mathematics-class-12-unit-03-chapter-11-matrix-and-determinant-l-5-5-24nnzmodx1m

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-5-success-1"><Success style="font-size:25px"> Matrix And Determinant L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-primary"><primary style="font-size:24px"> Problem </primary></h3>
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<div style='float: left; width:100%; font-size: 30px; text-align: left;'>
<ul>
<li>
<p>let $S$ be the set of all column matrices $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ such that $b_1, b_2, b_3 \in \mathbb{R}$, and the system of equations (in real variables)</p>
</li>
<li>
<p>$
\begin{aligned}
& -x+2 y+5 z=b_1 \\
& 2 x-4 y+3 z=b_2 \\
& x-2 y+2 z=b_3
\end{aligned}
$</p>
</li>
<li>
<p>has at least one solution. Then, which of the following systems (in real variables)?</p>
</li>
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<footer style = "font-size: 18px"> $\rightarrow$ Martix and determinant  lecture - 11 $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-5-success-4"><Success style="font-size:25px"> Matrix And Determinant L-5 </Success></h3>
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<div style='float: left; width:100%;font-size 25px; text-align: left;'>
<ul>
<li>
<p>Consider the system of linear equations</p>
</li>
<li>
<p>$ -x+2 y+5 z=b_1 $</p>
</li>
<li>
<p>$ 2 x-4 y+3 z=b_2 $</p>
</li>
<li>
<p>$ x-2 y+2 z=b_3 $</p>
</li>
<li>
<p>$  R_2 \rightarrow R_2+2 R_1, R_3 \rightarrow R_3+R_1 $</p>
</li>
<li>
<p>$ \left[\begin{array}{ccc|c}-1 & 2 & 5 & b_1 \\ 2 & -4 & 3 & b_2 \\ 1 & -2 & 2 & b_3\end{array}\right]$</p>
</li>
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<li>
<p>$ \backsim\left[\begin{array}{ccc|c}-1 & 2 & 5 & b_1 \\ 0 & 0 & 13 & b_2+2 b_1 \\ 0 & 0 & 7 & b_3+b_1\end{array}\right] \quad R_3 \rightarrow R_3-\frac{7}{13} R_2 $</p>
</li>
<li>
<p>$ \backsim\left[\begin{array}{ccc|c}-1 & 2 & 5 & b_1 \\ 0 & 0 & 13 & b_2+2 b_1 \\ 0 & 0 & 0 & b_3+b_1-\frac{7}{13}\left(b_2+2 b_1\right)\end{array}\right] $</p>
</li>
<li>
<p>$ \left[\begin{array}{ccc|c}-1 & 2 & 5 & b_1 \\ 0 & 0 & 13 & b_2+2 b_1 \\ 0 & 0 & 0 & \frac{13 b_3+23 b_1-7 b_2-14 b_1}{13}\end{array}\right]$</p>
</li>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Matrix And Determinant L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $=\left[\begin{array}{ccc|c}-1 & 2 & 5 & b_1 \\\0 & 0 & 13 & b_2+2 b_1 \\\0 & 0 & 0 & \frac{-b_1-7 b_2+13 b_3}{13} \end{array}\right] $

- Given that the system has of least one solution. Therefore,

- $ -b_2-7 b_2+13 b_3=0 $

- $ \Rightarrow \quad 13 b_3=b_2+7 b_2 $

- $ S =\\{{\left[\begin{array}{c}b_1 \\\ b_2 \\\ b_3\end{array}\right] \in \mathbb{R}^3} 13 b_3=b_1+7 b_2\\}$

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### <Success style="font-size:25px"> Matrix And Determinant L-5 </Success>
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- (a) $x+2 y+3 z =b_1 $

- $ 4 y+5 z  =b_2 $

- $x+2 y+6 z  =b_3$

- $ R_3 \rightarrow R_3-R_1 $

- $ {\left[\begin{array}{ccc|c}1 & 2 & 3 & b_1 \\\ 0 & 4 & 5 & b_2 \\\ 1 & 2 & 6 & b_3\end{array}\right]} $

- ${\sim\left[\begin{array}{ccc|c}1 & 2 & 3 & b_1 \\\0 & 4 & 5 & b_2 \\\0 & 0 & 3 & b_3-b_1\end{array}\right] } $

- $\operatorname{Rank}(A \mid b)=\operatorname{Rank}(A)=3 $
- $\Rightarrow$ system (1) has unique solution for all $b_1, b_2, b_3 \in \mathbb{R}$.

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

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<ul>
<li>
<p>$\left[\begin{array}{ccc|c}1 & 1 & 3 & b_1 \\ 5 & 2 & 6 & b_2 \\ -2 & -1 & -3 & b_3\end{array}\right] \sim\left[\begin{array}{ccc|c}1 & 1 & 3 & b_1 \\ 0 & -3 & -9 & b_2-5 b_2 \\ 0 & 1 & 3 & b_3+2 b_1\end{array}\right] $</p>
</li>
<li>
<p>$ \begin{aligned} & R_2 \rightarrow \ & R_2-5 R_1 ,\ & R_3 \rightarrow R_3 \ & +2 R_1\end{aligned}$</p>
</li>
</ul>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

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<div style='float: left; width:99%; font-size: 30px; text-align: left;'>
<ul>
<li>
<p>$ R_3 \rightarrow R_3+\frac{1}{3} R_2 $</p>
</li>
<li>
<p>$ \sim\left[\begin{array}{ccc|c}1 & 1 & 3 & b_1 \\0 & -3 & -9 & b_2-5 b_1 \\0 & 0 & 0 & b_3+2 b_1+\frac{1}{3}\left(b_2-5 b_1\right)\end{array}\right] $</p>
</li>
<li>
<p>$ \sim\left[\begin{array}{rrr|r}1 & 1 & 3 & b_1 \\0 & -3 & -9 & b_2-5 b_1 \\0 & 0 & 0 & \frac{b_1+b_2+3 b_3}{3}
\end{array}\right] $</p>
</li>
<li>
<p>$\Rightarrow$ For system (2) to have at least one solution for each $\left[\begin{array}{l}b_1 \ b_2 \ b_3\end{array}\right] \in S$</p>
</li>
</ul>
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<div style='float: left; width:99%; font-size: 30px; text-align: left;'>
<ul>
<li>
<p>we should have</p>
</li>
<li>
<p>$b_1+b_2+3 b_3=0 \quad \forall\left[\begin{array}{l}b_1 \\b_2 \\b_3\end{array}\right] \in S $</p>
</li>
<li>
<p>But this condition in not true for all $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] \in S$, became</p>
</li>
<li>
<p>$\left[\begin{array}{l}6 \\2 \\1\end{array}\right] \in S $</p>
</li>
<li>
<p>$ 13 b_3=b_1+7 b_2 $ but condition (3) does not hold.</p>
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### <Success style="font-size:25px"> Matrix And Determinant L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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<main>

(c) $\left.\begin{array}{lll}
 -x+2 y-5 z  =b_1 \\\\ 
 2 x-4 y+10 z  =b_2 \\\\ 
 x-2 y+5 z  =b_3
\end{array}\right\\}$ (4)

- $ \left[\begin{array}{ccc|c}-1 & 2 & -5 & b_1 \\\2 & -4 & 10 & b_2 \\\1 & -2 & 5 & b_3\end{array}\right] $

- $ \sim\left[\begin{array}{ccc|c}-1 & 2 & -5 & b_1 \\\0 & 0 & 0 & b_2+2 b_1 \\\0 & 0 & 0 & b_3+b_1\end{array}\right] $

- $ R_2 \rightarrow R_2+2 R_1, R_3 \rightarrow R_3+R_1 $

- $\Rightarrow \operatorname{rank}(A)=1$
- For (9) to have at least one solution we need $b_2+2 b_2=0 \quad 4 b_3+b_1=0$

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<div style='float: left; width:99%; font-size: 30px; text-align: left;'>
<ul>
<li>
<p>But</p>
</li>
<li>
<p>$\left[\begin{array}{l}6 \\ 1 \\ 1\end{array}\right] $</p>
</li>
<li>
<p>$ \in S $ doesn’t satisfy both these condition. Therefore system (4) will not have atleant one solution for all $ \left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] $ $\in S$</p>
</li>
<li>
<p>d)</p>
</li>
<li>
<p>$x+2 y+5 z  =b_1 $</p>
</li>
<li>
<p>$2 x+3 z  =b_2 $</p>
</li>
<li>
<p>$x+4 y-5 z  =b_3 $</p>
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</ul>
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<ul>
<li>
<p>$ R_3 \rightarrow R_3+1 / 2 R_2 $</p>
</li>
<li>
<p>$ \sim\left[\begin{array}{ccc|c}1 & 2 & 5 & b_1 \\0 & -4 & -7 & b_2-2 b_1 \\0 & 0 & -\frac{27}{2} & b_3-b_1+\frac{1}{2}(b_2-2 b_1)\end{array}\right] $</p>
</li>
<li>
<p>$\operatorname{Ran} K(A)=\operatorname{Ran} K(A \mid b)=3 \quad \forall b_1, b_2, b_3$ $\Rightarrow$ (5) has at least one solution for all $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] \in S$</p>
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<ul>
<li>
<p>$ \operatorname{Rank}(A)=\operatorname{Rank}(A \mid B) $</p>
</li>
<li>
<p>$ A=\left[\begin{array}{ccc}3 & -1 & -1 \\ -3 & 0 & 1 \\ -3 & 2 & 1\end{array}\right] $</p>
</li>
<li>
<p>$ \sim\left[\begin{array}{ccc}3 & -1 & -1 \\ 0 & -1 & 0 \\ 0 & 1 & 0\end{array}\right] $</p>
</li>
<li>
<p>$ \quad \begin{array}{l}R_3 \rightarrow R_2+R_1 \\ R_3 \rightarrow R_3+R_1\end{array} $</p>
</li>
<li>
<p>$ \sim\left[\begin{array}{ccc}3 & -1 & -1 \\ 0 & -1 & 0 \\ 0 & 0 & 0\end{array}\right] \quad R_3 \rightarrow R_3+R_2 $</p>
</li>
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<ul>
<li>
<p>$ \operatorname{Rank}(A)=2<3 $</p>
</li>
<li>
<p>$\Rightarrow$ system has in finitely many solutions</p>
</li>
<li>
<p>$ \left[\begin{array}{ccc}3 & -1 & -1 \\0 & -1 & 0 \\0 & 0 & 0\end{array}\right] $</p>
</li>
<li>
<p>$ \left[\begin{array}{l}x \\y \\z\end{array}\right] $= $\left[\begin{array}{l}0 \\0 \\0\end{array}\right] $</p>
</li>
<li>
<p>$  3 x-y-z=0, \quad y=0 $</p>
</li>
<li>
<p>$ \Rightarrow \quad z=3 x $</p>
</li>
<li>
<p>$={(\alpha, 0,3 \alpha) \mid \alpha }$ integers</p>
</li>
<li>
<p>in set of integer solution of system of equations</p>
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### <Success style="font-size:25px"> Matrix And Determinant L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- For $(\alpha, 0,3 \alpha)$ to satisfy
- $ x^2+y^2+z^2 \leq 100 $
- we have $\alpha^2+9 \alpha^2 \leq 100$
- $
\begin{aligned}
& \Rightarrow \quad 10 \alpha^2 \leq 100 \\
& \Rightarrow \quad \alpha^2 \leq 10 \\
&
\end{aligned}
$

- Given $\alpha$ is integer
- $ \Rightarrow \alpha \in\{-3,-2,-1,0,1,2,-3\} $

- The integer solutions satisfying
- $ - x^2+y^2+z^2 \leq 100 $

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<li>
<p>Consider the following linear equations</p>
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<li>
<p>$ a x+b y+c z=0 $</p>
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<li>
<p>$ b x+c y+a z=0 $</p>
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<li>
<p>$ c x+a y+b z=0 $</p>
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<li>
<p>i) If $a+b+c \neq 0$ and $a^2+b^2+c^2=a b+b c+c a$ Show that the equations represent identical planes</p>
</li>
<li>
<p>ii) If $a+b+c \neq 0, \quad a^2+b^2+c^2 \neq a b+b c+c a$ Then, show that the equations $r$ present planes meeting at a single point.</p>
</li>
<li>
<p>iii) If $a+b+c=0$ and $a^2+b^2+c^2=a b+b c+c a$,</p>
</li>
<li>
<p>show that equations represent the whole of $\mathbb{R}^3$.</p>
</li>
</ul>
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<div style='float: left; width:99%; font-size: 30px; text-align: left;'>
<ul>
<li>i)</li>
<li>$ a+b+c \neq 0 $</li>
<li>and</li>
<li>$ a^2+b^2+c^2=a b+b c+c a $</li>
<li>$ \Rightarrow  2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a=0 $</li>
<li>$\Rightarrow \quad  (a-b)^2+(b-c)^2+(c-a)^2=0 $</li>
<li>$\Rightarrow \quad  a-b=0, b-c=0, c-a=0 $</li>
<li>$\Rightarrow \quad  a=b=c \neq 0 $</li>
</ul>
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<div style='float: left; width:99%; font-size: 30px; text-align: left;'>
<ul>
<li>
<p>$\Rightarrow$ equations reperent the identical planes.</p>
</li>
<li>
<p>ii)</p>
</li>
<li>
<p>$ a+b+c \neq 0, a^2+b^2+c^2 \neq a b+b c+c a $</p>
</li>
<li>
<p>$ A=\left[\begin{array}{lll}a & b & c \\b & c & a \\c & a & b\end{array}\right] $</p>
</li>
<li>
<p>$ \operatorname{det} A=a\left(b c-a^2\right)-b\left(b^2-a c\right)+c\left(a b-c^2\right) $</p>
</li>
<li>
<p>$=3 a b c-a^3-b^3-c^3 $</p>
</li>
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### <Success style="font-size:25px"> Matrix And Determinant L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- We know that

- $ a^3+b^3+c^3=(a+b+c)(-a^2-b^2-c^2 -a b-b c-c a+3 abc) $

- $ \Rightarrow \operatorname{det} A=-(a+b+c)(-a^2-b^2-c^2 -a b-b c-c a) $

- $ \Rightarrow \operatorname{det} A \neq 0 $

- $ \Rightarrow$ system of homogeneous equations will have unique solution

- $ x=0, y=0, z=0 $

- $\Rightarrow$ equations represent planes meeting at single print $(0,0,0)$

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<ul>
<li>
<p>(iii) $ a+b+c=0 \quad 4 \quad a^2+b^2+c^2=a b+b c+c a $</p>
</li>
<li>
<p>$ \Rightarrow  a^2+b^2+c^2=a b+b c+c a $</p>
</li>
<li>
<p>gives us $a=b=c$</p>
</li>
<li>
<p>$\Rightarrow a+a+a=0 \Rightarrow a=0$</p>
</li>
<li>
<p>$\Rightarrow \quad a=b=c=0 $</p>
</li>
<li>
<p>$\Rightarrow$ Any $\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \in \mathbb{R}^3$ will be the solution of system of equations</p>
</li>
<li>
<p>$\Rightarrow$ Equations represent the whole of $ \ R^3$.</p>
</li>
</ul>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>