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  • Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 24 maths

Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 24 maths

Answers (1)

Answer:

Correct option (c)

Hint:

Using properties of determinant.

Given:

Let\: \:\;\Delta =\begin{vmatrix} a-b &b+c &a \\ b-c &c+a &b \\ c-a &a+b &c \end{vmatrix}

We have to find the value of \Delta.

Solution:

        \Delta =\begin{vmatrix} a-b &b+c &a \\ b-c &c+a &b \\ c-a &a+b &c \end{vmatrix}

Using properties,

        \Rightarrow \Delta =\begin{vmatrix} a &b &a \\ b &c &b \\ c &a &c \end{vmatrix}+\begin{vmatrix} -b &c &a \\ -c &a &b \\ -a &b &c \end{vmatrix}

Here two columns are identical

        \Rightarrow \begin{vmatrix} a &b &a \\ b &c &b \\ c &a &c \end{vmatrix}=0

        \Rightarrow \Delta =0+(-1)\begin{vmatrix} b &c &a \\ c &a &b \\ a &b &c \end{vmatrix}

Applying R1→R1+R2+R3

        \Rightarrow \Delta =-\begin{vmatrix} a+b+c &a+b+c &a+b+c \\ c &a &b \\ a &b &c \end{vmatrix}

Taking (a+b+c) common from R1

        \Rightarrow \Delta =-(a+b+c)\begin{vmatrix} 1 &1 &1 \\ c &a &b \\ a &b &c \end{vmatrix}

Expanding along R1, we have

        \Rightarrow \Delta =-(a+b+c)[(ac-b^{2})-(c^{2}-ab)+(bc-a^{2})]

        \Rightarrow \Delta =-(a+b+c)[ac-b^{2}-c^{2}+ab+bc-a^{2}]

        \Rightarrow \Delta =(a+b+c)[a^{2}+b^{2}+c^{2}-ab-bc-ca]

[\because a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)]

        \Rightarrow \Delta =a^{3}+b^{3}+c^{3}-3abc

Hence, this is the required answer.

Posted by

Gurleen Kaur

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