mathematics-class-12-unit-04-chapter-07-determinants-l-7-11-yw4wvi2mzxk

<h3 id="success-stylefont-size25px-determinants-l-7-success-1"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-4-primary"><primary style="font-size:24px"> Property 4 </primary></h3>
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<ul>
<li>
<p>If $A=\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)$</p>
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<li>
<p>Then if $B=\left(\begin{array}{lll}k a_{11} & k a_{12} & k a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)$</p>
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<li>
<p>Then $|B|=k|A|$</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Determinants lecture - 7 $\rightarrow$ <span style = "color:blue"> Property 4 </span> $\rightarrow$ Problem $\rightarrow$ Property 5 </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-2"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary"><primary style="font-size:24px"> Problem </primary></h3>
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<li>
<p>Question: What if all the elements of A are multiplied with $k$. How does it affect the determinant?</p>
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<li>
<p>Soln. $B=\left(\begin{array}{lll}k a_{11} & k a_{12} & k a_{13} \\ k a_{21} & k a_{22} & k a_{23} \\ k a_{31} & k a_{32} & k a_{33}\end{array}\right)$</p>
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<li>
<p>$ =k a_{11}\left(k a_{22} \cdot k a_{33}-k a_{23} \cdot k a_{32}\right) $</p>
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<li>
<p>$  - k a_{12}\left(k a_{21} \cdot k a_{33}\right. - k a_{31} \cdot k a_{23})$</p>
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<li>
<p>$ +k a_{13}(k a_{21}. k a_{32} $ $-ka_{22} \cdot ka_{31})$</p>
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<li>
<p>$=k^3\left(|B|=k^3|A|\right.$ if all element of A are multiple with k )</p>
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<footer style = "font-size: 18px">Determinants lecture - 7 $\rightarrow$ Property 4 $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Property 5 $\rightarrow$ Property 6 </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-3"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-5-primary"><primary style="font-size:24px"> Property 5 </primary></h3>
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<ul>
<li>
<p>If two rows or column of a matrix are identical then the determinant of the matrix in 0</p>
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<li>
<p>Consider $ A=\left(\begin{array}{lll}a & b & c \\a & b & c \\ x & y & z\end{array}\right) $</p>
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<li>
<p>$ |A|=a(b z-c y)-b(a z-c x) +c(a y-b x) $</p>
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<li>
<p>$=a b z-a c y-a b z+b c x+a c y-b c x =0 $</p>
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<footer style = "font-size: 18px">Property 4 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Property 5 </span> $\rightarrow$ Property 6 $\rightarrow$ Property 6 </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-4"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-6-primary"><primary style="font-size:24px"> Property 6 </primary></h3>
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<ul>
<li>
<p>Suppose $A$ is a given matrix. And we obtain a new matrix $B$ by interchanging two rows (or two column) then $|B|=-|A|$</p>
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<li>
<p>Ex $\left(\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & k\end{array}\right)$</p>
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<li>
<p>$ |A|=a(e k-f h)-b(d k-f g) =c(d h-e g) $</p>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Property 5 $\rightarrow$ <span style = "color:blue"> Property 6 </span> $\rightarrow$ Property 6 $\rightarrow$ Property 7 </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-5"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-6-primary-1"><primary style="font-size:24px"> Property 6 </primary></h3>
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<ul>
<li>
<p>Let B=$\left(\begin{array}{lll}g & h & k \\ d & e & f \\ a & b & f\end{array}\right)$</p>
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<li>
<p>$\therefore|B|={g(e c}-b f) -h(d e-a f) +k(d b-a e)$</p>
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<li>
<p>$ \therefore |B| = - |A|$</p>
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<li>
<p>Ex. $\left(\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & k\end{array}\right) $</p>
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<li>
<p>$|A|=a(e k-f h)-b(d k-f g)+ e(d h - e g)$</p>
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<footer style = "font-size: 18px">Property 5 $\rightarrow$ Property 6 $\rightarrow$ <span style = "color:blue"> Property 6 </span> $\rightarrow$ Property 7 $\rightarrow$ Property 7 </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-6"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-7-primary"><primary style="font-size:24px"> Property 7 </primary></h3>
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<li>
<p>$|A \cdot B|=|A||B| $</p>
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<li>
<p>Illustration Let $ A=\left(\begin{array}{ll}a & b \\c & d\end{array}\right)$ &  $B=\left(\begin{array}{ll}m & n \\p & q\end{array}\right) $</p>
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<li>
<p>$\therefore |A|=a d-b c$  & $|B|=m q-n p $</p>
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<p>Now  $A B=\left(\begin{array}{ll}a m+b p & a x+b q \\c m+d p & c x+d q\end{array}\right) $</p>
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<li>
<p>$\therefore |A B|=(a m+b p)(c x+d q)-(c m+d p)(a x + bq)$</p>
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<li>
<p>$=a m c x+b p c x+a m d q + b p d q -a x c m  - a x d p - b c m q - b d p  q $</p>
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<footer style = "font-size: 18px">Property 6 $\rightarrow$ Property 6 $\rightarrow$ <span style = "color:blue"> Property 7 </span> $\rightarrow$ Property 7 $\rightarrow$ Property 8 </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-7"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-7-primary-1"><primary style="font-size:24px"> Property 7 </primary></h3>
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<li>
<p>Let us now consider</p>
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<li>
<p>$|A| \cdot|B|=(a d-b c)(m q-n p) $</p>
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<li>
<p>$=a d m q-b c m q-a d n p+b c n p $</p>
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<footer style = "font-size: 18px">Property 6 $\rightarrow$ Property 7 $\rightarrow$ <span style = "color:blue"> Property 7 </span> $\rightarrow$ Property 8 $\rightarrow$ Property 8 </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-9"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-8-primary-1"><primary style="font-size:24px"> Property 8 </primary></h3>
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<ul>
<li>
<p>Illustration</p>
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<li>
<p>consider $A=\left|\begin{array}{ccc}a+k & b+m & c+n \\ d & e & f \\ x & y & z\end{array}\right|$</p>
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<li>
<p>Claim $|A|$</p>
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<li>
<p>$=\left|\begin{array}{lll}a & b & c \\d & e & f \\x & y & z\end{array}\right|+\left|\begin{array}{lll}k & m & n \\d & e & f \\x & y & z\end{array}\right|$</p>
</li>
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<footer style = "font-size: 18px">Property 7 $\rightarrow$ Property 8 $\rightarrow$ <span style = "color:blue"> Property 8 </span> $\rightarrow$ Property 8 $\rightarrow$ Problem </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-10"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-property-8-primary-2"><primary style="font-size:24px"> Property 8 </primary></h3>
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<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>$|A|=(a+k)\left|\begin{array}{ll}e & f \\ y & z\end{array}\right|-(b+m)\left|\begin{array}{ll}d & f \\ x & z\end{array}\right|$</p>
</li>
<li>
<p>$+(c+n)\left|\begin{array}{ll}d & e \\ x & y\end{array}\right|$</p>
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<li>
<p>$=a \left|\begin{array}{ll}e & f \\ y & z\end{array}\right|$ $-b\left|\begin{array}{ll}d & f \\ x & z\end{array}\right|$</p>
</li>
<li>
<p>$+c\left|\begin{array}{ll}d & e \\ x & y\end{array}\right|$</p>
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<li>
<p>$=k \left|\begin{array}{ll}e & f \\ y & z\end{array}\right|$ $-m\left|\begin{array}{ll}d & f \\ x & z\end{array}\right|$</p>
</li>
<li>
<p>$+n\left|\begin{array}{ll}d & e \\ x & y\end{array}\right|$</p>
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<li>
<p>$=\left|\begin{array}{lll}a & b & c \\ d & e & f \\ x & y & z\end{array}\right|+\left|\begin{array}{lll}k & m & n \\ d & n & n \\ x & y & z\end{array}\right|$</p>
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<footer style = "font-size: 18px">Property 8 $\rightarrow$ Property 8 $\rightarrow$ <span style = "color:blue"> Property 8 </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-11"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary-1"><primary style="font-size:24px"> Problem </primary></h3>
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<ul>
<li>
<p>Find the determinant of</p>
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<li>
<p>$\left(\begin{array}{lll}1 & a & a^2 \\1 & b & b^2 \\1 & c & c^2\end{array}\right)$</p>
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<li>
<p>$\therefore|A|=1\left(b c^2-c b^2\right)-a(c^2-b^2) +a^2(c-b) $</p>
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<li>
<p>Soln. $=b c(c-b)-a(c-b)(c+b) +a^2(c-b)  $</p>
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<li>
<p>$ =(c-b)\left(b c-a c-a b+a^2\right) $</p>
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<li>
<p>$=(c-b)(c(b-a)-a(b-a)$</p>
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<li>
<p>$=(c-b)(b-a)(c-a)=((a-b)(b-c)(c-a) $</p>
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<footer style = "font-size: 18px">Property 8 $\rightarrow$ Property 8 $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-12"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary-2"><primary style="font-size:24px"> Problem </primary></h3>
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<ul>
<li>
<p>If in a matrix the $i^{th}$ rows in replaced by the sum of $i^{th}$ rite $j^{th}$ row then the determinant does not change</p>
</li>
<li>
<p>Illustration -</p>
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<li>
<p>$\text { consider } A=\left(\begin{array}{lll}a & b & c \\m & N & p \\x & y & z\end{array}\right)$</p>
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<p>Soln. let $B= R_1\leftrightarrow R_1+R_2\left(\begin{array}{ccc}a+m & b+n & c+p \\m & n & p \\x & y & z\end{array}\right)$</p>
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<footer style = "font-size: 18px">Property 8 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-13"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary-3"><primary style="font-size:24px"> Problem </primary></h3>
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<ul>
<li>
<p>Then $|B|=|A|$</p>
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<p>$\ |B|=\left|\begin{array}{lll}a & b & c \\m & n & p \\x & y & z\end{array}\right|+\left|\begin{array}{ccc}m & n & p \\m & n & p \\x & y & z\end{array}\right| $</p>
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<li>
<p>$=|A|+0$</p>
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<li>
<p>$\therefore|B|=|A|$ Since two rows are identical</p>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-14"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<p>More generalization</p>
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<li>
<p>If $R_i \longleftrightarrow R_i+k R_j$</p>
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<li>
<p>Then also $|B|=|A|$</p>
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<p>or If $B=\left|\begin{array}{ccc}a+k m & b+k n & c+k p \\ m & n & b \\ x & y & z\end{array}\right|$</p>
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<p>Then $|B|=|A|$</p>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-15"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
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<li>
<p>This is because</p>
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<li>
<p>$ \left|\begin{array}{ccc}a+k m & b+k n & c+k p \\ m & n & p \\ x & y & z   \end{array}\right| $</p>
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<li>
<p>$ =\left|\begin{array}{lll}a & b & c \\m & n & p \\x & y & z\end{array}\right| +\left|\begin{array}{ccc}k m & k n & k p \\m & n & p \\x & y & z\end{array}\right| $</p>
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<li>
<p>$ =|A|+k\left|\begin{array}{lll}m & n & p \\m & n & p \\x & y & z\end{array}\right| $</p>
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<p>$ \therefore|B|=|A|$</p>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-16"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary-4"><primary style="font-size:24px"> Problem </primary></h3>
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<ul>
<li>
<p>Find the determinant of</p>
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<li>
<p>$ \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| $</p>
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<p>Soln. = $ \left|\begin{array}{lll} 1 & b c & a b+a c \\1 & c a & b c+b a \\1 & a b & c a+c b\end{array}\right| $</p>
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<li>
<p>= $ \left|\begin{array}{lll} 1 & b c & b c+a b+a c \\1 & c a & c a+b c+b a \\1 & a b & a b+c a+c b\end{array}\right|$</p>
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<li>
<p>$c_3 \leftarrow c_2+c_3 $</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Problem </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-17"><Success style="font-size:25px"> Determinants L-7 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary-5"><primary style="font-size:24px"> Problem </primary></h3>
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<ul>
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<p>Find the determinants of</p>
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<li>
<p>$ \left|\begin{array}{lll}x+y+2z & x & y \\z & y+z+2 x & y \\z & x & z+x+2 y\end{array}\right| $</p>
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<li>
<p>Soln. $ S1:  C_1 \leftarrow C_1+C_2 $</p>
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<li>
<p>$ \left|\begin{array}{lll}2 x+y+2 z & x & y \\2 x+y+2 z & y+z+2 x & y \\x+z & x & z+x+2 y\end{array}\right| $</p>
</li>
<li>
<p>$  S2:  c_1 \leftarrow C_1+C_3 $</p>
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<li>
<p>$ \left|\begin{array}{lll}2 x+2 y+2 z & x & y \\2 x+2 y+2 z & y+z+2 x & y \\2x+2 z+2 y & x & z+x+2 y\end{array}\right| $</p>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Problem </footer>

<h3 id="success-stylefont-size25px-determinants-l-7-success-18"><Success style="font-size:25px"> Determinants L-7 </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>$ = (2x+2y+2z)\left|\begin{array}{lll}1 & x & y \\1 & y+z+2x & y \\1 & x & z+x+2 y\end{array}\right| $</p>
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<p>$ R_2 \leftarrow R_2-R_1 $</p>
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<li>
<p>$ 2(x+y+z)\left|\begin{array}{lll}1 & x & y \\0 & y+z+x & 0 \\1 & x & z+x+2 y\end{array}\right| $</p>
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<li>
<p>$R_3 \longleftarrow R_3-R_1 \quad \therefore$ Determinant</p>
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<li>
<p>$ 2(x+y+z)\left|\begin{array}{lll}1 & x & y \\0 & x+y+z & 0 \\0 & 0 & x+y+ z\end{array}\right| $</p>
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<li>
<p>$ 2(x+y+z)^3 $</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Determinants L-7 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- What is the determinant of

- $ \left|\begin{array}{lll}2 & 7 & 65 \\\3 & 8 & 75 \\\5 & 9 & 86\end{array}\right| $

- Soln. $ =  \left|\begin{array}{lll}2 & 7 & 9 \times 7+2 \\\3 & 8 & 9 \times 8+3 \\\5 & 9 & 9 \times 9+5\end{array}\right| $

- $ =  \left|\begin{array}{lll}2 & 7 & 9 \times 7 \\\3 & 8 & 9 \times 8 \\\5 & 9 & 9 \times 9 \end{array}\right|+\left|\begin{array}{lll}2 & 7 & 2 \\\3 & 8 & 3 \\\5 & 9 & 5\end{array}\right| $

- $ \therefore$ Answer is 0

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Property 7