#### Need Solution for R.D.Sharma Maths Class 12 Chapter 5 determinants  Exercise 5.4 Question 26  Maths Textbook Solution.

Answer: $\mathrm{x}=\frac{5}{3}, \mathrm{y}=k-\frac{4}{3} \text { and } \mathrm{z}=\mathrm{k}$

Hint: Use Cramer’s rule for system of linear equations.

Given:

\begin{aligned} &x-y+z=3 \\ &2 x+y-z=2 \\ &-x-2 y+2 z=1 \end{aligned}

Solution:

$\begin{gathered} \mathrm{AX}=\mathrm{B} \\ \qquad\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ \text { Where } \mathrm{A}=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right], \mathrm{X}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \end{gathered}$

Solving determinant,

$|\mathbf{A}|=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right|$

Expanding along $1^{st}$ row,

\begin{aligned} &=1(2-2)+1(4-1)+1(-4+1) \\ &=0+3-3 \end{aligned}

|A| = 0 $\Rightarrow$System of linear equations have infinite number of solutions.

Let z = k

$\Rightarrow x-y+k=3$                                                                            ...(1)

$\Rightarrow 2 \mathrm{x}+\mathrm{y}-\mathrm{k}=2$                                                                        .....(2)

From (1) and (2),

$\Rightarrow x-y=3-k$                                                                          ......(3)

$\Rightarrow 2 \mathrm{x}+\mathrm{y}=2+\mathrm{k}$                                                                        .......(4)

\begin{aligned} &\Rightarrow 3 \mathrm{x}=5 \\ &\therefore x=\frac{5}{3} \end{aligned}

From (3),

$\Rightarrow \frac{5}{3}-y=3-k$

\begin{aligned} &y=\frac{5}{3}-3+k \\ &y=\frac{5-9+3 k}{3} \\ &y=\frac{3 k-4}{3} \\ &y=k-\frac{4}{3} \\ &\therefore z=k \end{aligned}

Concept: Solving matrix of order 3x3 by solving linear equations

Note: When D = 0, there is either no solution or infinite solutions.