mathematics-class-12-unit-04-chapter-08-determinants-l-8-11-xjj55gydvr8

### <Success style="font-size:25px"> Determinants L-8 </Success>
### <primary style="font-size:24px"> Problem 1 </primary>
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- Find the determinant of

- $A=\left(\begin{array}{ccc}1 & x & x^2 \\\ x^2 & 1 & x \\\ x & x^2 & 1\end{array}\right)$

- Soln. $ R_1 \leftarrow R_1+R_2$

- $\therefore|A|= \left(\begin{array}{ccc} 1+x^2 & x+1 & x^2+x \\\  x^2 & 1 & x \\\ x & x^2 & 1 \end{array}\right)$

- $ R_1 \leftarrow R_1+R_3$

- Then $|A|=\left|\begin{array}{ccc}1+x+x^2 & 1+x+x^2 & 1+x+1 x^2 \\\ x^2 & 1 & x \\\ x & x^2 & 1\end{array}\right|$

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<footer style = "font-size: 18px"> $\rightarrow$ Determinants lecture - 8 $\rightarrow$ <span style = "color:blue"> Problem 1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-1"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>$ \begin{array}{l}\therefore|A|=\left(1+x+x^2\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ x^2 & 1 & x \\ x & x^2 & 1\end{array}\right|\end{array}$</p>
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<li>
<p>Now $   C_1 \leftarrow C_1-C_2$</p>
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<li>
<p>$ \therefore |A|=\left(1+x+x^2\right)\left| \begin{array}{ccc}0 & 1 & 1 \\ x^2-1 & 1 & x \\ x-x^2 & x^2 & 1 \end{array}\right|$</p>
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<li>
<p>$   C_2 \leftarrow C_2-C_3$</p>
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<li>
<p>$ \therefore|A|=\left(1+x+x^2\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ x^2-1 & 1-x & x \\ x(1-x) & x^2-1 & 1\end{array}\right|$</p>
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<footer style = "font-size: 18px">Determinants lecture - 8 $\rightarrow$ Problem 1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem 2 </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-2"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
<li>
<p>$ \therefore|A|=\left(1+x+x^2\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ x^2-1 & 1-x & x \\ x(1-x) & x^2-1 & 1\end{array}\right|$</p>
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<li>
<p>$= \left(1+x+x^2\right)\left(\left(x^2-1\right)^2-x(1-x)^2\right)$</p>
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<li>
<p>$ = \left(1+x+x^2\right)(x-1)^2\left((x+1)^2-x\right)$</p>
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<li>
<p>$ = \left(1+x+x^2\right)(x-1)^2\left(x^2+x+1\right)$</p>
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<li>
<p>$ = (x-1)\left(x^2+x+1\right) (x-1)\left(x^2+x+1\right)$</p>
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<li>
<p>$ = \left(1-x^3\right)^2$</p>
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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-8 </Success>
### <primary style="font-size:24px"> Problem 2 </primary>
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- Find the determinant of

- $  A=\left(\begin{array}{ccc}\sqrt{11}+\sqrt{3} & \sqrt{20} & \sqrt{5} \\\ \sqrt{15}+\sqrt{22} & \sqrt{25} & \sqrt{10} \\\ 3+\sqrt{55} & \sqrt{15} & \sqrt{25}\end{array}\right)$

- Soln.  $ |A|=\left(\begin{array}{lll}\sqrt{11} & \sqrt{20} & \sqrt{5} \\\ \sqrt{22} & \sqrt{25} & \sqrt{10} \\\ \sqrt{55} & \sqrt{15} & \sqrt{25} \end{array}\right)+\left|\begin{array}{lll}\sqrt{3} & \sqrt{20} & \sqrt{5} \\\ \sqrt{15} & \sqrt{25} & \sqrt{10} \\\ \sqrt{3} \cdot \sqrt{3} & \sqrt{15} & \sqrt{25}\end{array}\right|$

- $= \left|A_1\right|+\left|A_2\right|$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-3"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-2"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
<li>
<p>$ \left|A_1\right|=\sqrt{11} \cdot \sqrt{5} \cdot \sqrt{5} \left|\begin{array}{ccc}1 & 2 & 1 \\ \sqrt{2} & \sqrt{5} & \sqrt{2} \\ \sqrt{5} & \sqrt{3} & \sqrt{5}\end{array}\right|$</p>
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<li>
<p>Since two columns are identical the determinant $=0$</p>
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<li>
<p>$\therefore\left|A_1\right|=0$</p>
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<li>
<p>$ \therefore|A|=\left|A_2\right|=\left|\begin{array}{lll}\sqrt{3} & \sqrt{20} & \sqrt{5} \\ \sqrt{15} & \sqrt{25} & \sqrt{10} \\ \sqrt{3} \cdot \sqrt{3} & \sqrt{15} & \sqrt{25}\end{array}\right|$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem 3 </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-4"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-3"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
<li>
<p>$ \therefore|A|=\sqrt{3} \cdot \sqrt{5} \cdot \sqrt{5}\left|\begin{array}{ccc}1 & 2 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5}\end{array}\right|$</p>
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<p>Its determinants</p>
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<li>
<p>$  =1 \cdot(5-\sqrt{6})-2(5-\sqrt{6})$</p>
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<li>
<p>$ +1(\sqrt{15}-\sqrt{15})$</p>
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<li>
<p>$ =-1(5-\sqrt{6})=(\sqrt{6}-5)$</p>
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<li>
<p>$ \therefore|A|=5 \sqrt{3} \times (\sqrt{6}-5)$</p>
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<footer style = "font-size: 18px">Problem 2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 3 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-5"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-3-primary"><primary style="font-size:24px"> Problem 3 </primary></h3>
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<p>Find the determinant of</p>
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<li>
<p>$  A=\left(\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right)$</p>
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<p>Soln.  $  R_1 \leftarrow R_1+R_2+R_3$</p>
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<li>
<p>$|A|= \left|\begin{array}{ccc}a+b+c & b+c+a & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 3 </span> $\rightarrow$ Solution $\rightarrow$ Problem 4 </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-6"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-4"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>$\therefore |A|=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$</p>
</li>
<li>
<p>$ |A|=(a+b+c)\left|\begin{array}{ccc}0 & 1 & 1 \\ 0 & b-c-a & 2b \\ a+b+c & 2c & c-a-b\end{array}\right|$</p>
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<li>
<p>$ \begin{array}{l}\therefore|A|=(a+b+c)(-1)^{3+1}(a+b+c)(2 b-(b-c-a))\end{array}$</p>
</li>
<li>
<p>$  =(a+b+c)^3 Ans$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 4 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-7"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-4-primary"><primary style="font-size:24px"> Problem 4 </primary></h3>
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<li>
<p>$ A=\left|\begin{array}{ccc}x & x^2 & y z \\ y & y^2 & z x \\ z & z^2 & x y\end{array}\right|$</p>
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<li>
<p>Soln.  $ R_1=R_1-R_2$</p>
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<li>
<p>$|A|=\left|\begin{array}{ccc}x-y & (x-y)(x+y) & z(y-x) \\ y & y^2 & z x \\ z & z^2 & x y\end{array}\right|$</p>
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<footer style = "font-size: 18px">Problem 3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 4 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-8"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-5"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
<li>
<p>$ \therefore|A|=(x-y) \cdot\left|\begin{array}{ccc}1 & x+y & -z \\ y & y^2 & z x \\ z & z^2 & x y\end{array}\right|$</p>
</li>
<li>
<p>$ \quad R_2 \leftarrow R_2-R_3$</p>
</li>
<li>
<p>$ |A|=(x-y)\left|\begin{array}{ccc}1 & x+y & -z \\ y-z & (y-z)(y+z) & x(z-y) \\ z & z^2 & x y\end{array}\right|$</p>
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<li>
<p>$ \therefore|A|=(x-y)(y-z)\left|\begin{array}{ccc}1 & x+y & -z \\ 1 & y+z & -x \\ z & z^2 & x y\end{array}\right|$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-9"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-6"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>$\therefore R_2=R_2-R_1$   We have</p>
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<li>
<p>$ \therefore|A|=(x-y)(y-z)\left|\begin{array}{ccc}1 & x+y & -z \\ 0 & z-x & z-x \\ z & z^2 & x y\end{array}\right|$</p>
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<li>
<p>$= (x-y)(y-z)(z-x)\\ =(x-y)(y-z)(z-x) |B|\left|\begin{array}{ccc}1 & x+y & -z \\ 0 & 1 & 1 \\ z & z^2 & x y\end{array}\right|$</p>
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<footer style = "font-size: 18px">Problem 4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem 5 </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-10"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-7"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>$  |B|=1 \cdot\left(x y-z^2\right)+z(x+y-(-z))$</p>
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<p>$   =\left(x y-z^2\right)+(x+y+z) z$</p>
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<p>$  =x y-z^2+x z+y z+z^2$</p>
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<p>$  =x y+y z+z x $</p>
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<p>$ \therefore|A| =(x-y)(y-z)(z-x)(x y+y z+z x)$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 5 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-11"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-5-primary"><primary style="font-size:24px"> Problem 5 </primary></h3>
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<li>
<p>$  A \cdot\left(\begin{array}{ccc}a^2+1 & a b & a c \\ a b & b^2+1 & b c \\ c a & c b & c^2+1\end{array}\right)$</p>
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<p>Soln.   $ \therefore|A|=\frac{1}{a b c}\left|\begin{array}{ccc}a\left(a^2+1\right) & a^2 b & a^2 c \\ a b^2 & b\left(b^2+1\right) & b^2 c \\ c^2 a & c^2 b & c\left(c^2+1\right)\end{array}\right|$</p>
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<p>$ \therefore|A|=\frac{a b c}{a b c}\left|\begin{array}{lll}a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{array}\right|$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 5 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-12"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-8"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>$ \therefore  |A|=\left|\begin{array}{ccc}a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{array}\right|$</p>
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<p>$ \therefore  R_1 \leftarrow R_2+R_3+R_1$</p>
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<p>$  |A|=\left|\begin{array}{ccc}a^2+b^2+c^2+1 & a^2+b^2+c^2+1 & a^2+b^2+1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{array}\right|$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem 6 </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-14"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-6-primary"><primary style="font-size:24px"> Problem 6 </primary></h3>
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<p>Find the determinant of</p>
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<p>$ A= \left(\begin{array}{ccc} n & ^np_n & n_{C_n} \\ n+1 & ^{n+1}P_{n+1} & ^{n+1}C_{n+1} \\ n+2 & ^{n+2}P_{n+2} & ^{n+2}C_{n+2}\end{array}\right)$</p>
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<p>Soln.   $ \therefore A=\left(\begin{array}{ccc}n & n ! & 1 \\ n+1 & (n+1) ! & 1 \\ n+2 & (n+2) ! & 1\end{array}\right)$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 6 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-15"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-10"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>$ |A|=n !\left|\begin{array}{ccc}n & 1 & 1 \\ n+1 & (n+1) & 1 \\ n+2 & (n+1)(n+2) & 1\end{array}\right|$</p>
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<p>$  R_3=R_3-R_2$</p>
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<li>
<p>$ |A|=n !\left|\begin{array}{ccc}n & 1 & 1 \\ n+1 & n+1 & 1 \\ 1 & (n+1)^2 & 0\end{array}\right|$</p>
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<p>$  R_2=R_2-R_1$</p>
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<li>
<p>$ |A|=n !\left|\begin{array}{ccc}n & 1 & 1 \\ 1 & n & 0 \\ 1 & (n+1)^2 & 0\end{array}\right|$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-16"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-11"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
<li>
<p>$\therefore|A|=n !(-1)^{1+3}\left((n+1)^2-n\right)$</p>
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<li>
<p>$=n !\left(n^2+n+1\right)$</p>
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<li>
<p>Verification: $\quad n=2$</p>
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<li>
<p>$A=\left|\begin{array}{lll}2 & 2 ! & 1 \\ 3 & 3 ! & 1 \\ 4 & 4 ! & 1\end{array}\right|$</p>
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<li>
<p>$\therefore|A|=\left|\begin{array}{lll}2 & 2 & 1 \\ 3 & 6 & 1 \\ 4 & 24 & 1\end{array}\right|$</p>
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<footer style = "font-size: 18px">Problem 6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-8-success-17"><Success style="font-size:25px"> Determinants L-8 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-12"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>$|A|=\left|\begin{array}{ccc}2 & 2 & 1 \\ 3 & 6 & 1 \\ 1 & 18 & 0\end{array}\right|$</p>
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<li>
<p>$ R_3=R_3-R_2$</p>
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<li>
<p>$=\left|\begin{array}{ccc}2 & 2 & 1 \\ 1 & 4 & 0 \\ 1 & 18 & 0\end{array}\right|$</p>
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<li>
<p>$ R_2=R_2-R_1$</p>
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<li>
<p>$=|A|=1\cdot(18-4)=14$</p>
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<li>
<p>We got for general $n$</p>
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<li>
<p>$|A|=n !\left(n^2+n+1\right) \text {. }$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>