|A| = d₁,d₂.d₃......d_n

For a non-singular matrix of order n > 1,
Adj(AdjA) = (|A|)n-2. A
∴  if order of matrix A is 3 x 3 and |A| = 5, then
Adj(AdjA) = 5A

|A| = ±10

k = 10/3

0 (Using properties)

We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

0

\(\text { If } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right], \text { then adj } A=\left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right] \text { . }\)
So, Adj \(A=\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}\)

\(2 x^{2}-15 =32-15 \Rightarrow x^{2}=16 \)
\(\Rightarrow x =\pm 4 \Rightarrow x=4( > 0)\)

\(|3 A|=3^{3}|A|=k|A| \Rightarrow k=27 \)

Inconsistent

Here, \(\left| A \right| =\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
= 1(0+0) + 1(9+2) + 2(0-0)
= 11
\(\Rightarrow \left| A \right| =\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ...(i)\)
\(adj\quad A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \quad \)
\(A(adj\quad A)=\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| \)
\((adj\quad A)A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ....(i)\)
Thus, it is verified that A(adj A) = (adj A)A = |A|I

\(\Delta=\left| \begin{matrix} { yz }-x^{ 2 } & { zx }-y^{ 2 } & { xy-z }^{ 2 } \\ { zx-y }^{ 2 } & { xy }-z^{ 2 } & { yz-x }^{ 2 } \\ { xy-z }^{ 2 } & { yz }-x^{ 2 } & { zx-x }^{ 2 } \end{matrix} \right| \)
\(=\begin{vmatrix} -(x^2+y^2+z^2-xy-yz-zx)&zx-y^2&xy-z^2\\-(x^2+y^2+z^2-xy-yz-zx)&xy-z^2&yz-x^2\\-(x^2+y^2+z^2-xy-yz-zx)&yz-x^2&zx-y^2\end{vmatrix}\) [Operating C₁-->C₁+C₂+C₃]
\(= -(x^2+y^2+z^2-xy-yz-zx)\begin{vmatrix}1&zx-y^2&xy-z^2\\1&xy-z^2&yz-x^2\\1&yz-x^2&zx-y^2\end{vmatrix}\)  [Taking -(x²+y²+z²-xy-yz-zx) common from C₁]
=-(x²+y²+z²-xy-yz-zx)
\(\begin{vmatrix}0&(x-y)(x+y+z)&(y-z)(x+y+z)\\0&(x-y)(x+y+z )&(y-z)(x+y+z)\\1&yz-x^2&zx-y^2 \end{vmatrix}\) [Operating R₁-->R₁-R₃ and R₂+R₂-R₃]
\(=-(x^2+y^2+z^2-xy-yz-zx)(x+y +z)\)
\(=\begin{vmatrix} 0&x-y&y-z\\0&x-z&y-x\\1&yz-x^2&zx-y^2\end{vmatrix}\) [Taking (x+y+z)common from C₂and C₃]
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)\begin{vmatrix}0&x-y&y-z\\0&x-z&y-x \\1&yz-x^2&zx-y^2\end{vmatrix}\)
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(z-y)(y-x)-(x-z)(x-2)]\)
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(yz-zx-y^2+xy-x^2-zx+zx-z^2)]\) [Expanding by C₁]
 \(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(x^2+y^2+z^2-xy-yz-zx)]\)
Hence, △ is divisible by (x+y+z) and quotient is (x³+y³+z³-3xyz)(x²+y²+z²-x-yz-zx)

\(\text { Applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} \text { to the given determinant, we get }\)
\(\Delta=\left|\begin{array}{lll} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{array}\right|=(a+b+c)\left|\begin{array}{lll} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array}\right|\)
\(=(a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{array}\right|\left(\text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\right)\)
= (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C₁)
= (a + b + c)(– a² – b² – c² + ab + bc + ca)
\(=\frac{-1}{2}(a+b+c)\left(2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a\right) \)
\(=\frac{-1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \)
which is negative (since a + b + c > 0 and (a – b)² + (b – c)²+ (c – a)²> 0)

LHS: \(\begin{vmatrix}a+b+2c&a&b\\c&b+c+2a&b\\c&c&c+a+2b \end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix} 1&-1&0\\0&1&-1\\c&a&c+a+2b\end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix}1&0&0\\0&1&-1\\ c&a+c&c+a+2b\end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix} 1&-1\\a+c&c+a+2b\end{vmatrix}\)
= (a+b+c)²[(c+a+2b)+(a+c)]
= (a+b+c)²(2a+2b+2c)
= 2(a+b+c)²(a+b+c)
= 2(a+b+c)³ = RHS

(i)\(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4 \)
\(\Rightarrow 2-20=2 x^{2}-24 \)
\(\Rightarrow 2 x^{2}=6 \)
\(\Rightarrow x^{2}=3 \)
\(\Rightarrow x=\pm \sqrt{3} \)
(ii)We have:
\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\2x & 5 \end{vmatrix}\)
\(\Rightarrow 2 \times 5-3 \times 4=x \times 5-3 \times 2 x \)
\(\Rightarrow 10-12=5 x-6 x \)
\(\Rightarrow-2=-x \)
\(\Rightarrow x=2\)

We have \(A^2=A \cdot A=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]\)
Hence \(\mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]-\left[\begin{array}{cc} 8 & 12 \\ 4 & 8 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=\mathrm{O}\)
\(\begin{array}{l} \mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\mathrm{O} \\ \mathrm{A} \mathrm{A}-4 \mathrm{~A}=-\mathrm{I} \end{array}\)
or \( \mathrm{A} \mathrm{A}\left(\mathrm{A}^{-1}\right)-4 \mathrm{AA}^{-1}=-\mathrm{IA}^{-1} \) (Post multiplying by A–1 because |A| ≠ 0)
or \( {A}\left(\mathrm{~A} \mathrm{~A}^{-1}\right)-4 \mathrm{I}=-\mathrm{A}^{-1} \)
or \(\mathrm{AI}-4 \mathrm{I}=-\mathrm{A}^{-1} \)
or \(\mathrm{A}^{-1}=4 \mathrm{I}-\mathrm{A}=\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]-\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\)
Hence \(A^{-1}=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\)

Expanding R₁, we get:
\(\Delta =0\left| \begin{matrix} 0 & sin\beta \\ -sin\beta & 0 \end{matrix} \right| -sin\alpha \begin{vmatrix} -sin\alpha & sin\beta \\ cos\alpha & 0 \end{vmatrix}-cos\alpha \begin{vmatrix} -sin\alpha & 0 \\ cos\alpha & -sin\beta \end{vmatrix}\)
= 0 - sin \(\alpha\) (-0-sin \(\beta\) cos \(\alpha\))-cos \(\alpha\) (sin \(\alpha\) sin \(\beta\) - 0)
= sin \(\alpha\) sin \(\beta\) cos \(\alpha\) - cos \(\alpha\) sin \(\alpha\) sin \(\beta\) = 0

First find cofactors of each element.
a₁₁, A₂₁ + a₁₂ A₂₂ + a₁₃ A₂₃ = 0 as cofactors of a₁₁, a₁₂ and a₁₃ are not A₂₁, A₂₂ and A₂₃