\(LHS=\left| \begin{matrix} -{ a }^{ 2 } & ab & ac \\ ab & -{ b }^{ 2 } & bc \\ ac & bc & -{ c }^{ 2 } \end{matrix} \right| ={ 4a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\) 
Taking a from R₁ b from R₂ and c from R₃ as common factors we get,
= \(\left( abc \right) \left| \begin{matrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{matrix} \right| \) 
Again taking a from C₁' b from C₂ and e from C₃ we get,
= \(\left( abc \right) \left( abc \right) \left| \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right| \) 
=  \(\left( abc \right) ^{ 2 }\left| \begin{matrix} 0 & 0 & 2 \\ 2 & 0 & 0 \\ 1 & 1 & -1 \end{matrix} \right| \begin{matrix} { R }_{ 1 }\rightarrow { { R }_{ 1 }+{ R }_{ 2 } } \\ { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 } \end{matrix}\) 
= \(\left( abc \right) { c }^{ 2 }\left[ 0+0+2\left| \begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix} \right| \right] \) 
= (abc)² (4) = 4a²b²c² = RHS

LHS = \(\left| \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| \) 
= \(\left| \begin{matrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\) 
= \(\left( a+b+c \right) \left| \begin{matrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| \) 
= \(\left( a+b+c \right) \left| \begin{matrix} 1 & 1 & 1 \\ 2b-b+c+a & b-c-a-2b & 2b \\ 0 & 2c-c+a+b & c-a-b \end{matrix} \right| \begin{matrix} { C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 2 } \\ { C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 } \end{matrix}\)
= \(\left( a+b+c \right) \left| \begin{matrix} 0 & 0 & 1 \\ a+b+c & -(a+b+c) & 2b \\ 0 & a+b+c & c-a-b \end{matrix} \right| \) 
Expanding along R_I
= \(\left( a+b+c \right) \left\{ 0()-0()+1\left| \begin{matrix} a+b+c & -(+b+c) \\ 0 & a+b+ \end{matrix} \right| \right\} \) 
= (a + b + c) [1 (a + b + c)² - 0]
= (a + b + c)³ = RHS

Let \(\triangle\) = \(\left| \begin{matrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \end{matrix} \right| \)
Expanding along R₁ we get,
\(\Rightarrow \triangle =x\left| \begin{matrix} -x & 1 \\ 1 & x \end{matrix} \right| -sin\theta \left| \begin{matrix} -sin\theta & 1 \\ cos\theta & x \end{matrix} \right| +cos\theta \left| \begin{matrix} -sin\theta & -x \\ cos\theta & 1 \end{matrix} \right| \)
\(\Rightarrow \triangle \) = x(-x² - 1)-sin\(\theta\)(-x sin\(\theta\)-cos\(\theta\)) + cos\(\theta\) (-sin\(\theta\)  + x cos\(\theta\))

\(\Rightarrow \triangle \)=-x³ - x + x(sin²\(\theta\) + cos²\(\theta\))
\(\Rightarrow \triangle \)=-x³ - x + x(1)    [sin²\(\theta\) + cos²\(\theta\) = 1]
\(\Rightarrow \) = 
\(\Rightarrow \triangle \) = -x³ which is independent of \(\theta\).

Given A² - B² = (A - B)(A + B)
\(\Rightarrow\) A² - B² =(A - B)A + (A - B)B
[Distributive property of matrix multiplication over addition]
\(\Rightarrow\) A² - B² = A² - BA + AB - B^{2
\(\Rightarrow\)
\(\Rightarrow\)} BA - AB = 0
^{\(\Rightarrow\)} AB = BA
Thus, the given matrix is true if the matrices A and B are commutative with each other.

Given \(x\left[ \begin{matrix} 2x & 2 \\ 3 & x \end{matrix} \right] +2\left[ \begin{matrix} 8 & 5x \\ 4 & 4x \end{matrix} \right] =\left[ \begin{matrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { 2x }^{ 2 } & 2x \\ 3x & { x }^{ 2 } \end{matrix} \right] +\left[ \begin{matrix} 16 & 10x \\ 8 & 8x \end{matrix} \right] =\left[ \begin{matrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 6x \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 2{ { x }^{ 2 }+16 } & 2x+10x \\ 3x+8 & { x }^{ 2 }+8x \end{matrix} \right] =\left[ \begin{matrix} 2{ x }^{ 2 }+16 & 48 \\ 20 & 12x \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 2{ { x }^{ 2 }+16 } & 12x \\ 3x+8 & { x }^{ 2 }+8x \end{matrix} \right] =\left[ \begin{matrix} 2{ x }^{ 2 }+16 & 48 \\ 20 & 12x \end{matrix} \right] \)
Equating the corresponding entries on both sides, we get
12x = 48 \(\Rightarrow\) x = 4
and x² + 8x = 12x \(\Rightarrow\) x² - 4x = 0 \(\Rightarrow\) x(x - 4) = 0
\(\Rightarrow\) x = 0,4
Since x = 0 is not possible \(\Rightarrow\) x = 4.