mathematics-class-12-unit-03-chapter-08-matrix-and-determinant-l-2-5-lcvhlvb0ule

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- If the adjoint of a $3 \times 3$ matrix $P$ is $\left[\begin{array}{lll}1 & 4 & 4 \\\\ 2 & 1 & 7 \\\\ 1 & 1 & 3\end{array}\right]$, then what are the possible values of the determinant of $P$.

- ${\text{Solution:}}$

- For an $n \times n$ matrix $A$ $
\operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{n-1}$

- we have,

- $
\text { adj } P=\left[\begin{array}{ccc}
 1 & 4 & 4 \\\\
 2 & 1 & 7 \\\\
 1 & 1 & 3
\end{array}\right] $

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<footer style = "font-size: 18px"> $\rightarrow$ Matrix and determination lecture - 8 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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-  i) suppose $M$ is a symmetric matrix  
- $\Rightarrow M^{\top}=M$

- $\left(N^{\top} M N\right)^{\top} $

- $=\left(N^{\top} \underline{M N}\right)^{\top} $

- $=(M N)^{\top}\left(N^{\top}\right)^{\top} $

- $=N^{\top} M^{\top} N =N^{\top} M N$

- $\Rightarrow N^{\top} M N$ is a symmetric matrix.  
- Let $M$ be a skew -symmetric $ \Rightarrow \quad M^{\top}=-M $

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

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\left(N^{\top} M N\right)^{\top}  = N^{\top} M^{\top} N =-N^{\top} M N$

- $\Rightarrow N^{\top} M N$ is a skew-symmetric matrix.

- ii) Let, $M$ and $N$ be symmetric matrices.    
- $\Rightarrow M^{\top}=M\\\\,N^{\top}=N$

- $(M N-N M)^{\top} \\\\=\left(M N)^{\top}-(N M)^{\top}\right.\\\\=N^{\top}M^{\top}-M^{\top} N^{\top}\\\\=NM-MN\\\\=-(M N-N M)$

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

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- let $X$ and $Y$ be two arbitrary $3 \times 3$, non-zero, skew-symmetric matrices and $Z$ be an arbitrary $3 \times 3$. non - zero, symmetric matrix.

- Then, show that

- a) $Y^3 Z^4-Z^4 Y^3$ is a symmetric matrix

- b) $X^{44}+Y^{44}$ is a symmetric matrix

- c) $X^4 Z^3-Z^3 X^4$ is a skew -symmetric matrix

- d) $X^{23}+Y^{23}$ is a skew-symmetric matrix.

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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-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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-  It is given that
- $X^{\top}=-X,\\\\ Y^{\top}=-Y, Z^{\top}=Z$

- $\left(X^n\right)^{\top},\left(Y^n\right)^{\top},\left(Z^n\right)^{\top}$ for any natural number $n$.

- $\left(X^n\right)^{\top}=(X \cdot X \cdot \cdot \cdot X)_{n~\text{times}}^{\top} $
 
 - $=\left(X^{\top} \cdot X^{\top} \cdot \cdots X^{\top}\right) $

- $=\left(X^{\top}\right)^n=(-X)^n $

- $\left(X^n\right)^{\top}=(-1)^n X^n $

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### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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$(X^n)^T = \left\\{\begin{array}{lll}
 X^n & n ~ even \\\\ 
-X^n & n~ odd 
\end{array}\right.$,
 
$ (Y^n)^T = \left\\{\begin{array}{lll}
 Y^n & n ~ even \\\\ 
-Y^n & n~ odd 
\end{array}\right.$,
   
- $
\left(Z^n\right)^{\top}=(Z \cdot Z \cdot \cdots Z)_{n~\text{times}}^{\top}\\\\
 =\left(Z^T \cdot Z^{\top} \cdot \cdots Z^{\top}\right) \\\\
 =\left(Z^{\top}\right)^n=Z^n \\\\
\left(Z^n\right)^{\top}=Z^n \quad \quad \forall n
 $

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

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- ${(a)}$ 
- $
\begin{aligned}
& \left(Y^3 Z^4-Z^4 Y^3\right)^{\top} \\\\
 = & \left(Y^3 Z^4\right)^{\top}\\ -\left(Z^4 Y^3\right)^{\top} \\\\
 = & \left(Z^4\right)^{\top}\left(Y^3\right)^{\top}-\left(Y^3\right)^{\top}\left(Z^4\right)^{\top} \\\\
 = & Z^4\left(-Y^3\right)-\left(-Y^3\right) Z^4 \\\\
 = & Y^3 Z^4-Z^4 Y^3 \\\\
\Rightarrow & Y^3 Z^4-Z^4 Y^3 \text { is a symmetric }
\end{aligned}
 $

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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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### <primary style="font-size:24px"> Solution </primary>
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- $\{(b)}$ $\left(X^{44}+Y^{44}\right)^{\top}\\\\=\left(X^{44}\right)^{\top}+\left(Y^{44}\right)^{\top} $

- $ =X^{44}+Y^{44}$  $\Rightarrow X^{44}+Y^{44}$ is a symmetric matrix.

- ${(c)}$

- $\left(X^4 Z^3-Z^3 X^4\right)^{\top} $

- $=  \left(X^4 Z^3\right)^{\top} -\left(Z^3 X^4\right)^{\top} $

- $=  \left(Z^3\right)^{\top}\left(X^4\right)^{\top}-\left(X^4\right)^{\top}\left(Z^3\right)^{\top} $

- $=  Z^3 (-X)^4 - (-X)^4 Z^3 $

- $=  -(X^4 Z^3-Z^3 X^4) \Rightarrow  X^4 Z^3-Z^3 X^4 \text { is a skew-symmetric .}$

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

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Let $k$ be a positive real number and let

$ A=\left[\begin{array}{ccc}
 2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\\\
 2 \sqrt{k} & 1 & -2 k \\\\
-2 \sqrt{k} & 2 k & -1
\end{array}\right] $

- and

$
 B=\left[\begin{array}{ccc}
 0 & 2 k-1 & \sqrt{k} \\\\
 1-2 k & 0 & 2 \sqrt{k} \\\\
-\sqrt{k} & -2 \sqrt{k} & 0
\end{array}\right] $

- If $\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6$, then what is the value of $[k]$, where$[k]$ denotes the largest integer less than or equal to $k$.

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### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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$ \operatorname{det} A=\left|\begin{array}{ccc}
 2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\\\
 2 \sqrt{k} & 1 & -2 k \\\\
-2 \sqrt{k} & 2 k & -1
\end{array}\right|$

- $ =(2 k-1)\left(-1+4 k^2\right) -2 \sqrt{k}(-2 \sqrt{k}-4 k \sqrt{k})  +2 \sqrt{k}(4 k \sqrt{k}+2 \sqrt{k}) $

- $ =1-2 k+8 k^3-4 k^2+4 k+8 k^2+8 k^2+4 k $

- $ =8 k^3+12 k^2+6 k+1 $
- $\because \Rightarrow k>0\\\\ \Rightarrow \operatorname{det} A>0$

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### <primary style="font-size:24px"> Solution </primary>
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$
\begin{aligned}
& \operatorname{det} B=\left|\begin{array}{ccc}
 0 & 2 k-1 & \sqrt{k} \\\\
 1-2 k & 0 & 2 \sqrt{k} \\\\
-\sqrt{k} & -2 \sqrt{k} & 0
\end{array}\right| 
\end{aligned}
$

- $=0-(2 k-1)(2 k)+\sqrt{k}(-2 \sqrt{k}+4 k \sqrt{k})\\\\ =-4 k^2+2 k-2 k+4 k^2=0  \Rightarrow \operatorname{det} B=0 $

- $ \operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6$

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

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $
\begin{aligned}
& p(1)=8+12+6-999<0 \\\\
& p(2)=64+12 \times 4+6 \times 2-999<0 \\\\
& p(3)=8 \times 27+12 \times 9+18-999<0 \\\\
& p(4)=8 \times 64+12 \times 16+24-999 \\\\
&=512+192+24-999<0 \\\\
& p(5)=8 \times 125+12 \times 25+30-999 \\\\
&=1000+300+30-999>0 \\\\
& \Rightarrow p(k)=0 ~ \text{for some} ~ 4 < k < 5 \\\\
& \Rightarrow [k]=4 
\end{aligned} $
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Matrix And Determinant L-2 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- Let,

- $
 P=\left[\begin{array}{ccc}
 3 & -1 & -2 \\\\
 2 & 0 & \alpha \\\\
 3 & -5 & 0
\end{array}\right], \quad \alpha \in \mathbb{R.}
 $

- Suppose $Q=\left[q_ {i j}\right]$ is a matrix such that  $P Q=k I$, when $k \in R, k \neq 0$ and $I$ is a $3 \times 3$ identity matrix. If $q_{23}=-\frac{k}{8}$ and $\operatorname{det}Q=\frac{k^2}{2}$, then show that

- a) $\operatorname{det}(P$ adj $Q)=2^9 \qquad$  
- b) $4 \alpha-k+8=0$

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### <primary style="font-size:24px"> Solution </primary>
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- It is given that

- $P Q= k \cdot I $

- $\operatorname{det}(P Q) =\operatorname{det}(k \cdot I) $

- $=k^3 \operatorname{det} I $

- $=k^3 \neq 0 \quad  {{\because k \neq 0}}$

- $\operatorname{det} P \cdot \operatorname{det} Q  \neq 0$

- $\Rightarrow \operatorname{det} P \neq 0 \quad \text{and} \quad \operatorname{det} Q \neq 0$

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- $\left[\begin{array}{ccc}
 3 & -1 & -2 \\\\ 
 2 & 0 & \alpha \\\\ 
 3 & -5 & 0\end{array}\right]$

- $\operatorname{det} P=-2(-10)-\alpha(-15+3) \\\\$
- $ \operatorname{det} P=12 \alpha+20 \\\\$
- $ \Rightarrow \quad 12 \alpha+20 \neq 0 \quad $ $\\{\because\operatorname{det} P \neq 0  \\\\$
- $ PQ = K \cdot I \\\\$
- $ \Rightarrow \quad \operatorname{det} P \cdot \operatorname{det} Q=k^3\\\\$
- $ \Rightarrow \quad \operatorname{det} Q = \frac{k^3}{12 \alpha+20} \\\\$

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- $\quad P Q  =k I $

- $\Rightarrow \quad Q  =k \cdot P^{-1}$

- $=k \cdot \frac{\operatorname{adj} P}{\operatorname{det} P} $

$\operatorname{adj} P  =\left[\begin{array}{ccc}
 5 \alpha & 10 & -\alpha \\\\
 3 \alpha & 6 & -(3 \alpha+4) \\\\
-10 & 12 & 2
\end{array}\right]$

$
 Q  =  \frac{k}{12 \alpha+20} \left[\begin{array}{ccc}
 5 \alpha & 10 & -\alpha \\\\
 3 \alpha & 6 & -(3 \alpha+4) \\\\
-10 & 12 & 2
\end{array}\right]$

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- a) $\operatorname{det} (P \cdot \operatorname{adj} Q)$

- $ =\operatorname{det} P \cdot \operatorname{det}(\operatorname{adj} Q)  =(12 \alpha+20)\times(\operatorname{det} Q)^2 $

- $ =(12 \alpha+20) \left(\frac{k^3}{12 \alpha+20}\right)^2$

- $
 =\frac{k^6}{12 \alpha+20}
 =\frac{4^6}{8}
 =\frac{2^{12}}{2^3}=2^9\\\\$

- $\operatorname{det}(P \cdot \operatorname{adj} Q)=2^9 $

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