mathematics-class-12-unit-03-chapter-10-matrix-and-determinant-l-4-5-i1mln-sr5jc

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-1-primary"><primary style="font-size:24px"> Problem-1 </primary></h3>
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<p>Let $P_1=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$</p>
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<p>$P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 4\end{array}\right], P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]$</p>
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<p>$P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right], P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$. Then</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Matrix and determinant lecture - 10 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-1 contd... $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-1"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-1-contd-primary"><primary style="font-size:24px"> Problem-1 contd… </primary></h3>
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<p>$\text { and } \quad X=\sum_{k=1}^6 P_k\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^T$</p>
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<p>Then,</p>
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<p>a) Ib $X\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha=30$</p>
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<p>b) X is symmetric matrix</p>
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<p>c) X - 30 I is not an invertible matrix</p>
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<footer style = "font-size: 18px">Matrix and determinant lecture - 10 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Problem-1 contd... </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Matrix And Determinant L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $B=\left[\begin{array}{lll}2 & 1 & 3 \\\  1 & 0 & 2 \\\ 3 & 2 & 1\end{array}\right], B^T=B$
- $X=\sum_{k=1}^6 P_k B P_k^T, X\left[\begin{array}{l}1 \\\  1 \\\ 1\end{array}\right]=\sum_{k=1}^6 P_k B P_k^T\left[\begin{array}{l}1 \\\  1 \\\ 1\end{array}\right]$
- $ =\sum_{k=1}^6 P_k B \left[\begin{array}{l}1 \\\  1 \\\ 1\end{array}\right] =\sum_{k=1}^6 P_k\left[\begin{array}{l}6 \\\  3 \\\ 6\end{array}\right]$
- $\because P_k^T\left[\begin{array}{l}1 \\\  1 \\\  1\end{array}\right]=\left[\begin{array}{l}1 \\\  1 \\\  1\end{array}\right], k = 1, 2, ....., 6 $

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

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-2"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$X\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right] = (P_1+P_2+P_3+P_4+P_5+P_6)\left[\begin{array}{l}6 \\  3 \\  6\end{array}\right]$</p>
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<p>$=2\left[\begin{array}{lll}1 & 1 & 1 \\  1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right] \left[\begin{array}{l}6 \\  3 \\  6\end{array}\right]$</p>
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<p>$=2\left[\begin{array}{l}15 \\  15 \\  15\end{array}\right] =30\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right]$</p>
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<p>$\Rightarrow X\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right] =30\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right]$</p>
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<footer style = "font-size: 18px">Problem-1 contd... $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-3"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$\Rightarrow \left[\begin{array}{l}1 \\  1 \\  1\end{array}\right] =30\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right]$</p>
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<p>$(30)\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right]=0$</p>
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<p>$\Rightarrow X = 30$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-4"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-2"><primary style="font-size:24px"> Solution </primary></h3>
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<p>c) (X - 30 I) is not an invertible matrix</p>
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<p>from part we have</p>
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<p>$X\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right] =30\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right]$</p>
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<p>$\Rightarrow (X-30I)\left[\begin{array}{l}1 \\  1 \\  1\end{array}\right] =\left[\begin{array}{l}0 \\  0 \\  0\end{array}\right]$</p>
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<p>$\Rightarrow \left[\begin{array}{l}1 \\  1 \\  1\end{array}\right]$ is non-trivial solution of (X - 30I)y = 0</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ System of linear equations </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-5"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-3"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$\Rightarrow$ X - 30 I is not invertible because if X - 30 I is invertible then (X - 30 I) y = 0</p>
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<p>has only zero solution.</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ System of linear equations $\rightarrow$ System of linear equations </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-6"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-system-of-linear-equations-primary"><primary style="font-size:24px"> System of linear equations </primary></h3>
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<p>Let A be an $n \times n$ matrix, X be an $n \times 1$ vector, and b be an $n \times 1$ vector. Then system of $n$ linear equations in variable</p>
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<p>$x=\left[\begin{array}{c}x_1 \\  x_2 \ \vdots \\  x_n\end{array}\right]$</p>
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<p>can be written as Ax = b  _____ (1)</p>
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<p>Any $x$ which satisfies (1) is $\wedge$ aid to be solution of system of Linear equations.</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> System of linear equations </span> $\rightarrow$ System of linear equations $\rightarrow$ System of linear equations </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-7"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-system-of-linear-equations-primary-1"><primary style="font-size:24px"> System of linear equations </primary></h3>
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<p>A system of linear equations can have</p>
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<p>i) Unique solution</p>
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<p>ii) Infinitely many solutions</p>
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<p>iii) No solution</p>
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<p><strong>Example</strong></p>
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<p>i) $x_1+2 x_2=4$</p>
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<p>$2 x_1+x_2 =2–(1)$</p>
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<p>$x_1=0, x_2=2$ is a unique solution of (1)</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ System of linear equations $\rightarrow$ <span style = "color:blue"> System of linear equations </span> $\rightarrow$ System of linear equations $\rightarrow$ System of linear equations </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-8"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-system-of-linear-equations-primary-2"><primary style="font-size:24px"> System of linear equations </primary></h3>
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<p>ii) $x_1+ x_2=2$</p>
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<p>$2x_1+2x_2 = 4$—(2)</p>
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<p>It is easy to we that second equation can be obtained by multiplying first equation by 2.</p>
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<p>$(\alpha, 2-\alpha)$ where $\alpha \in R$ is solution of (2).</p>
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<p>That is, (2) has infinitely many solutions.</p>
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<footer style = "font-size: 18px">System of linear equations $\rightarrow$ System of linear equations $\rightarrow$ <span style = "color:blue"> System of linear equations </span> $\rightarrow$ System of linear equations $\rightarrow$ Remarks </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-9"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-system-of-linear-equations-primary-3"><primary style="font-size:24px"> System of linear equations </primary></h3>
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<p>iii) $x_1+ x_2=2$</p>
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<p>$2 x_1+2x_2 =3$–(3)</p>
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<p>This system has no solution.</p>
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<p>The question is – How do we decide about the solution of a system of linear equation</p>
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<p>Ax = b</p>
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<p>$a \rightarrow n \times n$</p>
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<p>$x \rightarrow n \times 1$</p>
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<p>$b \rightarrow n \times 1$</p>
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<footer style = "font-size: 18px">System of linear equations $\rightarrow$ System of linear equations $\rightarrow$ <span style = "color:blue"> System of linear equations </span> $\rightarrow$ Remarks $\rightarrow$ Remarks </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-10"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-remarks-primary"><primary style="font-size:24px"> Remarks </primary></h3>
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<li>If rank(A) = rank (A | b) = n.</li>
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<p>Then system of equations has unique solution.</p>
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<p>In this case $\operatorname{det} A \neq 0$</p>
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<li>If rank (A) = rank (A | b)</li>
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<p>= m < n</p>
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<p>Then, system of equations has infinitely many solutions.</p>
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<footer style = "font-size: 18px">System of linear equations $\rightarrow$ System of linear equations $\rightarrow$ <span style = "color:blue"> Remarks </span> $\rightarrow$ Remarks $\rightarrow$ Example </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-11"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-2-primary"><primary style="font-size:24px"> Problem-2 </primary></h3>
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<p>Let $\alpha, \lambda, \mu \in \mathbb{R}$. Consider the system of linear equations</p>
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<p>$\alpha x+2 y=\lambda$</p>
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<p>$3 x-2 y=\mu$</p>
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<p>For what values of $\alpha, \lambda, \mu$ the system of linear equation’s have</p>
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<p>a) Unique solution</p>
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<p>b) In finitely many solutions</p>
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<p>c) No solution</p>
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<footer style = "font-size: 18px">Remarks $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-12"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-4"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$A=\left[\begin{array}{cc}\alpha & 2 \\  3 & -2\end{array}\right], b=\left[\begin{array}{l}\lambda \\  \mu\end{array}\right]$</p>
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<p>a) For unique solution</p>
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<p>det A $\neq$ 0</p>
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<p>rank (A | b) = rank (A) = 2</p>
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<p>det A = $-2\alpha - b \neq$ 0</p>
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<p>$\Rightarrow \alpha \neq -3$</p>
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<footer style = "font-size: 18px">Example $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-13"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-5"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$\Rightarrow$ For $\alpha \neq-3$ and $\lambda, \mu \in \mathbb{R}$ systems will have unique solution</p>
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<p>$b) A=\left[\begin{array}{cc}\alpha & 2 \\  3 & -2\end{array}\right], b=\left[\begin{array}{l}\lambda \\  \mu\end{array}\right]$</p>
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<p>$[A | b]=\left[\begin{array}{cc}\alpha & 2 \\  3 & -2\end{array}\right] \left[\begin{array}{l}\lambda \\  \mu\end{array}\right]$</p>
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<p>$~\left[\begin{array}{cc}\alpha & 2 \\  3+\alpha & 0\end{array}\right], \left[\begin{array}{l}\lambda \\  \lambda+\mu\end{array}\right]$</p>
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<p>$R_2 \rightarrow R_2+R_1$</p>
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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-14"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-6"><primary style="font-size:24px"> Solution </primary></h3>
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<p>For infinitely many solution</p>
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<p>rank (A | b) = rank (A) < 2</p>
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<p>If $\alpha$ + 3 = 0 and $\lambda + \mu = 0$</p>
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<p>ie $\alpha=-3 \quad 4 \quad \lambda=-\mu$</p>
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<p>Then rank (A | b) =  rank (A) = 1</p>
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<p>$\Rightarrow$ system of equations will have infinitely many solutions</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-15"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-7"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$[A \mid b] \sim\left[\begin{array}{cc|c}\alpha & 2 & 1 \\  3+\alpha & 0 & \lambda+\mu\end{array}\right]$</p>
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<p>If $\alpha=-3 ~ and ~ \lambda \neq-\mu$</p>
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<p>$\Rightarrow$ rank (A | b) = 2</p>
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<p>rank (A) = 1</p>
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<p>i.e. rank (A) $\neq$ rank (A | b)</p>
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<p>$\Rightarrow$ systam has no solution.</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-16"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-3-primary"><primary style="font-size:24px"> Problem-3 </primary></h3>
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<p>For a real number $\alpha$ of the system</p>
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<p>$\left[\begin{array}{ccc}1 & \alpha & \alpha^2 \\  \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}1 \\ -1 \\ 1\end{array}\right]$</p>
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<p>of linear equations has infinitely many solutions, then what will be the value of $1+\alpha+\alpha^2$ ?</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Matrix And Determinant L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Consider the augmented matrix

- $\left[\begin{array}{lll}A & \mid & b\end{array}\right]=\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\\  \alpha & 1 & \alpha & -1 \\\ \alpha^2 & \alpha & 1 & 1\end{array}\right]$

- $\sim\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\\  0 & 1-\alpha^2 & \alpha-\alpha^3 & -1-\alpha \\\ 0 & \alpha-\alpha^3 & 1-\alpha^4 & 1-\alpha^2\end{array}\right]$

- $\sim\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\\  0 & 1-\alpha^2 & \alpha-\alpha^3 & -1-\alpha \\\ 0 & 0 & 1-\alpha^2 & 1+\alpha\end{array}\right]$

- $R_2 \rightarrow R_2-\alpha R_1$

- $R_3 \rightarrow R_3-\alpha^2 R_1$

- $R_3 \rightarrow R_3-\alpha R_2$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-17"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-8"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>$\left[\begin{array}{ll}A |b\end{array}\right]=\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\ 0 & 1-\alpha^2 & \alpha-\alpha^3 & -1-\alpha \\ 0 & 0 & 1-\alpha^2 & 1+\alpha\end{array}\right]$</p>
</li>
<li>
<p>Since the system has infinitely many solutions</p>
</li>
<li>
<p>$\Rightarrow$ rank (A | b) = rank (A) < 3</p>
</li>
<li>
<p>If 1 - $\alpha^2 = 0$ and  1 + $\alpha = 0$</p>
</li>
<li>
<p>$\Rightarrow \alpha = -1$</p>
</li>
<li>
<p>$\Rightarrow 1+\alpha+\alpha^2 = 1$</p>
</li>
<li>
<p>$\Rightarrow$ rank (A | b) = rank k(A) = 2 < 3</p>
</li>
<li>
<p>$\Rightarrow$ for $\alpha$ = -1 system has infinite solutions.</p>
</li>
</ul>
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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-18"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-4-primary"><primary style="font-size:24px"> Problem-4 </primary></h3>
<hr>
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<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>x - 2y + 3z = -1</p>
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<li>
<p>-x + y - 2z = K</p>
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<li>
<p>x - 3y + 4z = 1</p>
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<li>
<p>For what values of K system of equations has no solution.</p>
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<li>
<p>Soln. $[A \mid b]=\left[\begin{array}{ccc|c}1 & -2 & 3 & -1 \\ -1 & 1 & -2 & k \\ 1 & -3 & 4 & 1\end{array}\right]$</p>
</li>
<li>
<p>$R_2 \rightarrow R_2+R_1, R_3 \rightarrow R_3-R_1$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>

<h3 id="success-stylefont-size25px-matrix-and-determinant-l-4-success-19"><Success style="font-size:25px"> Matrix And Determinant L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-9"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>$\sim\left[\begin{array}{ccc|c}1 & -2 & 3 & -1 \\ 0 & -1 & 1 & k-1 \\ 0 & -1 & 1 & 2\end{array}\right] R_3 \rightarrow R_3-R_2$</p>
</li>
<li>
<p>$\sim\left[\begin{array}{ccc|c}1 & -2 & 3 & -1 \\ 0 & -1 & 1 & k-1 \\ 0 & 0 & 0 & 3-k\end{array}\right]$</p>
</li>
<li>
<p>rank (A) = 2, For system to have ho solution we need rank (A | b) = 3</p>
</li>
<li>
<p>$\Rightarrow k \neq 3$, rank (A | b) = 3</p>
</li>
<li>
<p>$\Rightarrow$ For $k \neq 3$ system will have no solution.</p>
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</li>
</ul>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>