Determinants

\(A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|\)

\(2 A_{10}-A_8=\left|\begin{array}{ccc}20 & 1 & \frac{n^2}{2}+a \\ 40 & 2 & n^2-\beta \\ 56 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8 & 1 & \frac{n^2}{2}+a \\ 16 & 2 & n^2-\beta \\ 22 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|\)

\(=\left|\begin{array}{ccc}12 & 1 & \frac{n^2}{2}+a \\ 24 & 2 & n^2-\beta \\ 34 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|\)

\(=\left|\begin{array}{ccc}0 & 1 & \frac{n^2}{2}+a \\ 0 & 2 & n^2-\beta \\ -2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|\)

\(=-2\left(\left(n^2-\beta\right)-\left(n^2+2 \alpha\right)\right)\)

\(=-2(-\beta-2 \alpha)\)

\(=4 \alpha+2 \beta\)

\(A=\left[\begin{array}{lll}\alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5\end{array}\right]\)

\(\Rightarrow|A|=15 \alpha-10+16=15 \alpha+6\)

Now \(\operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc}15 & 3 & -6 \\ -10 & 5 \alpha & 4 \\ 8 & -4 \alpha & 3 \alpha-2\end{array}\right]\)

\(\therefore \mathrm{B}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -5 \alpha & 0 \\ 0 & 4 \alpha & -2 \alpha\end{array}\right]+\left[\begin{array}{ccc}15 & 3 & -6 \\ -10 & 5 \alpha & 4 \\ 8 & -4 \alpha & 3 \alpha-2\end{array}\right]\)

$\Rightarrow B=\left[\begin{array}{ccc}16& 3& -6\\ -10& 0& 4\\ 8& 0& \alpha -2\end{array}\right]$

\(|\mathrm{B}|=30 \alpha+36=66\)

\(\Rightarrow \alpha=1\)

\(\Rightarrow|\mathrm{A}|=21\)

\(\Rightarrow|\operatorname{adj}(A)|=|A|^2=(21)^2=441\)

Since $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$,

we subtract the identity matrix $I$ to get $A$:

$$A = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}$$

We compute the determinant of $A$:

$$\det(A) = \begin{vmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{vmatrix}$$

$$-2a + 2 = -4 \implies -2a = -6 \implies a = 3$$

$$\det((a + 1) \text{adj}(aI) A) = \left[(a + 1)a^2\right]^3 \det(A)$$

Substitute $a = 3$ and $\det(A) = -4$:

$$(a + 1)a^2 = (3 + 1) \cdot 3^2 = 4 \cdot 9 = 36$$ $$\left[36\right]^3 \cdot (-4) = 36^3 \cdot (-4)$$

$$36^3 \cdot (-4) = 2^6 \cdot 3^6 \cdot (-2^2) = -2^{8} \cdot 3^6$$

$$|\det((a + 1) \text{adj}(aI) A)| = 2^8 \cdot 3^6$$

Hence, $m = 8$ and $n = 6$.

$\left|Q\right|=\left|\begin{array}{ccc}2{\text{p}}_{11}& 2^2{\text{p}}_{12}& 2^3{\text{p}}_{13}\\ 2^2{\text{p}}_{21}& 2^3{\text{p}}_{22}& 2^4{\text{p}}_{23}\\ 2^3{\text{p}}_{31}& 2^4{\text{p}}_{32}& 2^5{\text{p}}_{33}\end{array}\right|=2^{10}$

$\Rightarrow 2^2\cdot 2\cdot 2^3\left|\begin{array}{ccc}{\text{p}}_{11}& {\text{p}}_{12}& {\text{p}}_{13}\\ 2{\text{p}}_{21}& 2{\text{p}}_{22}& 2{\text{p}}_{23}\\ 2^2{\text{p}}_{31}& 2^2{\text{p}}_{32}& 2^2{\text{p}}_{33}\end{array}\right|=2^{10}$

$\Rightarrow 2^9\left|\begin{array}{lll}{\text{p}}_{11}& {\text{p}}_{12}& {\text{p}}_{13}\\ {\text{p}}_{21}& {\text{p}}_{22}& {\text{p}}_{23}\\ {\text{p}}_{31}& {\text{p}}_{32}& {\text{p}}_{33}\end{array}\right|=2^{10}$

$\Rightarrow \left|\text{P}\right|=2$

\(|\operatorname{adj}(\operatorname{adj}(P))|=|P|^{(n-1)^2}\)

$=|P{|}^4=2^4=16$

$$\begin{gathered} f\left(x\right)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x}& 1& \mathrm{b}\\ \mathrm{a}& 1+\frac{\sin x}{x}& \mathrm{b}\\ \mathrm{a}& 1& \mathrm{b}+\frac{\sin x}{x}\end{array}\right| \\ \underset{x\to 0}{\lim }f\left(x\right)=\left|\begin{array}{ccc}a+1& 1& b\\ a& 1+1& b\\ a& 1& b+1\end{array}\right| \\ =\left(a+1\right)\left(2\left(b+1\right)-b\right)+1\left(a b-a\left(b+1\right)\right)-b a \\ =\left(a+1\right)\left(b+2\right)-a-a b \\ =b+a+2=\lambda +\mu a+vb \\ \text{On comparing, we get} \\ \lambda =2,\mu =1,v=1 \\ \Rightarrow {\left(\lambda +\mu +v\right)}^2=16 \end{gathered}$$

$$\begin{gathered} \text{ As }\left|adj\right(adj\mathrm{A}\left)\right|=|\mathrm{A}{|}^{{(\mathrm{n}-1)}^2}\text{,} \\ \text{where }\mathrm{n}\text{ is the order of the determinant. } \\ \text{ So, }\left|adj\left(adj{\mathrm{A}}^2\right)\right|={\left|{\mathrm{A}}^2\right|}^{(3-1{)}^2} \\ ={\left|{\mathrm{A}}^2\right|}^4=|\mathrm{A}{|}^8 \end{gathered}$$

\(\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right|=\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right|\)

\(=\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right|\)

\(=\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}|\)

\(=|\mathrm{A}-2 \mathrm{I}|\)

\(=\left|\begin{array}{ccc}0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0\end{array}\right|\)

\(=69\)

So, Prime factor of 69 is 3 and 23

So, sum \(=26\)

\(f(0)=\left|\begin{array}{ccc}0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{array}\right|\)

\(=-2(-2-4)=12\)
Now

\(f^{\prime}(x)=\left|\begin{array}{ccc}3 x^2 & 4 x & 3 \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2\end{array}\right|\)\(+\left|\begin{array}{ccc}x^3 & 2 x^2+1 & 3 x+1 \\ 6 x & 2 & 3 x^2 \\ x^3-x & 4 & x^2-2\end{array}\right|\)\(+\left|\begin{array}{ccc}x^3 & 2 x^2+1 & 3 x+1 \\ 3 x^2+2 & 2 x & x^3+6 \\ 3 x^2-1 & 0 & 2 x\end{array}\right|\)

$f^{\text{'}}\left(0\right)=\left|\begin{array}{ccc}0& 0& 3\\ 2& 0& 6\\ 0& 4& -2\end{array}\right|+\left|\begin{array}{ccc}0& 1& 1\\ 0& 2& 0\\ 0& 4& -2\end{array}\right|+\left|\begin{array}{ccc}0& 1& 1\\ 2& 0& 6\\ -1& 0& 0\end{array}\right|$

\(=-2(0-12)+0+0-1(0+6)+1(0-0)\)
\(=24-6=18\)

Now

\(2 f(0)+f^{\prime}(0)= 2.12+18\)
\(= 24+18=42\)

Solution of determinant by using operation of determinant

$$\begin{gathered} R_1\to R_1-R_2 \\ {R}_2\to R_2-R_3 \end{gathered}$$

$\left|\begin{array}{ccc}\alpha -a& b-\beta & 0\\ 0& \beta -b& c-\gamma \\ a& b& \gamma \end{array}\right|=0$

$\left(\alpha -a\right)\left(\beta -b\right)\left(\gamma -c\right)\left|\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ \frac{a}{\alpha -a}& \frac{b}{\beta -b}& \frac{\gamma }{\gamma -c}\end{array}\right|=0$

$\frac{a}{\alpha -a}\left(1-0\right)-\frac{b}{\beta -b}\left(-1-0\right)+\frac{\gamma }{\gamma -c}\left(1-0\right)=0$

$\frac{a}{\alpha -a}+\frac{b}{\beta -b}+\frac{\gamma }{\gamma -c}=0$

Solution of determinant by using operation of determinant

$$\begin{gathered} R_1\to R_1-R_2 \\ {R}_2\to R_2-R_3 \end{gathered}$$

$\left|\begin{array}{ccc}\alpha -a& b-\beta & 0\\ 0& \beta -b& c-\gamma \\ a& b& \gamma \end{array}\right|=0$

$\left(\alpha -a\right)\left(\beta -b\right)\left(\gamma -c\right)\left|\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ \frac{a}{\alpha -a}& \frac{b}{\beta -b}& \frac{\gamma }{\gamma -c}\end{array}\right|=0$

$\frac{a}{\alpha -a}\left(1-0\right)-\frac{b}{\beta -b}\left(-1-0\right)+\frac{\gamma }{\gamma -c}\left(1-0\right)=0$

$\frac{a}{\alpha -a}+\frac{b}{\beta -b}+\frac{\gamma }{\gamma -c}=0$

First, use property (3) for $\text{adj}(2A)$: Since $n=3$, $\text{adj}(2A) = 2^{3-1}\text{adj}(A) = 4\ \text{adj}(A)$

Now substitute back:

$$3A \cdot \text{adj}(2A) = 3A \cdot 4\ \text{adj}(A) = 12 \cdot \left(A \cdot \text{adj}(A)\right)$$

Using property (4): $A \cdot \text{adj}(A) = |A|I = 5I$

$$\implies 3A \cdot \text{adj}(2A) = 12 \times 5I = 60I$$

For scalar identity matrix $kI$, $\text{adj}(kI) = k^{n-1}I$ (for $n=3$, this is $k^2 I$):

$$\text{adj}(60I) = 60^2 I$$

First, $2 \cdot \text{adj}(60I) = 2 \times 60^2 I$ Using property (1), determinant of $mI$ (3x3) is $m^3$:

$$\left|2 \times 60^2 I\right| = \left(2 \times 60^2\right)^3$$

Write $60 = 2^2 \times 3 \times 5$, so:

$$\left(2 \times (2^2 \times 3 \times 5)^2\right)^3 = 2^3 \times (2^2 \times 3 \times 5)^6$$ $$= 2^3 \times 2^{12} \times 3^6 \times 5^6 = \boldsymbol{2^{15} \cdot 3^6 \cdot 5^6}$$

Comparing with $2^\alpha \cdot 3^\beta \cdot 5^\gamma$:

$\alpha=15$, $\beta=6$, $\gamma=6$

$$\alpha +\beta +\gamma =15+6+6=27$$

Given:- \(A\) is order 3 square matrix \(|A|=2\)
To find: \(|(\operatorname{adj}\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A)))|\)
we know \(\mid\) adj \(A\left|=|A|^{n-1}\right.\)

\(\begin{aligned}
&|A|=2 \quad and \quad n=3 \\
& \mid \operatorname{adj}(\operatorname{adj}(\operatorname{adi}(\operatorname{adj} A) \mid=|A|^{(n-1)^4} \\
& =|A|^{2^4} \\
&=2^{2^4} \\
&=2^{16}
\end{aligned}\)

\(\begin{equation}
\begin{gathered}
A\left[(\operatorname{adj} A^{-1})\left(\operatorname{adj}\left(B^{-1}\right)\right)\right]^{-1} \quad|A|,|B| \neq 0 \\
A B=B A \\
A^{-1} B=B A^{-1} \\
A^{-1} B^{-1}=B^{-1} A^{-1} \\
(\operatorname{adj} A)=A^{-1}|A| \\
(\operatorname{adj} A)^{-1}=\left(A^{-1}|A|\right)^{-1} \\
=\frac{A}{|A|} \\
A\left(\operatorname{adj} B^{-1}\right)^{-1}(\operatorname{adj} A^{-1})^{-1} \cdot B \\
\end{gathered}
\end{equation}\)

$A\left(\frac{B^{-1}}{\left|B^{-1}\right|}.\frac{A^{-1}}{\left|A^{-1}\right|}\right)B=I_n\left|B\right|\left|A\right|$

Let $\cos \alpha =x$

$\cos \beta =y$

$\cos \gamma =z$

$\left|\begin{array}{lll}0& x& y\\ x& 0& z\\ y& z& 0\end{array}\right|=\left|\begin{array}{lll}1& x& y\\ x& 1& z\\ y& z& 1\end{array}\right|$

Expending both sides, we get

$x^2+y^2+z^2=1$

i.e. ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

Statement 1 is false

Now, $\left|\begin{array}{ccc}{x}^2+x& 1+x& x-2\\ 2x^2+3x-1& 3x& 3x-3\\ x^2+2x+3& 2x-1& 2x-1\end{array}\right|=px+q$

Put $x=0$ both sides

$q=\left|\begin{array}{ccc}0& 1& -2\\ -1& 0& -3\\ 3& -1& -1\end{array}\right|$

$\Rightarrow q=-12$

Now put $x=1$ both sides

$p+q=\left|\begin{array}{ccc}2& 2& -1\\ 4& 3& 0\\ 6& 1& 1\end{array}\right|=12$

$\Rightarrow p=24$

$\text{ Now }\frac{p^2}{q^2}={\left(\frac{24}{-12}\right)}^2\ne 196$

$\Rightarrow p^2\ne 196q^2$

Statement (2) is false

$AdjA=\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]$

${\left(AdjA\right)}^2=\left[\begin{array}{cc}1& -4\\ 0& 1\end{array}\right]$

${\left(AdjA\right)}^{10}=\left[\begin{array}{cc}1& -20\\ 0& 1\end{array}\right]$

$B=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& -4\\ 0& 1\end{array}\right]+\dots +\left[\begin{array}{cc}1& -20\\ 0& 1\end{array}\right]$

$B=\left[\begin{array}{cc}11& -110\\ 0& 11\end{array}\right]$

$\Rightarrow$ sum of elements of $B$$=-88$

$AdjA=\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]$

${\left(AdjA\right)}^2=\left[\begin{array}{cc}1& -4\\ 0& 1\end{array}\right]$

${\left(AdjA\right)}^{10}=\left[\begin{array}{cc}1& -20\\ 0& 1\end{array}\right]$

$B=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& -4\\ 0& 1\end{array}\right]+\dots +\left[\begin{array}{cc}1& -20\\ 0& 1\end{array}\right]$

$B=\left[\begin{array}{cc}11& -110\\ 0& 11\end{array}\right]$

$\Rightarrow$ sum of elements of $B$$=-88$

Key concept ) Differentiation of determinant Now \(f(0)=\left|\begin{array}{ccc}0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2\end{array}\right|\)

\[
\begin{aligned}
&=-2(-2-4)=12\\
&\text { Now } f^{\prime}(x)=\left|\begin{array}{ccc}
3 x^2 & 4 x & 3 \\
3 x^2+2 & 2 x & x^3+6 \\
x^3-x & 4 & x^2-2
\end{array}\right|+\left\lvert\, \begin{array}{ccc}
x^3 & 2 x^2+1 & 3 x+1 \\
6x & 2 & 3 x^2 \\
x^3-x & 4 & x^2-2
\end{array}\right|.
\end{aligned}
\]

$+\left|\begin{array}{ccc}{x}^3& 2x^2+1& 3x+1\\ 3x^2+2& 2x& x^3+6\\ 3x^2-1& 0& 2x\end{array}\right|$

\[
f^{\prime}(0)=\left|\begin{array}{ccc}
0 & 0 & 3 \\
2 & 0 & 6 \\
0 & 4 & -2
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 1 \\
0 & 2 & 0 \\
0 & 4 & -2
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 1 \\
2 & 0 & 6 \\
-1 & 0 & 0
\end{array}\right|
\]

\[
\begin{aligned}
& =-2(0-12)+0+0-1(0+6)+1(0-0) \\
& =24-6=18
\end{aligned}
\]

Mow

\[
\begin{aligned}
2 f(0)+f^{\prime}(0)= & 2.12+18 \\
= & 24+18
\end{aligned}
\]

\(|\mathrm{A}|=3\)

\(\left|\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right)\right)\right|\)

\(\mid-4 \operatorname{adj}\left(-\left.3 \operatorname{adj}\left(3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right)\right|^{2}\right.\)

\(4^{6}\left|\operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right)\right)\right|^{2}\)

\(2^{12} \cdot 3^{12}\left|3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right|^{8}\)

\(2^{12} \cdot 3^{12} \cdot 3^{24}\left|\operatorname{adj}(2 \mathrm{~A})^{-1}\right|^{8}\)

\(2^{12} \cdot 3^{36}\left|(2 \mathrm{~A})^{-1}\right|^{16}\)

\(2^{12} \cdot 3^{36} \frac{1}{|2 \mathrm{~A}|^{16}}\)

\(2^{12} \cdot 3^{36} \frac{1}{2^{48}|\mathrm{~A}|^{16}}\)

\( 2^{12} \cdot 3^{36} \frac{1}{2^{48} \cdot 3^{16}}\)

\(\frac{3^{20}}{2^{36}}=2^{-36} \cdot 3^{20}\)

\(\mathrm{m}=-36 \quad \mathrm{n}=20\)

\(\mathrm{m}+2 \mathrm{n}=4\)

$\left|A\right|=\left|\begin{array}{ccc}2& 2+p& 2+p+q\\ 4& 6+2p& 8+3p+2q\\ 6& 12+3p& 20+6p+3q\end{array}\right|$

$\begin{array}{l}{R}_2\to R_2-2R_1;R_3\to R_3-3R_1\\ \left|A\right|=\left|\begin{array}{ccc}2& 2+p& 2+p+q\\ 0& 2& 4+p\\ 0& 6& 14+3p\end{array}\right|\\ =2(28+6p-24-6p)\\ =2\left(4\right)=8\end{array}$

Now, $adj\mathrm{.}\left(adj\right(3A\left)\right)=\mathrm{\mid }3A{\mathrm{\mid }}^1\left(3A\right)$

$\begin{array}{l}=3^3\mathrm{\mid }A\mathrm{\mid }\left(3A\right)\\ =3^3\times 2^3\times 3^{1}A\end{array}$

$=2^3{3}^{4}A$

$\Rightarrow \det \left(adj\mathrm{.}\right(adj\mathrm{.}\left(3A\right)\left)\right)=\det \left(2^3{3}^{4}A\right)$

$\begin{array}{l}{\left(2^3{3}^4\right)}^3\mathrm{\mid }A\mathrm{\mid }\\ =2^9{3}^{12}\times 2^3=2^{12}{3}^{12}\\ \Rightarrow m=12,n=12\Rightarrow m+n=24\end{array}$

Given:- \(A\) is order 3 square matrix \(|A|=2\)
To find: \(|(\operatorname{adj}\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A)))|\)
we know \(\mid\) adj \(A\left|=|A|^{n-1}\right.\)

\(\begin{aligned}
&|A|=2 \quad and \quad n=3 \\
& \mid \operatorname{adj}(\operatorname{adj}(\operatorname{adi}(\operatorname{adj} A) \mid=|A|^{(n-1)^4} \\
& =|A|^{2^4} \\
&=2^{2^4} \\
&=2^{16}
\end{aligned}\)

$X=A^{-1}B=\left(\frac{\text{adj}A}{\left|A\right|}\right)B$

$\left|adjA\right|=\left|\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right|=100$

$$\begin{gathered} \left|adjA\right|={\left|A\right|}^2=100 \\ \Rightarrow \left|A\right|=\pm 10 \end{gathered}$$

$=\pm \frac{1}{10}\left(\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right)\left(\begin{array}{l}4\\ 0\\ 2\end{array}\right)$

$=\pm \frac{1}{10}\left(\begin{array}{c}20\\ -10\\ 10\end{array}\right)=\pm \left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)$

$\therefore \left|x+y+z\right|=2$

\(\Rightarrow|2 A|^3=2^9 \cdot|A|^3=2^{21}\)

\(\Rightarrow|A|=16 \Rightarrow \alpha^2-\beta^2=16\)

\(\Rightarrow \alpha \text { can be } 5 \text { for } \beta= \pm 3\)

\(\Rightarrow|2 A|^3=2^9 \cdot|A|^3=2^{21}\)

\(\Rightarrow|A|=16 \Rightarrow \alpha^2-\beta^2=16\)

\(\Rightarrow \alpha \text { can be } 5 \text { for } \beta= \pm 3\)

$\left|\begin{array}{ccc}1& \frac{3}{2}& \alpha +\frac{3}{2}\\ 1& \frac{1}{3}& \alpha +\frac{1}{3}\\ 2\alpha +3& 3\alpha +1& 0\end{array}\right|=0$

\(\Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0\)

\(\Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0\)

\(\Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0\)

\(\Rightarrow 2 \alpha^2+6 \alpha+1=0\)

\(\Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2}\)

$\left|\begin{array}{ccc}1& \frac{3}{2}& \alpha +\frac{3}{2}\\ 1& \frac{1}{3}& \alpha +\frac{1}{3}\\ 2\alpha +3& 3\alpha +1& 0\end{array}\right|=0$

\(\Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0\)

\(\Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0\)

\(\Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0\)

\(\Rightarrow 2 \alpha^2+6 \alpha+1=0\)

\(\Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2}\)

$$\begin{gathered} \sin 2x=2\sin x \cos x \\ {R}_2\to R_2-R_1,R_3\to R_3-R_1 \\ f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3& 0& -3\\ 0& 3& -3\end{array}\right| \\ f\left(x\right)=2{\cos }^{4}x\left(0+9\right)-3\left(-6{\sin }^{4}x-9-3{\sin }^{2}2x\right) \\ =18{\cos }^{4}x+18{\sin }^{4}x+9{\sin }^{2}2x+27 \\ =18\left({\left({\cos }^{2}x+{\sin }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\right)+9{\sin }^{2}2x+27 \\ =18-9{\sin }^{2}2x+9{\sin }^{2}2x+27 \\ f\left(x\right)=45 \\ {f}^{\text{'}}\left(x\right)=0 \\ \frac{1}{5}{f}^{\text{'}}\left(0\right)=0 \end{gathered}$$

$$\begin{gathered} \sin 2x=2\sin x \cos x \\ {R}_2\to R_2-R_1,R_3\to R_3-R_1 \\ f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3& 0& -3\\ 0& 3& -3\end{array}\right| \\ f\left(x\right)=2{\cos }^{4}x\left(0+9\right)-3\left(-6{\sin }^{4}x-9-3{\sin }^{2}2x\right) \\ =18{\cos }^{4}x+18{\sin }^{4}x+9{\sin }^{2}2x+27 \\ =18\left({\left({\cos }^{2}x+{\sin }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\right)+9{\sin }^{2}2x+27 \\ =18-9{\sin }^{2}2x+9{\sin }^{2}2x+27 \\ f\left(x\right)=45 \\ {f}^{\text{'}}\left(x\right)=0 \\ \frac{1}{5}{f}^{\text{'}}\left(0\right)=0 \end{gathered}$$

We have $f\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$
$f^{-1}\left(x\right)=\frac{Adjf\left(x\right)}{\left|f\left(x\right)\right|}=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]=f\left(-x\right)$
Now, $f\left(x\right)f\left(y\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos y& -\sin y& 0\\ \sin y& \cos y& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}\cos (x+y)& -\sin (x+y)& 0\\ \sin (x+y)& \cos (x+y)& 0\\ 0& 0& 1\end{array}\right]=f\left(x+y\right)$

We have $f\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$
$f^{-1}\left(x\right)=\frac{Adjf\left(x\right)}{\left|f\left(x\right)\right|}=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]=f\left(-x\right)$
Now, $f\left(x\right)f\left(y\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos y& -\sin y& 0\\ \sin y& \cos y& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}\cos (x+y)& -\sin (x+y)& 0\\ \sin (x+y)& \cos (x+y)& 0\\ 0& 0& 1\end{array}\right]=f\left(x+y\right)$

$$\begin{gathered} \text{Given, }\mathrm{A}, \mathrm{B}\text{ and }\left(adj\left({\mathrm{A}}^{-1}\right)+adj\left({\mathrm{B}}^{-1}\right)\right) \\ \text{ are non-singular matrix} \\ \text{so}, \mathrm{A}, \mathrm{B}\text{ and }\left(adj\left({\mathrm{A}}^{-1}\right)+adj\left({\mathrm{B}}^{-1}\right)\right) \\ \text{ are invertible.} \\ \text{Now}, {\left[\mathrm{A}{\left(adj\left({ \mathrm{A}}^{-1}\right)+adj\left({\mathrm{B}}^{-1}\right)\right)}^{-1}· \mathrm{B}\right]}^{-1} \\ ={\mathrm{B}}^{-1}·\left(adj\left({ \mathrm{A}}^{-1}\right)+adj\left({\mathrm{B}}^{-1}\right)\right)·{\mathrm{A}}^{-1} \\ \mathrm{On} \mathrm{simplification}, \mathrm{we} \mathrm{get} \\ {\mathrm{B}}^{-1}adj\left({ \mathrm{A}}^{-1}\right){\mathrm{A}}^{-1}+{\mathrm{B}}^{-1}\left(adj\left({ \mathrm{B}}^{-1}\right)\right)·{\mathrm{A}}^{-1} \\ {\mathrm{B}}^{-1}\left|{ \mathrm{A}}^{-1}\right|\mathrm{I}+\left|{\mathrm{B}}^{-1}\right|\mathrm{I}{ \mathrm{A}}^{-1} \\ \Rightarrow \frac{adj\mathrm{B}}{| \mathrm{B}|\left|\mathrm{A}\right|}+\frac{adj\mathrm{A}}{| \mathrm{A}|\left|\mathrm{B}\right|} \\ =\frac{1}{\left|\mathrm{A}\right|\left|\mathrm{B}\right|}(adjB+adjA) \end{gathered}$$

\(|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))|=81\)

\(\Rightarrow |A|^2=3\)

Also \(|\operatorname{adj}(\operatorname{adj} A)|^{\frac{(n-1)^2}{2}}=|A|^{3 n^2-5 n-4}\)

\(\Rightarrow 2(n-1)^2=3 n^2-5 n-4\)

\(\Rightarrow n^2-n-6=0, n=3,-2\)

\(\sum_{n \in s}\left|A^{n^2+n}\right|=|A|^{6+3}+|A|^{4-2}=3^6+3=732\)

$$\begin{gathered} \left|\mathrm{A}\right|=\frac{1}{2},trace( \mathrm{A})=3 \\ \mathrm{B}=adj(adj(2\mathrm{A}\left)\right)=\left|2\mathrm{A}{|}^{n-2}\right(2\mathrm{A}) \\ \mathrm{n}=3, \mathrm{B}=|2\mathrm{A}\left|\right(2\mathrm{A}) \\ \Rightarrow B=2^3·|\mathrm{A}\left|\right(2\mathrm{A})=8\mathrm{A} \\ |\mathrm{B}|=|8\mathrm{A}|=8^3·| \mathrm{A}|=2^8=256 \\ trace(\mathrm{B})=8trace( \mathrm{A})=24 \\ |\mathrm{B}|+trace(\mathrm{B})=280 \end{gathered}$$

\(\left|\begin{array}{ccc}1-\alpha & 2 & 7 \\ 4 & -2-\alpha & 8 \\ 3 & 8 & -7-\alpha\end{array}\right|=0\)

\( \Rightarrow(1-\alpha)[(\alpha+2)(\alpha+7)-64]-2[-28-4 \alpha-24]+7[32+6+3 \alpha]=0\)

\(\Rightarrow \alpha^3+8 \alpha^2-88 \alpha-320=0\)

\((\alpha-8)\left(\alpha^2+16 \alpha+40\right)=0\)

\( \alpha=8, \alpha=-8 \pm 2 \sqrt{6}\)

So \( \mathrm{p}=8\)

\(\Rightarrow(\mathrm{x}-8)^2+(\mathrm{y}-16)^2=320=8^2+16^2\)

Put \( \mathrm{y}=0 \Rightarrow \mathrm{x}=16,0\)

Put \( x=0 \Rightarrow y=32,0\)

\(\Rightarrow(16,0),(0,32),(0,0)\)

\(\Rightarrow 3\) points of intersection

$$\begin{gathered} \text{ As }\left|adj\right(adj\mathrm{A}\left)\right|=|\mathrm{A}{|}^{{(\mathrm{n}-1)}^2}\text{,} \\ \text{where }\mathrm{n}\text{ is the order of the determinant. } \\ \text{ So, }\left|adj\left(adj{\mathrm{A}}^2\right)\right|={\left|{\mathrm{A}}^2\right|}^{(3-1{)}^2} \\ ={\left|{\mathrm{A}}^2\right|}^4=|\mathrm{A}{|}^8 \end{gathered}$$

\(\operatorname{Adj}\left(A-2 A^{\top}\right) \operatorname{Adj}\left(2 A-A^{\top}\right) \mid=2^8\)

Let \(\mathrm{P}=\mathrm{A}-2 \mathrm{~A}^{\top}\)

\(Q=2 A-A^{\top} \Rightarrow Q^{\top}=2 A^{\top}-A=-P\)

\(|\operatorname{adj}(P) \operatorname{adj} Q|=2^8\)

\(\Rightarrow\left|Q^{\top}\right|=|-P| \Rightarrow|Q|=-|P|\)

\(|P|^2|Q|^2=2^8 \Rightarrow|P Q|=-2^4\)

\(\Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4\) and \(|Q|=-4\)

\(\left|A-2 A^T\right|=4\)

\(A-2 A^T=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]-2\left[\begin{array}{lll}1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}-1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2\end{array}\right]\)

\(\Rightarrow\left|A-2 A^{\top}\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4\)

\(\Rightarrow|A|^2=16\)

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{cc}{\log }_{5}128& {\log }_{4}5\\ {\log }_{5}8& {\log }_{4}25\end{array}\right] \\ \left|A\right|={\log }_{5}128\times {\log }_{4}25-{\log }_{4}5\times {\log }_{5}8 \\ =7{\log }_{5}2\times 2\left(\frac{1}{2}\right){\log }_{2}5-\frac{1}{2}{\log }_{2}5\times 3{\log }_{5}2 \\ =7-\frac{3}{2}=\frac{11}{2} \end{gathered}$$

Now, ${\mathrm{C}}_{11}=\sum _{\mathrm{k}=1}^2{\mathrm{a}}_{1\mathrm{k}}·{ \mathrm{A}}_{1\mathrm{k}}$

$$\begin{gathered} ={\mathrm{a}}_{11}{ \mathrm{A}}_{11}+{\mathrm{a}}_{12}{ \mathrm{A}}_{12}=\left|\mathrm{A}\right|=\frac{11}{2} \\ {\mathrm{C}}_{12}=\sum _{\mathrm{k}=1}^2{\mathrm{a}}_{1\mathrm{k}}·{ \mathrm{A}}_{2\mathrm{k}}=0 \\ {\mathrm{C}}_{21}=\sum _{\mathrm{k}=1}^2{\mathrm{a}}_{2\mathrm{k}}·{ \mathrm{A}}_{1\mathrm{k}}=0 \\ {\mathrm{C}}_{22}=\sum _{\mathrm{k}=1}^2{\mathrm{a}}_{2\mathrm{k}}·{ \mathrm{A}}_{2\mathrm{k}}=\left|\mathrm{A}\right|=\frac{11}{2} \end{gathered}$$

Therefore, $C=\left[\begin{array}{cc}\frac{11}{2}& 0\\ 0& \frac{11}{2}\end{array}\right]$

$$\begin{gathered} \left|C\right|=\frac{121}{4} \\ \Rightarrow 8\left|C\right|=242 \end{gathered}$$

$X=A^{-1}B=\left(\frac{\text{adj}A}{\left|A\right|}\right)B$

$\left|adjA\right|=\left|\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right|=100$

$$\begin{gathered} \left|adjA\right|={\left|A\right|}^2=100 \\ \Rightarrow \left|A\right|=\pm 10 \end{gathered}$$

$=\pm \frac{1}{10}\left(\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right)\left(\begin{array}{l}4\\ 0\\ 2\end{array}\right)$

$=\pm \frac{1}{10}\left(\begin{array}{c}20\\ -10\\ 10\end{array}\right)=\pm \left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)$

$\therefore \left|x+y+z\right|=2$

\( A=\left[\begin{array}{ll}
1 & 2 \\
1 & \alpha
\end{array}\right] \)
\( A^2-4 A+I=0\)
\(A^2-4 A+I=\left[\begin{array}{cc}0 & 2 \alpha-6 \\ \alpha-3 & \alpha^2-4 \alpha+3\end{array}\right]=O\)
\( \alpha=3 \)
\( B=\left[\begin{array}{ll}
3 & 3 \\
\beta & 2
\end{array}\right] \)
\( \mathrm{B}^2-5 \mathrm{~B}-6 \mathrm{I}=0 \)
\(B^2-5 B-6 I=\left[\begin{array}{cc}3 \beta-12 & 0 \\ 0 & 3 \beta-12\end{array}\right]=O\)
\(\beta=4 \)
\( \mathrm{B}-\mathrm{A}=\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right] \)

\(\mathrm{B}+\mathrm{A}=\left[\begin{array}{ll}
4 & 5 \\
5 & 5
\end{array}\right] \)
\( (B-A)(B+A)=\left[\begin{array}{cc}
13 & 15 \\
7 & 10
\end{array}\right]\)

\([(B-A)(B+A)]^T=\left[\begin{array}{cc}13 & 7 \\ 15 & 10\end{array}\right]\),

so \(S_1\) is incorrect

\(\operatorname{adj}(A)=\left[\begin{array}{cc}3 & -2 \\ -1 & 1\end{array}\right]\)

\(\operatorname{adj \ (B)}=\left[\begin{array}{cc}
2 & -3 \\
-4 & 3
\end{array}\right] \)
\(\operatorname{adj}(A+B)=\left[\begin{array}{cc}5 & -5 \\ -5 & 4\end{array}\right]\)

\(\operatorname{det}(\operatorname{adj}(A+B))=5(4)-(-5)(-5)=20-25=-5\),

so \(S_2\) is correct

Let $\cos \alpha =x$

$\cos \beta =y$

$\cos \gamma =z$

$\left|\begin{array}{lll}0& x& y\\ x& 0& z\\ y& z& 0\end{array}\right|=\left|\begin{array}{lll}1& x& y\\ x& 1& z\\ y& z& 1\end{array}\right|$

Expending both sides, we get

$x^2+y^2+z^2=1$

i.e. ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

Statement 1 is false

Now, $\left|\begin{array}{ccc}{x}^2+x& 1+x& x-2\\ 2x^2+3x-1& 3x& 3x-3\\ x^2+2x+3& 2x-1& 2x-1\end{array}\right|=px+q$