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X+ Y = 5 Y + Z = 3 X + Z = 4

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Question

xy = 5
y + z = 3
x + z = 4

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Solution

These equations can be written as
x + + 0z = 5
0x + y + z = 3
x + 0y + z = 4

\[D = \begin{vmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{vmatrix}\] 
\[ = 1(1 - 0) - 1(0 - 1) + 0(0 - 1)\] 
\[ = 1(1) - 1( - 1) + 0\] 
\[ = 2\] 
\[ D_1 = \begin{vmatrix}5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1\end{vmatrix}\] 
\[ = 5(1 - 0) - 1(3 - 4) + 0(0 - 4)\] 
\[ = 5(1) - 1( - 1)\] 
\[ = 6\] 
\[ D_2 = \begin{vmatrix}1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1\end{vmatrix}\] 
\[ = 1(3 - 4) - 5(0 - 1) + 0(0 - 4)\] 
\[ = 1( - 1) - 5( - 1)\] 
\[ = 4\] 
\[ D_3 = \begin{vmatrix}1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4\end{vmatrix}\] 
\[ = 1(4 - 0) - 1(0 - 3) + 5(0 - 1)\] 
\[ = 1(4) - 1( - 3) + 5( - 1)\] 
\[ = 2\] 
Now, 
\[x = \frac{D_1}{D} = \frac{6}{2} = 3\] 
\[y = \frac{D_2}{D} = \frac{4}{2} = 2\] 
\[z = \frac{D_3}{D} = \frac{2}{2} = 1\] 
\[ \therefore x = 3, y = 2\text{ and }z = 1\] 

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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 14 | Page 84

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