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Explain solution RD Sharma class 12 chapter Determinants exercise 5.1 question 1 subquestion (iv) maths

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Answer:

\begin{aligned} &\mathrm{M}_{11}=\mathrm{a}\left(\mathrm{b}^{2}-\mathrm{c}^{2}\right), \mathrm{M}_{21}=\mathrm{b}\left(\mathrm{a}^{2}-\mathrm{c}^{2}\right), \mathrm{M}_{31}=\mathrm{c}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \\ &\mathrm{C}_{11}=\mathrm{a}\left(\mathrm{b}^{2}-\mathrm{c}^{2}\right), \mathrm{C}_{21}=\mathrm{b}\left(\mathrm{a}^{2}-\mathrm{c}^{2}\right), \mathrm{C}_{31}=\mathrm{c}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \end{aligned}

Hint:

Let Mij, Cij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.

Given:

\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right] \\ \end{aligned}

Solution:

\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right] \\ &\mathrm{M}_{11}=\left|\begin{array}{ll} b & c a \\ c & a b \end{array}\right|=\mathrm{ab}^{2}-\mathrm{ac}^{2}=\mathrm{a}\left(\mathrm{b}^{2}-\mathrm{c}^{2}\right) \\ &\mathrm{M}_{21}=\left|\begin{array}{ll} a & b c \\ c & a b \end{array}\right|=\mathrm{a}^{2} \mathrm{~b}-\mathrm{bc}^{2}=\mathrm{b}\left(\mathrm{a}^{2}-\mathrm{c}^{2}\right) \\ &\mathrm{M}_{31}=\left|\begin{array}{ll} a & b c \\ b & c a \end{array}\right|=\mathrm{a}^{2} \mathrm{c}-\mathrm{b}^{2} \mathrm{c}=\mathrm{c}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \\ &\mathrm{C}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \mathrm{M}_{\mathrm{ij}} \end{aligned}

\begin{aligned} &C_{11}=(-1)^{2} a\left(b^{2}-c^{2}\right)=a\left(b^{2}-c^{2}\right) \\ &C_{21}=(-1)^{3} b\left(a^{2}-c^{2}\right)=-b\left(a^{2}-c^{2}\right) \\ &C_{31}=(-1)^{4} c\left(a^{2}-b^{2}\right)=c\left(a^{2}-b^{2}\right) \\ &D=1 . a\left(b^{2}-c^{2}\right)-a(a b-c a)+b . c(c-b) \\ &=a b^{2}-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2} \\ &=a^{2}(c-b)+b^{2}(a-c)+c^{2}(b-a) \end{aligned}

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Gurleen Kaur

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