mathematics-class-12-unit-04-chapter-05-determinants-l-5-11-6zctdb9lpdq

<h3 id="success-stylefont-size25px-determinants-l-5-success-1"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-example-1-primary"><primary style="font-size:24px"> Example-1 </primary></h3>
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<p>Example 1: Show that $\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$</p>
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<p>$L H S=\left|\begin{array}{lll}1 & b c & a b+c a \\ 1 & c a & b c+a b \\ 1 & a b & c a+b c\end{array}\right| ~ \text{apply} ~ c_3 \rightarrow c_3+c_2$</p>
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<p>$  \Rightarrow \left|\begin{array}{lll}1 & b c & a b+b c+c a \\ 1 & c a & a b+b c+c a \\ 1 & a b & a b+b c+c a\end{array}\right| $</p>
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<p>$=(a b+b c+c a)\left|\begin{array}{lll}1 & bc & 1 \\ 1 & ca & 1 \\1 & ab & 1\end{array}\right|$</p>
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<p>$ =(a b+b c+ca)0 =0 \rightarrow$  because $C_ 1$ and $C_3$ are identical</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Determinants lecture - 5 $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-3 </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-2"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-example-2-primary"><primary style="font-size:24px"> Example-2 </primary></h3>
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<p>Example: Show that</p>
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<p>$\left|\begin{array}{lll}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{array}\right| = (a-b)(b-c)(c-a) LHS \quad \quad = \left|\begin{array}{lll}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{array}\right|$ apply</p>
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<p>$R_2 \rightarrow R_2-R_1$
,$R_3 \rightarrow R_3-R_1$</p>
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<p>$=\left|\begin{array}{ccc}1 & a & b c \\ 0 & b-a & ca-bc \\ 0 & c-a & ab-bc\end{array}\right| \quad \quad =\left|\begin{array}{ccc}1 & a & b c \\ 0 & b-a & ca-bc \\ 0 & c-a & ab-bc\end{array}\right|$</p>
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<p>$= (b-a)(c-a)$ $\left|\begin{array}{ccc}1 & a & b c \\ 0 & 1 & -c \\ 0 & 1 & -b\end{array}\right| \quad \quad = $(b-a)(c-a)$  \left|\begin{array}{cc}1 & -c \\ 1 & -b\end{array}\right|$</p>
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<p>$= (b-a)(c-a)(-b(-c))=(b-a)(c-a)(-b+c)= (a-b)(b-c)(c-a)= $RHS</p>
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<footer style = "font-size: 18px">Determinants lecture - 5 $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-3 $\rightarrow$ Example-3 </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-3"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-example-3-primary"><primary style="font-size:24px"> Example-3 </primary></h3>
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<p>Example. A triangle with vertices at</p>
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<p>$\left(x_1, y_1\right),\left({x}_2, y_2\right),\left(x_3, y_3\right)$ has area equal to</p>
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<p>$\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$.</p>
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<p>Show that this expression can be obtained by evaluating the determinant</p>
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<p>$\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right| \text {. }$</p>
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<p>$\therefore$ we have to show that</p>
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<p>$\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$</p>
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<p>=$\frac{1}{2}$ $[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3 (y_1 - y_2)]$</p>
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<footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Example-3 $\rightarrow$ Example-4 </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-4"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-example-3-primary-1"><primary style="font-size:24px"> Example-3 </primary></h3>
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<p>$\text { LHS }=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$</p>
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<p>$R_2 \rightarrow R_2-R_1 , R_3 \rightarrow R_3-R_1$</p>
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<p>$=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2-x_1 & y_2-y_1 & 0 \\ x_3-x_1 & y_3-y_1 & 0\end{array}\right|$</p>
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<p>$=\frac{1}{2}\left|\begin{array}{ll}x_2-x_1 & y_2-y_1 \\ x_3-x_1 & y_3-y_1\end{array}\right|$</p>
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<p>$=\frac{1}{2}\left(\left(x_2-x_1\right)\left(y_3-y_1\right)-\left(x_3-x_1\right)\left(y_2-y_1\right)\right)$</p>
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<p>= $\frac{1}{2}\left[x_2\left(y_3-y_1\right)-x_1\left(y_3-y_1\right)\right.$</p>
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<p>$\left.-x_3\left(y_2-y_1\right)+x_1\left(y_2-y_1\right)\right]=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] $ = RHS</p>
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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-3 $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Example-4 $\rightarrow$ Example-4 </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-7"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-example-4-primary-2"><primary style="font-size:24px"> Example-4 </primary></h3>
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<p>$\alpha=-1, \beta=2$</p>
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<p>$=\left|\begin{array}{ccc}3 & 2 & 6 \\ 2 & 6 & 8 \\ 6 & 8 & 18\end{array}\right|$</p>
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<p>$= 3(6 \times 18-64)$-2(36-48)+6(16-36) = 3(108-64)+24-120</p>
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<p>$ = 3 \times 44 + 24 - 120 = 132 + 24 - 120 = 36$</p>
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<p>$ k(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$</p>
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<p>$(\alpha=-1, \beta=2)$</p>
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<p>$=k \cdot 2^2(-1)^2(-3)^2$</p>
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<p>$=k \times 4 \times 1 \times 9 =36 k$</p>
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<p>LHS=36</p>
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<p>RHS =36 k $\Rightarrow k=1$</p>
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<footer style = "font-size: 18px">Example-4 $\rightarrow$ Example-4 $\rightarrow$ <span style = "color:blue"> Example-4 </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-8"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary"><primary style="font-size:24px"> Problem </primary></h3>
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<p>Find K from the equality</p>
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<p>=$\left|\begin{array}{ccc}3 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \\ 1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \\  1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4\end{array}\right|$</p>
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<p>= $k(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$</p>
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<p>Soln. LHS = $\left|\begin{array}{ccc}1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \\ 1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \\  1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4\end{array}\right|$</p>
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<footer style = "font-size: 18px">Example-4 $\rightarrow$ Example-4 $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-9"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<p>After simplifying and noting that many determinants are zero, we are left with the following 6</p>
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<p>$ =\left|\begin{array}{lll}1 & 1 & 1 \\ \alpha & \alpha^2 & \alpha^3 \\ \beta^2 & \beta^3&\beta^4\end{array}\right|+\left|\begin{array}{ccc}1 & 1 & 1 \\ \beta & \beta^2 & \beta^3 \\ \alpha^2 & \alpha^3 & \alpha^4\end{array}\right|$<br>
$+\left|\begin{array}{ccc}1 & \alpha & \alpha^2 \\ 1 & 1 & 1 \\ \beta^2 & \beta^3 & \beta^4\end{array}\right|+\left|\begin{array}{ccc}1 & \alpha & \alpha^2 \\ \beta & \beta^2 & \beta^3 \\ 1 & 1 & 1\end{array}\right|$<br>
$+\left|\begin{array}{ccc}1 & \beta & \beta^2 \\ 1 & 1 & 1 \\ \alpha^2 & \alpha^3 & \alpha^4\end{array}\right|+\left|\begin{array}{ccc}1 & \beta & \beta^2 \\ \alpha & \alpha^2 & \alpha^3 \\ 1 & 1 & 1\end{array}\right|$</p>
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<p>$ =\left|\begin{array}{ccc}1 & 1 & 1 \\ \beta & \beta^2 & \beta^3 \\ \alpha^2 & \alpha^3 & \alpha^4\end{array}\right|$</p>
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<footer style = "font-size: 18px">Example-4 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-10"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$=\alpha^2\beta\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & \beta & \beta^2 \\ 1 & \alpha & \alpha^2\end{array}\right|$</p>
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<p>$=\alpha^2 \beta\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<p>$\cdot\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \alpha^2 & \alpha^3 \\ \beta^2 & \beta^3 & \beta^4\end{array}\right|=\alpha \beta^2\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<p>$\cdot\left|\begin{array}{ccc}1 & 1 & 1 \\ \beta & \beta^2 & \beta^3 \\ \alpha^2 & \alpha^3 & \alpha^4\end{array}\right|=\alpha^2 \beta\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & \beta & \beta^2 \\ 1 & \alpha & \alpha^2\end{array}\right|=-\alpha^2 \beta\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<p>$ \left|\begin{array}{ccc}1 & \alpha & \alpha^2 \\ 1 & 1 & 1 \\ \beta^2 & \beta^3 & \beta^4\end{array}\right|=\beta^2\left|\begin{array}{ccc}1 & \alpha & \alpha^2 \\ 1 & 1 & 1 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-5-success-11"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-2"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$=\beta^2\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<p>$\left|\begin{array}{lll}1 & \alpha & \alpha^2 \\ \beta & \beta^2 & \beta^3 \\ 1 & 1 & 1\end{array}\right|=\beta\left|\begin{array}{lll}1 & \alpha & \alpha^2 \\ 1 & \beta^2 & \beta^2 \\ 1 & 1 & 1\end{array}\right| =+\beta\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<p>$\cdot\left|\begin{array}{ccc}1 & \beta & \beta^2 \\ 1 & 1 & 1 \\ \alpha^2 & \alpha^3 & \alpha^4\end{array}\right|$</p>
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<p>$ =\alpha^2\left|\begin{array}{ccc}1 & \beta & \beta^2 \\ 1 & 1 & 1 \\ 1 & \alpha & \alpha^2\end{array}\right|$</p>
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<p>$=+\alpha^2\left|\begin{array}{lll}2 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</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>

<h3 id="success-stylefont-size25px-determinants-l-5-success-12"><Success style="font-size:25px"> Determinants L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-3"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$\cdot\left|\begin{array}{ccc}1 & \beta^2 & \beta^2 \\ \alpha & \alpha^2 & \alpha^3 \\ 1 & 1 & 1\end{array}\right| =\alpha\left|\begin{array}{ccc}1 & \beta & \beta^2 \\ 1 & \alpha & \alpha^2 \\ 1 & 1 & 1\end{array}\right| =-\alpha\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<p>$\therefore\left(\alpha \beta^2-\alpha^2 \beta-\beta^2+\beta+\alpha^2-\alpha\right) x$</p>
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<p>$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right|$</p>
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<p>$=\left(\alpha \beta^2-\alpha^2 \beta-\beta^2+\alpha^2+\beta-\alpha\right)^2$</p>
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<p>$=(\alpha \beta(\beta-\alpha)-(\beta+\alpha)(\beta-\alpha)+(\beta-\alpha))^2$</p>
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<p>$=(\beta-\alpha)^2(\alpha \beta-\beta-\alpha+1)$</p>
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<p>$=(\beta-\alpha)^2(1-\alpha)^2(1-\beta)^2 $</p>
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<p>$=(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$ $\Rightarrow K=1$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>