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Please Solve R.D.Sharma class 12 Chapter 5 determinants Exercise  5.4 Question 1 Maths textbook Solution.

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Answer: \mathrm{x}=-6 \text { and } \mathrm{y}=-5

Hint: Use Cramer’s rule to solve a system of two equations in two variables.

Given: \begin{aligned} &x-2 y=4 \\ &-3 x+5 y=-7 \end{aligned}

Solution:First D: determinant of the coefficient matrix

\mathrm{D}=\left|\begin{array}{cc} 1 & -2 \\ -3 & 5 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)

\begin{aligned} &=(1)(5)-(-3)(-2) \\ &=5-6 \\ &=-1 \end{aligned}

Now,D\neq \: 0

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

\begin{aligned} &\mathrm{D}_{1}=\left|\begin{array}{cc} 4 & -2 \\ -7 & 5 \end{array}\right| \\ &=(4)(5)-(-7)(-2) \\ &=20-14 \end{aligned}

=6

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

\mathrm{D}_{2}=\left|\begin{array}{cc} 1 & 4 \\ -3 & -7 \end{array}\right|

 

Now,\mathrm{x}=\frac{D_{1}}{D}=\frac{6}{-1}=-6

\mathrm{y}=\frac{D_{2}}{D}=\frac{5}{-1}=-5

    Hence,\mathrm{x}=-6 \text { and } \mathrm{y}=-5

Concept: Cramer’s rule for system of two equations.

Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.

 

 

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