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Using properties of determinants prove that: $\begin{array}{l} \left|\begin{array}{ccc} (\mathrm{m}+\mathrm{n})^{2} & 1^{2} & \mathrm{mn} \\ (\mathrm{n}+1)^{2} & \mathrm{~m}^{2} & \ln \\ (1+\mathrm{m})^{2} & \mathrm{n}^{2} & \operatorname{lm} \end{array}\right|=\left(1^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)(1-\mathrm{m}) \\ (\mathrm{m}-\mathrm{n})(\mathrm{n}-1) \end{array}$
Using properties of determinants prove that: $\begin{array}{l} \left|\begin{array}{ccc} (\mathrm{m}+\mathrm{n})^{2} & 1^{2} & \mathrm{mn} \\ (\mathrm{n}+1)^{2} & \mathrm{~m}^{2} & \ln \\ (1+\mathrm{m})^{2} & \mathrm{n}^{2} & \operatorname{lm} \end{array}\right|=\left(1^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)(1-\mathrm{m}) \\ (\mathrm{m}-\mathrm{n})(\mathrm{n}-1) \end{array}$
CBSE | Class 12 | Determinants | Exercise 6B | Maths | RS Aggarwal
Using properties of determinants prove that: $\begin{array}{l} \left|\begin{array}{ccc} (\mathrm{m}+\mathrm{n})^{2} & 1^{2} & \mathrm{mn} \\ (\mathrm{n}+1)^{2} & \mathrm{~m}^{2} & \ln \\ (1+\mathrm{m})^{2} & \mathrm{n}^{2} & \operatorname{lm} \end{array}\right|=\left(1^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)(1-\mathrm{m}) \\ (\mathrm{m}-\mathrm{n})(\mathrm{n}-1) \end{array}$

Solution:

$\left|\begin{array}{ccc}(\mathrm{m}+\mathrm{n})^{2} & \mathrm{l}^{2} & \mathrm{mn} \\ (\mathrm{n}+\mathrm{l})^{2} & \mathrm{~m}^{2} & \mathrm{ln} \\ (1+\mathrm{m})^{2} & \mathrm{n}^{2} & \mathrm{~lm}\end{array}\right|$
$=\left(\frac{1}{2}\right)\left|\begin{array}{ccc}m^{2}+2 m n+n^{2} & 1^{2} & 2 m n \\ n^{2}+2 n l+1^{2} & m^{2} & 2 \ln \\ 1^{2}+2 l m+m^{2} & n^{2} & 2 l m\end{array}\right|\left[C_{3}^{\prime}=2 C_{3}\right]$
$=\left(\frac{1}{2}\right)\left|\begin{array}{ccc}m^{2}+n^{2} & 1^{2} & 2 m n \\ n^{2}+1^{2} & m^{2} & 2 \ln \\ 1^{2}+m^{2} & n^{2} & 2 \operatorname{lm}\end{array}\right|\left[C_{1}^{\prime}=C_{1}-C_{3}\right]$
$=\left(\frac{1}{2}\right)\left|\begin{array}{lcc}1^{2}+m^{2}+n^{2} & 1^{2} & 2 m n \\ 1^{2}+m^{2}+n^{2} & m^{2} & 2 \ln \\ 1^{2}+m^{2}+n^{2} & n^{2} & 2 \operatorname{lm}\end{array}\right|\left[C_{1}^{\prime}=C_{1}+C_{2}\right]$
$=\left(\frac{1}{2}\right)\left(l^{2}+m^{2}+n^{2}\right)\left|\begin{array}{ccc}1 & 1^{2} & 2 m n \\ 1 & m^{2} & 2 \ln \\ 1 & n^{2} & 2 \operatorname{lm}\end{array}\right|\left[C_{1}^{\prime}=C_{1} /\left(I^{2}+m^{2}+n^{2}\right)\right]$
$=\left(\frac{1}{2}\right)\left(1^{2}+m^{2}+n^{2}\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1^{2} & m^{2} & n^{2} \\ 2 m n & 2 \ln & 2 l m\end{array}\right|$ [transforming row and column]
$=\left(\frac{1}{2}\right)\left(1^{2}+m^{2}+n^{2}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ 1^{2}-m^{2} & m^{2}-n^{2} & n^{2} \\ -2 n(1-m) & -2 l(m-n) & 2 l m\end{array}\right|\left[C_{1}^{\prime}=C_{1}-C_{2} \& C_{2}^{\prime}=C_{2}-C_{3}\right]$
$=\left(1^{2}+m^{2}+n^{2}\right)(1-m)(m-n)\left|\begin{array}{ccc}0 & 0 & 1 \\ 1+m & m+n & n^{2} \\ -n & -1 & l m\end{array}\right|\left[C_{1}^{\prime}=C_{1} /(1-m) \& R_{2}^{\prime}=C_{2} /(I-m)\right]$
$=\left(l^{2}+m^{2}+n^{2}\right)(l-m)(m-n)\{0+0-l(l+m)+n(m+n)\}$ [expansion by first row]
$=\left(1^{2}+m^{2}+n^{2}\right)(1-m)(m-n)\{0+0-I(I+m)+n(m+n)\}$
$=\left(1^{2}+m^{2}+n^{2}\right)(1-m)(m-n)\left(-l^{2}-m l+m n+n^{2}\right)$
$=\left(1^{2}+m^{2}+n^{2}\right)(1-m)(m-n)\left\{\left(n^{2}-1^{2}\right)+m(n-1)\right\}$
$=\left(1^{2}+m^{2}+n^{2}\right)(1-m)(m-n)(n-1)(1+m+n)$

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