mathematics-class-12-unit-03-chapter-06-problem-of-matrices-l-6-6-2qjrpnb9oii

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-1"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-1-primary"><primary style="font-size:24px"> Problem-1 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px;'>
<ul>
<li>
<p>Problem:</p>
</li>
<li>
<p>Let $P$ be an invertible matrix with $I+p+p^2+\ldots+p^n=0$ (i.e, the zero matrix).</p>
</li>
<li>
<p>Find $P^{-1}$ in terms of $P$.</p>
</li>
<li>
<p>Solution:</p>
</li>
<li>
<p>$I+P+p^2+\cdots+P^n=0$.</p>
</li>
<li>
<p>Multiply this equation by</p>
</li>
<li>
<p>$p^{-1}$.</p>
</li>
<li>
<p>$p^{-1}+ I +p+\cdots+p^{n-1}=0  $</p>
</li>
<li>
<p>$\Rightarrow p^{-1} =-I-p-\cdots+p^{n-1}  =-\sum_0^{n-1} p^i .$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Matrices lecture-6 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-2"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-2-primary"><primary style="font-size:24px"> Problem-2 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px;'>
<ul>
<li>
<p>Problem:</p>
</li>
<li>
<p>If $ A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$, find $A^3$.</p>
</li>
<li>
<p>Solution:</p>
</li>
<li>
<p>$ A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$
$
\begin{aligned}
& A^2=\left[\begin{array}{ccc}
1 & 2 & 1 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 1 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right] \\
&
\end{aligned}
$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px"> Matrices lecture-6 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution contd... </footer>

### <Success style="font-size:25px"> Problem Of Matrices L-6 </Success>
### <primary style="font-size:24px"> Solution contd... </primary>
<hr>
<main>

$
\begin{aligned}
& A^3=A^2 \cdot A=\left[\begin{array}{ccc}
 9 & 5 & 10 \\\\
 3 & 10 & -1 \\\\
-1 & -3 & 1
\end{array}\right]\left[\begin{array}{ccc}
 1 & 2 & 1 \\\\
 5 & 2 & 6 \\\\
-2 & -1 & -3
\end{array}\right] \\\\
& =\left[\begin{array}{ccc}
 9+25-20  &  18+10-20  &  9+30-30 \\\\
 3+50+2  &  6+20+1  &  3+60+3 \\\\
-1+15-2  &  -2-6-1  &  -1+18-3
\end{array}\right] \\
&
\end{aligned}
$
$
\begin{aligned}
& =\left[\begin{array}{ccc}
 14  &  8  &  9 \\\\
 55  &  29  &  66 \\\\
-18  &  -9  &  14
\end{array}\right] \\
&
\end{aligned}
$

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution contd... </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-4"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-3-primary"><primary style="font-size:24px"> Problem-3 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px;'>
<ul>
<li>
<p>Problem:</p>
</li>
<li>
<p>The set of natural numbers is partitioned into arrays of rows & column in the form of matrices an $ M_1=[1], M_2=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right],\\ M_3=\left[\begin{array}{lll}6 & 1 & 11 \\ 9 & 10 & 11 \\ 13 & \text { is } & 14\end{array}\right]$ and so on. Find the trace of $M_n$.</p>
</li>
<li>
<p>Solution:</p>
</li>
<li>
<p>$\{{1,2,6,15,…, t_n…}\}$</p>
</li>
<li>
<p>$ \therefore \text { Let } S_n= 1+2+6+15+\cdots+t_n $</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

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

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-5"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px;'>
<ul>
<li>
<p>$ \therefore\quad 0=S_ n-S_ n\\ $</p>
</li>
<li>
<p>$=\left(1+2+6+15+\cdots+t_ n\right)-\left(1+2+6+15+\cdots+t_n\right) $</p>
</li>
<li>
<p>$= 1+(2-1)+(6-2)+(15-6)+\cdots+\left(t_{n-1}-t_{n-2}\right)$</p>
</li>
<li>
<p>$+\left(t_n-t_{n-1}\right)+t_n$</p>
</li>
<li>
<p>$
\begin{aligned}
& 0=1+1^2+2^2+3^2+\cdots+\left(t_n-t_{n-1}\right)-t_n \\
& \therefore \quad t_n=1+1^2+2^2+3^2+\cdots+\left(t_n-t_{n-1}\right) \\
&=1+\sum_{i=1}^{n-1} i^2 =1+\frac{n(n-1)(2 n-1)}{6} .
\end{aligned}
$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution contd... $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution contd... $\rightarrow$ Solution contd... </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-6"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-contd-primary"><primary style="font-size:24px"> Solution contd… </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:28px;'>
<ul>
<li>
<p>Let " $a_{11}$ denote the $1^{\text {st }}$ element of $M_n$.</p>
</li>
<li>
<p>Then $a_{11}=1+\frac{n(n-1)(2 n-1)}{6} $</p>
</li>
<li>
<p>$ a_{i i}=1+\frac{n(n-1)(2 n-1)}{6}+(i-1)(n+1), 1 \leqslant i \leqslant n .$</p>
</li>
<li>
<p>$\begin{aligned} \therefore & \operatorname{trace}\left(M_n\right) \end{aligned} $</p>
</li>
<li>
<p>$
\begin{aligned}
& =\sum_{i=1}^n a_{i i} \\
& =\sum_{i=1}^n\left[\frac{1+n(n-1)(2 n-1)}{6}+(i-1)(n+1)\right] \\
& =\sum_{i=1}^n \left(1+\frac{n(n-1)(2 n-1)}{6}\right)+\sum_{i=1}^n(i-1)(n+1)
\end{aligned}$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

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

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-7"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-contd-primary-1"><primary style="font-size:24px"> Solution contd… </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px;'>
<ul>
<li>
<p>$ =n\left[1+\frac{n(n-1)(2 n-1)}{6}\right]+(n+1) \sum_{i=1}^n i=1$</p>
</li>
<li>
<p>$A=n\left[1+\frac{n(n-1)(2 n-1)}{6}\right]+(n+1) \sum_{i=1}^n i$</p>
</li>
<li>
<p>$ \begin{aligned} & =n\left[1+\frac{n(n-1)(2 n-1)}{6}\right]+\frac{(n+1) \frac{(n-1) n}{2}}{6} \\
& =\frac{n}{6}\left[6+2 n^3-3 n^2+n+3 n^2-30\right] \\
& =\frac{n}{6}\left[2 n^3+2 n+6\right] \\
& \therefore \operatorname{tr}(M n)=\frac{n}{6}\left(2 n^3+n+3\right) .\end{aligned}$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

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

### <Success style="font-size:25px"> Problem Of Matrices L-6 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
<hr>
<main>

- Problem:

- If $A=\left[\begin{array}{cc}\frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{5}}{2 i}  \\\\\\  \frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{3}}{2 i}\end{array}\right]$ & $f(x)=x^2+1$,

- then, find $f(A)$.

- Solution:

- $f(x)=x^2+1$

- $\therefore \quad f(A)=A^2+I$

- $ A^2=\left[\begin{array}{cc}
\frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{3}}{2 i} \\\\
\frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{5}}{2 i}
\end{array}\right]\left[\begin{array}{cc}
\frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{5}}{2 i} \\\\
\frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{3}}{2 i}
\end{array}\right]
 $

</div>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution contd... $\rightarrow$ Solution contd... $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Solution cont... </footer>

### <Success style="font-size:25px"> Problem Of Matrices L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $\begin{aligned} & =\frac{-1}{4}\left[\begin{array}{cc}-1+i \sqrt{5} & -1-i \sqrt{5} \\\\ 1+i \sqrt{5} & 1-i \sqrt{5}\end{array}\right]\left[\begin{array}{cc}-1+i \sqrt{5} & -1-i \sqrt{5} \\\\ 1+i \sqrt{3} & 1-i \sqrt{3}\end{array}\right] \\\\ & =\frac{-1}{4}\left[\begin{array}{lc}(-1+i \sqrt{3})^2-(1+i \sqrt{3})^2 & 1+3+1+3 \\\\ -3-1+1+3 & -(1+i \sqrt{3})^2+(1-i \sqrt{3})^2\end{array}\right] \\\\ &\end{aligned}$

- $ =-\frac{1}{4}\left[\begin{array}{cc}
 1-2 i \sqrt{3}-3-1-2 i \sqrt{3}+3 & 0 \\\\
 0 & -1-2 i \sqrt{3}-3+1-2 i \sqrt{3}-3 \\\\
\end{array}\right]$

- $=-\frac{1}{4}\left[\begin{array}{cc}-4 i \sqrt{3} & 0 \\\\ 0 & -4 i \sqrt{3}\end{array}\right]=\left[\begin{array}{cc}i \sqrt{3} & 0 \\\\ 0 & i \sqrt{3}\end{array}\right]$

</div>

<div style='float: right; width:1%;'>
	

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution contd... $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution cont... $\rightarrow$ Problem-5 </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-10"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-2"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:28px;'>
<ul>
<li>
<p>A square matrix $A$ which satifies $A^2=A$ is called an idempotent matrix.</p>
</li>
<li>
<p>$
\begin{aligned}
A^2 & =\left[\begin{array}{ccc}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & x
\end{array}\right]\left[\begin{array}{ccc}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & x
\end{array}\right] \\
& =\left[\begin{array}{cccc}
4+2-4 & -4-6+2 & -8-8-4 x \\
-2  -3+4 & 2+9-8 & 4+12+4 x \\
2+2+x & -2-6-2 x & -4-8+x^2
\end{array}\right]
\end{aligned}
$</p>
</li>
<li>
<p>$=\left[\begin{array}{ccc}
2 & -2 & -16-4 x \\
-1 & 3 & 16+4 x \\
4+x & -8-2 x & -12+x^2
\end{array}\right]$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution cont... $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution contd... $\rightarrow$ Problem-6 </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-11"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-contd-primary-2"><primary style="font-size:24px"> Solution contd… </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px;'>
<ul>
<li>
<p>$
\begin{aligned}
& A^2=A ~  \text { i.e., }\left[\begin{array}{ccc}
2 & -2 & -16-4 x \\
-1 & 3 & 16+4 x \\
4+x & -8-2 x & -12+x^2
\end{array}\right] \\
&
\end{aligned}
$</p>
</li>
<li>
<p>$
\begin{aligned}
=\left[\begin{array}{ccc}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & x
\end{array}\right] \\
&
\end{aligned}
$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution contd... </span> $\rightarrow$ Problem-6 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-14"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-7-primary"><primary style="font-size:24px"> Problem-7 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:25px;'>
<ul>
<li>
<p>Problem:</p>
</li>
<li>
<p>$A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]$ & $A^2 = I$ then find $\alpha$</p>
</li>
<li>
<p>Solution:</p>
</li>
<li>
<p>$
\begin{aligned}
A^2 & =\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
\alpha^2 & 0 \\
\alpha+1 & 1
\end{array}\right] \\
A^2 & =I \text { i.e., }\left[\begin{array}{ll}
\alpha^2 & 0 \\
\alpha+1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
\alpha^2 & =1 \quad \alpha+1=0 \\
\Rightarrow \alpha & =-1 .
\end{aligned}
$</p>
</li>
</ul>
</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Problem-8 $\rightarrow$ Problem-9 </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-15"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-8-primary"><primary style="font-size:24px"> Problem-8 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px;'>
<ul>
<li>
<p>If $A=\left[\begin{array}{ll}\alpha & 0 \\ 2 & 3\end{array}\right]$ & $A^2=9 I$, then find $\alpha$.</p>
</li>
<li>
<p>Solution.</p>
</li>
<li>
<p>$
\begin{aligned}
A^2 & =\left[\begin{array}{ll}
\alpha & 0 \\
2 & 3
\end{array}\right]\left[\begin{array}{ll}
\alpha & 0 \\
2 & 3
\end{array}\right] \\
& =\left[\begin{array}{ll}
\alpha^2 & 0 \\
2 \alpha+6 & 9
\end{array}\right]
\end{aligned}
$</p>
</li>
<li>
<p>$
\begin{gathered}
A^2=9 I \text { i.e., }\left[\begin{array}{ll}
\alpha^2 & 0 \\
2 \alpha+6 & 9
\end{array}\right]=\left[\begin{array}{ll}
9 & 0 \\
0 & 9
\end{array}\right]
\end{gathered}
$</p>
</li>
<li>
<p>$\Rightarrow \alpha^2=9 \quad \Rightarrow 2 \alpha+6=0 \Rightarrow \alpha=-3 $</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Problem-8 </span> $\rightarrow$ Problem-9 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-problem-of-matrices-l-6-success-17"><Success style="font-size:25px"> Problem Of Matrices L-6 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-4"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:100%;font-size:30px;'>
<ul>
<li>
<p>$\left[\begin{array}{lll}x & x & x \\ x & x & x \\ x & x & x\end{array}\right]\left[\begin{array}{lll}y & y & y \\ y & y & y \\ y & y & y\end{array}\right]=\left[\begin{array}{lll}1 / 3 & 1 / 3 & 1 / 3 \\ 1 / 3 & 1 / 3 & 1 / 6 \\ 1 / 3 & 1 / 3 & 1 / 3\end{array}\right]$</p>
</li>
<li>
<p>$\begin{aligned} & {\left[\begin{array}{lll}x & x & x \\ x & x & x \\ x & x & x\end{array}\right]\left[\begin{array}{lll}y & y & y \\ y & y & y \\ y & y & y\end{array}\right]} =\left[\begin{array}{lll}3 x y & 3 x y & 3 x y \\ 3 x y & 3 x y & 3 x y \\ 3 x y & 3 x y & 3 x y\end{array}\right] \\ & 3 x y=1 / 3 \\ & \Rightarrow x y=1 / 9 \\ & \text { or } x=1 / 9 y .\end{aligned}$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Problem-8 $\rightarrow$ Problem-9 $\rightarrow$ <span style = "color:blue"> Solution</span> $\rightarrow$ Problem-10 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Problem Of Matrices L-6 </Success>
### <primary style="font-size:24px"> Problem-10 </primary>
<hr>
<main>

If the equations,
$\begin{aligned}
x - 2y + 3z &= 0 \\\\
-2x + 2y + 2z &= 0 \\\\
-8x + \lambda y &= 0
\end{aligned}$

- have non-trivial solution, then find $\lambda$.

$
\begin{aligned}
& \text { Solution: }\left[\begin{array}{ccc}
 1 & -2 & 3 \\\\
-2 & 3 & 2 \\\\
-8 & \lambda & 0
\end{array}\right] \\\\
& R_2 \rightarrow R_2+2 R_1\left[\begin{array}{ccc}
 1 & -2 & 3 \\\\
 0 & -1 & 1 \\\\
 0 & \lambda-16 & 24
\end{array}\right]
\end{aligned}
 $

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-9 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-10 </span> $\rightarrow$ Solution $\rightarrow$ Solution cont... </footer>

### <Success style="font-size:25px"> Problem Of Matrices L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $
\begin{aligned}
& R_2 \rightarrow-R_2\left[\begin{array}{ccc}
 1 & -2 & 3 \\\\
 0 & 1 & -1 \\\\
 0 & \lambda-16 & 24
\end{array}\right] \\\\
& R_1 \rightarrow R_1+2 R_2\left[\begin{array}{ccc}
 1 & 0 & -13 \\\\
 0 & 1 & -8 \\\\
 0 & 0 & 24+(16-\lambda)(-2)
\end{array}\right] \\\\
& R_3 \rightarrow R_3+(16-\lambda) R_2\left[\begin{array}{ccc}
 1 & 0 & -13 \\\\
 0 & 1 & -8 \\\\
 0 & 0 & \lambda-104
\end{array}\right]
\end{aligned}
$

- For the existence of a non-trivial solution, $8 \lambda-104$ should be equal to 0.

</div>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-10 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution cont... $\rightarrow$ Thank you </footer>