84

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