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  • need solution for rd sharma maths class 12 chapter determinants exercise 5.5 question 3

need solution for rd sharma maths class 12 chapter determinants exercise 5.5 question 3

Answers (1)

Answer: x = -7k, y = 8k, z = 13k, where k \in R

Hint: : If \left | A \right |= 0, then the system has an infinite number of solutions.

Given:

3x+y+z=0

x - 4y + 3z = 0

2 x + 5y - 2z = 0

Solution:

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

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

Now,

                \left | A \right | = \begin{vmatrix} 3 & 1& 1\\ 1 & -4 &3 \\ 2& 5& -2 \end{vmatrix}

                      = 3(8 -15) -1(-2 -6) + 1(5+8)

                      = 3(-7) + 8 +15

                      = - 21 + 21

                      = 0

So, the system of equations has non – trivial solutions.

Consider, first two equation and put z = k

        3 x + y = -k

        x - 4y = -3k

Now, By Crammer’s rule:

x = \frac{D_{1}}{D} = \frac{\begin{vmatrix} -k & 1\\ -3k & -4 \end{vmatrix}}{\begin{vmatrix} 3&1 \\ 1& -4 \end{vmatrix}} = \frac{7k}{-13} = \frac{-7k}{13}

y = \frac{D_{2}}{D} = \frac{\begin{vmatrix} 3 & -k\\ 1 & -3k \end{vmatrix}}{\begin{vmatrix} 3&1 \\ 1& -4 \end{vmatrix}} = \frac{-8k}{-13} = \frac{8k}{13}

x=\frac{-7k}{13}, y = \frac{8k}{13}  and z=k

or x = -7k, y = 8k and z = 13k, where k \in R

It will satisfy the third equation.

Hence, x = -7k, y = 8k and z = 13k, where k \in R

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