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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 5 determinants Exercise 5.4 Question 19 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 5 determinants  Exercise 5.4 Question 19 Maths Textbook Solution.

Answers (1)

Answer:\mathrm{x}=\frac{-(c-d)(d-b)}{(a-b)(c-a)}, \mathrm{y}=\frac{-(a-d)(d-c)}{(a-b)(b-c)} \text { and } \mathrm{z}=\frac{-(b-d)(d-a)}{(b-c)(c-a)}

Hint: Use Cramer’s rule to solve a system of linear equations

Given:

\begin{aligned} &x+y+z+1=0 \\ &a x+b y+c z+d=0 \\ &a^{2} x+b^{2} y+c^{2} z+d^{2}=0 \end{aligned}

Solution:

First take coefficient of variables x, y and z.

\begin{aligned} &|\mathrm{A}|=\left|\begin{array}{lll} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array}\right| \\ &\quad C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}

=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & b-a & c-a \\ a^{2} & b^{2}-a^{2} & c^{2}-a^{2} \end{array}\right|

Now taking (b-a) and (c-a) from C_{2} and C_{3} respectively,

=(b-a)(c-a)\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b+a & c+a \end{array}\right|

Expanding along R_{1},

\begin{aligned} &=(b-a)(c-a)[1(c+a-b-a)] \\ &=(b-a)(c-a)(c-b) \\ &=(a-b)(b-c)(c-a) \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ d & b & c \\ d^{2} & b^{2} & c^{2} \end{array}\right|

            \begin{aligned} &C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}

D_{x}=-\left|\begin{array}{ccc} 1 & 0 & 0 \\ d & b-d & c-d \\ d^{2} & b^{2}-d^{2} & c^{2}-d^{2} \end{array}\right|

Now taking (b-d) and (c-d) from C_{2} and C_{3} respectively,

=-(b-d)(c-d)\left|\begin{array}{ccc} 1 & 0 & 0 \\ d & 1 & 1 \\ d^{2} & b+d & c+d \end{array}\right|

Expanding along R_{1},

\begin{aligned} &=-(b-d)(c-d)[1(c+d-b-d)] \\ &=-(b-d)(c-d)(c-b) \\ &=-(b-c)(c-d)(d-b) \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{lll} 1 & 1 & 1 \\ a & d & c \\ a^{2} & d^{2} & c^{2} \end{array}\right|

            \begin{aligned} &C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}

\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & d-a & c-a \\ a^{2} & d^{2}-a^{2} & c^{2}-a^{2} \end{array}\right|

Now taking (d-a) and (c-a) from C_{2} and C_{3} respectively,

=-(d-a)(c-a)\left|\begin{array}{ccc} 1 & 0 & 0 \\ d & 1 & 1 \\ d^{2} & d+a & c+a \end{array}\right|

Expanding along R_{1},

\begin{aligned} &=-(d-a)(c-a)[1(c+a-d-a)] \\ &=-(d-a)(c-a)(c-d) \\ &=-(a-d)(d-c)(c-a) \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\mathrm{D}_{z}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & d \\ a^{2} & b^{2} & d^{2} \end{array}\right|

            \begin{aligned} &C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}

          \begin{aligned} &\mathrm{D}_{z}\\ &=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & b-a & d-a \\ a^{2} & b^{2}-a^{2} & d^{2}-a^{2} \end{array}\right| \end{aligned}

Now taking (b-a) and (d-a) from C_{2}and C_{3} respectively,

=-(b-a)(d-a)\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b+a & d+a \end{array}\right|

Expanding along R_{1},

\begin{aligned} &=-(b-a)(d-a)[1(d+a-b-a)] \\ &=-(b-a)(d-a)(d-b) \\ &=-(a-b)(b-d)(d-a) \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{-(b-c)(c-d)(d-b)}{(a-b)(b-c)(c-a)}=\frac{-(c-d)(d-b)}{(a-b)(c-a)} \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-(a-d)(d-c)(c-a)}{(a-b)(b-c)(c-a)}=\frac{-(a-d)(d-c)}{(a-b)(b-c)} \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{-(a-b)(b-d)(d-a)}{(a-b)(b-c)(c-a)}=\frac{-(b-d)(d-a)}{(b-c)(c-a)} \end{aligned}

Concept: Solving matrix of order 3x3 (Elementary row and column operations)

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