mathematics-class-12-unit-03-chapter-04-matrices-l-4-6-skevlhhpkao

<h3 id="success-stylefont-size25px-matrices-l-4-success"><Success style="font-size:25px"> Matrices L-4 </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;text-align:left'>
<ul>
<li>
<p>Find row reduced form of</p>
</li>
<li>
<p>$A=\left(\begin{array}{lllll}
0 & 3 & 0 & 1 \\
0 & 0 & 4 & 1 \\
0 & 0 & 0 & 0 \\
0 & 4 & 2 & 0
\end{array}\right) $</p>
</li>
<li>
<p>Solution:</p>
</li>
<li>
<p>$R_ 3 \leftrightarrow R_4\left(\begin{array}{llll}
0 & 3 & 0 & 1 \\
0 & 0 & 4 & 1 \\
0 & 4 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)$</p>
</li>
</ul>
</div>
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</div>
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<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Matrix lecture-4 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-2"><Success style="font-size:25px"> Matrices L-4 </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:18px; text-align:left'>
$\begin{aligned} & R_3 \rightarrow R_3-4 R_1\left(\begin{array}{cccc}0 & 1 & 0 & \frac{1}{3} \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 2 & \frac{-4}{3} \\ 0 & 0 & 0 & 0\end{array}\right) \\
& R_2 \rightarrow \frac{1}{4} R_2\left(\begin{array}{cccc}0 & 1 & 0 & \frac{1}{3} \\ 0 & 0 & 1 & \frac{1}{4} \\ 0 & 0 & 2 & \frac{-4}{3} \\ 0 & 0 & 0 & 0\end{array}\right) \\ 
& R_3 \longrightarrow R_3-2 R_2\left(\begin{array}{cccc}0 & 1 & 0 & \frac{1}{3} \\ 0 & 0 & 1 & \frac{1}{4} \\ 0 & 0 & 0 & -\frac{4}{3}-\frac{1}{2} \\ 0 & 0 & 0 & 0\end{array}\right)\end{aligned}$
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</main>
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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

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

<div style='float: left; width:99%; font-size:18px; text-align:left;'>
$\begin{aligned} &\left(\begin{array}{cccc}0 & 1 & 0 & \frac{1}{3} \\\ 0 & 0 & 1 & \frac{1}{4} \\ 0 & 0 & 0 & \frac{-11}{6} \\ 0 & 0 & 0 & 0\end{array}\right) \\ 
& R_3 \rightarrow-\frac{6}{11} R_3\left(\begin{array}{llll}0 & 1 & 0 & \frac{1}{3} \\ 0 & 0 & 1 & \frac{1}{4} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{array}\right) \\ 
& R_1 \rightarrow R_1-\frac{1}{3} R_3, R_2 \rightarrow R_2-\frac{1}{4} R_3\left(\begin{array}{llll}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1  \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{array}\right)\end{aligned}$
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</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-3"><Success style="font-size:25px"> Matrices L-4 </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:20px;text-align:left;'>
<ul>
<li>Thus, the row reduced echelon matrix of $A$</li>
<li>obtained by applying the row elementary operations is
$\left(\begin{array}{llll}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{array}\right)$.</li>
</ul>
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</div>		
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-5"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-rank-of-a-matrix-primary"><primary style="font-size:24px"> Rank of a matrix </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:30px;text-align:left;'>
<ul>
<li>
<p>The rank of a matrix is defined as the number of non-zero rows in its reduced  echelon matrix.</p>
</li>
<li>
<p>Note : Given a matrix $A$ then the row reduced echelon matrix associated with $A$ is unique.</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
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<hr>
<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Rank of a matrix </span> $\rightarrow$ Rank of a matrix example-1  $\rightarrow$ Rank of a matrix example-2,3 </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-6"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-rank-of-a-matrix-example-1-primary"><primary style="font-size:24px"> Rank of a matrix example-1 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:30px;text-align:left;'>
<ul>
<li>
<p>$ A=\left(\begin{array}{llll}0 & 0 & 4 & 1 \\ 0 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 4 & 2 & 0\end{array}\right)$</p>
</li>
<li>
<p>$\operatorname{RRE}=\left(\begin{array}{llll}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{array}\right)$</p>
</li>
<li>
<p>The number of non-zero rows in the RRE of $A$ is 3.</p>
</li>
<li>
<p>$\therefore \operatorname{Rank}(A)=3$.</p>
</li>
</ul>
</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Rank of a matrix $\rightarrow$ <span style = "color:blue"> Rank of a matrix example-1 </span> $\rightarrow$ Rank of a matrix example-2,3 $\rightarrow$ Rank of a matrix example-3 </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-7"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-rank-of-a-matrix-example-23-primary"><primary style="font-size:24px"> Rank of a matrix example-2,3 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:30px;text-align:left;'>
<ul>
<li>
<ol start="2">
<li></li>
</ol>
</li>
<li>
<p>$
\begin{aligned}
A & =\left[\begin{array}{ccc}
1 & -2 & 3 \\
4 & 2 & 5
\end{array}\right] \\
\operatorname{RRE} & =\left[\begin{array}{ccc}
1 & 0 & \frac{-2}{3} \\
0 & 1 & \frac{-11}{6}
\end{array}\right]
\end{aligned}
$</p>
</li>
<li>
<p>The number of non-zero rows in this RRE is 2 .</p>
</li>
<li>
<p>$\therefore$ Rank of $A$ is 2 .</p>
</li>
<li>
<ol start="3">
<li>$A=\left[\begin{array}{ll}2 & 3 \\ 4 & 5 \\ 2 & 1\end{array}\right] \quad R_1 \rightarrow \frac{1}{2} R_1  $</li>
</ol>
</li>
<li>
<p>$\left[\begin{array}{ll}1 & \frac{3}{2} \\ 4 & 5 \\ 2 & 1\end{array}\right]$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Rank of a matrix $\rightarrow$ Rank of a matrix example-1 $\rightarrow$ <span style = "color:blue"> Rank of a matrix example-2,3 </span> $\rightarrow$ Rank of a matrix example-3 $\rightarrow$ Rank of a matrix example-4 </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-8"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-rank-of-a-matrix-example-3-primary"><primary style="font-size:24px"> Rank of a matrix example-3 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:20px; text-align:left;'>
 $
\begin{aligned}
& \begin{array}{l}
 R_2 \longrightarrow R_2+(-4) R_1 \\
 R_3 \rightarrow R_3+(-2) R_1
\end{array}\left[\begin{array}{ll}
 1 & \frac{3}{2} \\
 0 & -1 \\
 0 & -2
\end{array}\right] \\
& R_2 \rightarrow R_2-(-1) R_2\left[\begin{array}{cc}
 1 & \frac{3}{2} \\
 0 & 1 \\
 0 & -2
\end{array}\right] \\
& \begin{array}{l}
 R_1 \rightarrow R_1+(\frac{-3}{2}) R_2 \\
 R_3 \longrightarrow R_3+(2) R_2
\end{array}\left[\begin{array}{ll}
 1 & 0 \\
 0 & 1 \\
 0 & 0
\end{array}\right] \\
&
\end{aligned}
$
<ul>
<li>The number of zero rows in this RRE matrix is 2 $\therefore \operatorname{Rank}(A)=2$.
  </div>
  <div style='float: right; width:1%;'>
  
</li>
</ul>
</main>
<hr>
<footer style = "font-size: 18px">Rank of a matrix example-1 $\rightarrow$ Rank of a matrix example-2,3 $\rightarrow$ <span style = "color:blue"> Rank of a matrix example-3 </span> $\rightarrow$ Rank of a matrix example-4 $\rightarrow$ Invertibility of a matrix </footer>

### <Success style="font-size:25px"> Matrices L-4 </Success>
### <primary style="font-size:24px"> Rank of a matrix example-4 </primary>
<hr>
<main>

<div style='float: left; width:99%; font-size:20px; text-align:left;'>
 $
\begin{aligned}
& \text { A) }\left(\begin{array}{llll}
 1 & 2 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 3
\end{array}\right) \\
& R_2 \longleftrightarrow R_3\left(\begin{array}{llll}
 1 & 2 & 0 & 0 \\
 0 & 0 & 1 & 3 \\
 0 & 0 & 0 & 0
\end{array}\right)
\end{aligned}
$

- This is the row reduced echelon matrix obtained from the given matrix.

- The number of non zero rows in this RRE is 2 .

- $\therefore$ Rank of the matrix is 2 .

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

</div>

</main>
<hr>
<footer style = "font-size: 18px">Rank of a matrix example-2,3 $\rightarrow$ Rank of a matrix example-3 $\rightarrow$ <span style = "color:blue"> Rank of a matrix example-4 </span> $\rightarrow$ Invertibility of a matrix $\rightarrow$ Problem-3 </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-9"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-invertibility-of-a-matrix-primary"><primary style="font-size:24px"> Invertibility of a matrix </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:30px;text-align:left;'>
<ul>
<li>
<p>A square matrix $A$ is said to be invertible if a square matrix $B$ with order $B=$ order $A$</p>
</li>
<li>
<p>such that $A B = B A = I$.</p>
</li>
<li>
<p>Result:</p>
</li>
<li>
<p>Suppose $R$ is the RRE matrix obtained from the given  matrix $A$. Then $A$ is invertible if and only if $R$ is the identity matrix.</p>
</li>
<li>
<p>The matrix $B$ is called as the inverse of $A$.</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Rank of a matrix example-3 $\rightarrow$ Rank of a matrix example-4 $\rightarrow$ <span style = "color:blue"> Invertibility of a matrix </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-10"><Success style="font-size:25px"> Matrices L-4 </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;text-align:left;'>
<ul>
<li>
<p>Is the inverse of a matrix unique?</p>
</li>
<li>
<p>Solution:</p>
</li>
<li>
<p>Suppose that $A$ is an invertible matrix.</p>
</li>
<li>
<p>Let, $B$ & $C$ be this inverses, i.e.,</p>
</li>
<li>
<p>$A$ $B$ & = $I = B A$ $\longrightarrow(1)$</p>
</li>
<li>
<p>$A$ $C$ & $ = I = C A$ $\longrightarrow(2)$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Rank of a matrix example-4 $\rightarrow$ Invertibility of a matrix $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-11"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-3"><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>Given any square matrix $A, A \cdot I=A$.</p>
</li>
<li>
<p>Let $A=\left[a_{i j}\right]$ & $I=\left[b b_{i j}\right],$</p>
</li>
<li>
<p>$b_{i j}= \begin{cases}1   ,\text{if} ~ ~ i=j \\ 0  ,\text{if} ~ ~ i ≠ j\end{cases}$.</p>
</li>
<li>
<p>$ A I=\left[a_{i j}\right]\left[b_{i j}\right] $</p>
</li>
<li>
<p>$=\left[\sum_{k=1}^n a_{i k} b_{k j}\right]   =\left[a_{i j} b_{j j}\right]  =\left[a_{i j}\right]=A . \quad \therefore A I=A \longrightarrow(3) $</p>
</li>
</ul>
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</main>
<hr>
<footer style = "font-size: 18px">Invertibility of a matrix $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Result $(A B)^{-1}=B^{-1} A^{-1}$ </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-13"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-result-a-b-1b-1-a-1-primary"><primary style="font-size:24px"> Result $(A B)^{-1}=B^{-1} A^{-1}$ </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:30px;text-align:left;'>
<ul>
<li>
<p>For any two square matrices $A$ & $B$ of same order, the Invertibility of $A$ & $B$ implies the Invertibility of $A B$ & $(A B)^{-1}=B^{-1} A^{-1}$.</p>
</li>
<li>
<p>Proof:</p>
</li>
<li>
<p>$(A B) \cdot\left(B^{-1} A^{-1}\right)=A\left(B B^{-1}\right) \cdot A^{-1}=(A \cdot I) A^{-1}$</p>
</li>
<li>
<p>$=A \cdot A^{-1}=I$</p>
</li>
<li>
<p>$\left(B^{-1} A^{-1}\right)(A B)=B^{-1}\left(A^{-1} A\right) B=B^{-1}(I \cdot B)$</p>
</li>
<li>
<p>$ =B^{-1} B=I \cdot$</p>
</li>
</ul>
</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Result $(A B)^{-1}=B^{-1} A^{-1}$  </span> $\rightarrow$ Observations $\rightarrow$ Observations contd... </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-14"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-observations-primary"><primary style="font-size:24px"> Observations </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:30px;text-align:left;'>
<ul>
<li>
<p>If $P$ is an elementary row operation on the set of all matrices, then $P$ is invertible as a function i.e., $P$ is both $1-1$ and onto.</p>
</li>
<li>
<p>Suppose $P_1, P_2, \ldots, P_m$ are the elementary row operations performed on an invertible matrix $A$ as obtain the identity matrix.</p>
</li>
<li>
<p>Then,$P_m P_{m-1}  \cdots  P_1(A)=I .$</p>
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  <div style='float: right; width:01%;'>
  
</li>
</ul>
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<hr>
<footer style = "font-size: 18px">Invertibility of a matrix $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Observations </span> $\rightarrow$ Observations contd... $\rightarrow$ Matrix inverse using row operations </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-15"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-observations-contd-primary"><primary style="font-size:24px"> Observations contd… </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:20px;text-align:left;'>
<ul>
<li>$
\begin{aligned}
& \text { Thus, }\left(P_n \circ P_{m-1}  \cdots  P_2  P_1\right)(I) \quad A=I \
& \Rightarrow A^{-1}=\left(P_m \cdot P_{m-1} \cdots \cdots \cdot P_1\right)(I) .
\end{aligned}
$</li>
</ul>
</div>
<div style='float: right; width:01%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Observations $\rightarrow$ <span style = "color:blue"> Observations contd... </span> $\rightarrow$ Matrix inverse using row operations $\rightarrow$ Matrix inverse using row operations contd... </footer>

<h3 id="success-stylefont-size25px-matrices-l-4-success-16"><Success style="font-size:25px"> Matrices L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-matrix-inverse-using-row-operations-primary"><primary style="font-size:24px"> Matrix inverse using row operations </primary></h3>
<hr>
<main>
<div style='float: left; width:99%; font-size:30px;text-align:left;'>
<ul>
<li>
<p>$\quad A=\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 1 \\  1 & 2 & 3\end{array}\right)$</p>
</li>
<li>
<p>Making an augumented matrix with the given matrix and identity matrix.</p>
</li>
<li>
<p>$=\left(\begin{array}{lll|lll} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 1 & 2 & 0 & 0 & 0 & 1 \end{array}\right)$</p>
</li>
<li>
<p>Now, applying row operations to it.</p>
</li>
</ul>
</div>
</main>
<hr>
<footer style = "font-size: 18px">Observations $\rightarrow$ Observations contd... $\rightarrow$ <span style = "color:blue"> Matrix inverse using row operations </span> $\rightarrow$ Matrix inverse using row operations contd... $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Matrices L-4 </Success>
### <primary style="font-size:24px"> Matrix inverse using row operations contd... </primary>
<hr>
<main>

<div style='float: left; width:99%; font-size:30px;text-align:left;'>
	
- $ R_2\rightarrow R_2-R_1 \\\  R_3 \rightarrow R_3-R_1 $

- $\left(\begin{array}{lll|ccc}1 & 1 & 1 & 1 & 0 & 0 \\\ 0 & 1 & 0 & -1 & 1 & 0 \\\ 0 & 1 & 2 & -1 & 0
& 1\end{array}\right)$

- Thus, $A^{-1}=\left(\begin{array}{ccc}2 & -1 / 2 & -1 / 2 \\\ -1 & 1 & 0 \\\ 0 & -1 / 2 & 1 / 2\end{array}\right)$.

</div>

</main>
<hr>
<footer style = "font-size: 18px">Observations contd... $\rightarrow$ Matrix inverse using row operations $\rightarrow$ <span style = "color:blue"> Matrix inverse using row operations contd... </span> $\rightarrow$ Thank you $\rightarrow$  </footer>