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provide solution for RD Sharma maths class 12 chapter determinant exercise 5.5 question 1

Answers (1)

x = k, y = k, z = k, where \: k \in R

Hint: The homogeneous system of linear equations has non-trivial solutions when the determinant is zero.

Given:

x + y - 2z =0

2x + y - 3z = 0

5x + 4y - 9z = 0

Solution:

The above linear equations can be represented in matrix form i.e.

A = \begin{bmatrix} 1& 1& -2\\ 2& 1& -3\\ 5& 4& -9 \end{bmatrix}, X= \begin{bmatrix} x\\ y\\ z\end{bmatrix},B= \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}

Now,        A = \begin{vmatrix} 1 & 1&-2 \\ 2& 1& -3\\ 5& 4& -9 \end{vmatrix}  

                   = 1(-9 + 12) -1(-18 + 15) - 2(8 -5)

                   =3 + 3 - 6

                   =0

So, the given system has non-trivial solutions.

Consider first two equations and put z=k.

x+ y - 2z =0

2x + y -3z = 0

Solving these equations by Crammer’s rule:

x = \frac{D_{1}}{D}=\frac{\begin{vmatrix} 2k&1 \\ 3k& 1 \end{vmatrix}}{\begin{vmatrix} 1 & 1\\ 2& 1 \end{vmatrix}} = \frac{-k}{-1}=k

 

y = \frac{D_{2}}{D}=\frac{\begin{vmatrix} 1&2k \\ 2& 3k \end{vmatrix}}{\begin{vmatrix} 1 & 1\\ 2& 1 \end{vmatrix}} = \frac{-k}{-1}=k

Thus, x = k, y = k, z = k satisfy the third equation.

Hence,x = k, y = k, z = k  where k \in R

 

 

Posted by

infoexpert26

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