Determinants

Given: $\left|\begin{array}{cc}x+1& x-1\\ x^2+x+1& x^2-x+1\end{array}\right|$

$=\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)$

\(=x^3-x^2+x+x^2-x+1-(x^3+x^2+x-x^2-x-1)\)

\(=\left(x^3+1\right)-\left(x^3-1\right)=2\)

$$\begin{gathered} {\mathrm{AA}}^{-1}=\mathrm{I} \\ \Rightarrow \left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 3& 3\\ 1& \mathrm{\lambda }& 3\\ 1& 3& 4\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 12-3\mathrm{\lambda }& 0\\ 0& -3+\mathrm{\lambda }& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Compare on both sides.

$12-3\lambda =0\Rightarrow \lambda =4$

Let, $\Delta =\left|\begin{array}{ccc}-a& b& c\\ a& -b& c\\ a& b& -c\end{array}\right|$

Apply $R_1\to R_1+R_2:$

$\Delta =\left|\begin{array}{ccc}0& 0& 2c\\ a& -b& c\\ a& b& -c\end{array}\right|$

$\Rightarrow \Delta =2c\left|\begin{array}{cc}a& -b\\ a& b\end{array}\right|$

$$\begin{gathered} \Rightarrow \Delta =2c(ab-(-b\left)a\right) \\ =2c\left(2ab\right)=4abc \end{gathered}$$

$=kabc$ (given)

$\Rightarrow k=4$

$$\begin{gathered} {\mathrm{AA}}^{-1}=\mathrm{I} \\ \left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 3& 3\\ 1& \mathrm{\lambda }& 3\\ 1& 3& 4\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right] \\ \left[\begin{array}{ccc}1& 12-3\mathrm{\lambda }& 0\\ 0& -3+\mathrm{\lambda }& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Compare on both sides.

$$\begin{gathered} 12-3\lambda =0 \\ \lambda =4 \end{gathered}$$

For any matrix $A$ of order $n$, we have:

$|adj·\mathrm{A}|=|\mathrm{A}{|}^{n-1}$

Substitute $n=3$ and $|adj·\mathrm{A}|=8$.

$$\begin{gathered} 8=|\mathrm{A}{|}^{3-1} \\ 8=|\mathrm{A}{|}^2 \\ \left|\mathrm{A}\right|=2\sqrt{2} \end{gathered}$$

Now,

$\left|{\mathrm{A}}^T\right|=\left|\mathrm{A}\right|=2\sqrt{2}$

For any matrix $A$ of order $n$, we have:

$|adj·\mathrm{A}|=|\mathrm{A}{|}^{n-1}$

Substitute $n=3$ and $|adj·\mathrm{A}|=8$.

$$\begin{gathered} 8=|\mathrm{A}{|}^{3-1} \\ \Rightarrow 8=|\mathrm{A}{|}^2 \\ \Rightarrow \left|\mathrm{A}\right|=2\sqrt{2} \end{gathered}$$

Now,

$\left|{\mathrm{A}}^T\right|=\left|\mathrm{A}\right|=2\sqrt{2}$

$\Delta =\left|\begin{array}{ccc}-a& b& c\\ a& -b& c\\ a& b& -c\end{array}\right|$

$R_1\to R_1+R_2: \Delta =\left|\begin{array}{ccc}0& 0& 2c\\ a& -b& c\\ a& b& -c\end{array}\right|$

$\Delta =2c(-1{)}^{1+3}\left|\begin{array}{cc}a& -b\\ a& b\end{array}\right|$

$\Delta =2c(ab-(-b\left)a\right)=2c\left(2ab\right)=4abc$

$k=4$

$\left|\begin{array}{cc}x+1& x-1\\ x^2+x+1& x^2-x+1\end{array}\right|=(x+1)\left(x^2-x+1\right)-(x-1)\left(x^2+x+1\right)$

\((x+1)\left(x^2-x+1\right)=x^3-x^2+x+x^2-x+1\)

$=x^3+1$

\((x-1)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1\)

$=x^3-1$

\(\left(x^3+1\right)-\left(x^3-1\right)=2\)