• 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,

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

• $\operatorname{det}(\operatorname{adj}P)= 1(3-7)\\-4(6-7)+4(2-1)\\=-4+4+4\\=4$

• Given $n=3$

• $\operatorname{det}(\operatorname{adj} P)=(\operatorname{det} P)^2 \\ \Rightarrow (\operatorname{det} P)^2=4,1\\ \Rightarrow \operatorname{det} P= \pm 2$

• For $3 \times 3$ matrices $M$ and $N$, show that

• i) $N^{\top} M N$ is symmetric or skew-symmetic according as $M$ is symmetric or skew-symmetric.

• ii) $M N-N M$ is skew-symmetric for all symmetric matrices $M$ and $N$.

• 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 $\Rightarrow \quad M^{\top}=-M$

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

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

• ii) Let, $M$ and $N$ be symmetric matrices.

• $\Rightarrow M^{\top}=M\\,N^{\top}=N$

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

• 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.

• 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 $n$.

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

• $=\left(X^{\top} \cdot X^{\top} \cdot \cdots X^{\top}\right)$

• $=\left(X^{\top}\right)^n=(-X)^n$

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

$(X^n)^T = \left\{\begin{array}{lll} X^n & n ~ even \\ -X^n & n~ odd \end{array}\right.$,

$(Y^n)^T = \left\{\begin{array}{lll} Y^n & n ~ even \\ -Y^n & n~ odd \end{array}\right.$,

• $\left(Z^n\right)^{\top}=(Z \cdot Z \cdot \cdots Z)_{n~\text{times}}^{\top}\\ =\left(Z^T \cdot Z^{\top} \cdot \cdots Z^{\top}\right) \\ =\left(Z^{\top}\right)^n=Z^n \\ \left(Z^n\right)^{\top}=Z^n \quad \quad \forall n$

• ${(a)}$
• $\begin{aligned} & \left(Y^3 Z^4-Z^4 Y^3\right)^{\top} \\ = & \left(Y^3 Z^4\right)^{\top}\ -\left(Z^4 Y^3\right)^{\top} \\ = & \left(Z^4\right)^{\top}\left(Y^3\right)^{\top}-\left(Y^3\right)^{\top}\left(Z^4\right)^{\top} \\ = & Z^4\left(-Y^3\right)-\left(-Y^3\right) Z^4 \\ = & Y^3 Z^4-Z^4 Y^3 \\ \Rightarrow & Y^3 Z^4-Z^4 Y^3 \text { is a symmetric } \end{aligned}$

• ${(b)}$ $\left(X^{44}+Y^{44}\right)^{\top}\\=\left(X^{44}\right)^{\top}+\left(Y^{44}\right)^{\top}$

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

• ${(c)}$

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

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

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

• $= Z^3 (-X)^4 - (-X)^4 Z^3$

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

• ${(d)}$

• $\left(X^{23}+Y^{23}\right)^{\top} \\ =\left(X^{23}\right)^{\top}+\left(Y^{23}\right)^{\top} \\ =-X^{23}-Y^{23} \\ =-\left(X^{23}+Y^{23}\right)$

• $\Rightarrow X^{23}+Y^{23}$ is a skew-symmetric matrix.

• 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 $k$.

$\operatorname{det} A=\left|\begin{array}{ccc} 2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1 \end{array}\right|$

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

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

• $=8 k^3+12 k^2+6 k+1$

• $\because \Rightarrow k>0\\ \Rightarrow \operatorname{det} A>0$

$\begin{aligned} & \operatorname{det} B=\left|\begin{array}{ccc} 0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0 \end{array}\right| \end{aligned}$

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

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

$(\operatorname{det} A)^2+(\operatorname{det} B)^2=10^6$

$(\operatorname{det} A)^2=10^6$

$\Rightarrow \operatorname{det} A=10^3$

$8 k^3+12 k^2+6 k+1=1000$

$\Rightarrow \quad 8 k^3+12 k^2+6 k-999=0$

$p(k)=8 k^3+12 k^2+6 k-999$

• $\begin{aligned} & p(1)=8+12+6-999<0 \\ & p(2)=64+12 \times 4+6 \times 2-999<0 \\ & p(3)=8 \times 27+12 \times 9+18-999<0 \\ & p(4)=8 \times 64+12 \times 16+24-999 \\ &=512+192+24-999<0 \\ & p(5)=8 \times 125+12 \times 25+30-999 \\ &=1000+300+30-999>0 \\ & \Rightarrow p(k)=0 ~ \text{for some} ~ 4 < k < 5 \\ & \Rightarrow [k]=4 \end{aligned}$

• 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) $4 \alpha-k+8=0$

• It is given that

• $P Q= k \cdot I$

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

• $=k^3 \operatorname{det} I$

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

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

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

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

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

• $\quad P Q =k I$

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

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

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

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

• $\because \quad q_{23}=-\frac{k}{8} \\ \Rightarrow \quad-\frac{(3 \alpha+4) k}{12 \alpha+20}\\ =-\frac{k}{8}$

• Given $k \neq 0$, we have, $\frac{(3 \alpha+4)}{12 \alpha+20}=\frac{1}{8}$

• $24 \alpha+32=12 \alpha+20 \\ \Rightarrow \quad 12 \alpha=-12 \\ \Rightarrow \quad {\alpha=-1}$

• $\operatorname{det} Q=\frac{k^2}{2}$

• $\Rightarrow \frac{k^3}{12 \alpha+20} =\frac{k^2}{2} \\ \Rightarrow 2k=12(-1) + 20\\ \Rightarrow 2 k =8, \\ \Rightarrow k=4 \\$

• $\text{(b)}\quad4 \alpha-k+8$ =

• $4(-1)-4+8 =0$

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

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

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

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

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