\(\text { Given, } A=\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\)
\(\Rightarrow \quad A^2  =A \times A=\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right] \times\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\)
\(=\left[\begin{array}{lll} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3-0 \end{array}\right]\)
\(\Rightarrow A^2  =\left[\begin{array}{rrr} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]\)
\( \text { Now, } A^2-5 A=\left[\begin{array}{lll} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-5\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-\left[\begin{array}{rrr} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 5-10 & -1-0 & 2-5 \\ 9-10 & -2-5 & 5-15 \\ 0-5 & -1+5 & -2-0 \end{array}\right]=\left[\begin{array}{rrr} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{array}\right]\)

Given, 2A - 3B + 5C = O
\(\Rightarrow 2 A=3 B-5 C\)
\(\Rightarrow 2 A=3\left[\begin{array}{rrr} -2 & 2 & 0 \\ 3 & 1 & 4 \end{array}\right]-5\left[\begin{array}{rrr} 2 & 0 & -2 \\ 7 & 1 & 6 \end{array}\right]\)
\(\Rightarrow 2 A=\left[\begin{array}{rrr} -6 & 6 & 0 \\ 9 & 3 & 12 \end{array}\right]-\left[\begin{array}{rrr} 10 & 0 & -10 \\ 35 & 5 & 30 \end{array}\right]\)
\(\Rightarrow 2 A=\left[\begin{array}{rrr} -6-10 & 6-0 & 0-(-10) \\ 9-35 & 3-5 & 12-30 \end{array}\right]\)
\( \Rightarrow A=\frac{1}{2}\left[\begin{array}{rrr} -16 & 6 & 10 \\ -26 & -2 & -18 \end{array}\right] \Rightarrow A=\left[\begin{array}{rrr} -8 & 3 & 5 \\ -13 & -1 & -9 \end{array}\right]\)

Given, \(A =\left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right] \)
LHS  = (A - 2I)(A - 3I) 
\( =\left\{\left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\right\}\left\{\left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]\right\}\)
\( =\left[\begin{array}{rr} 2 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{rr} 1 & 2 \\ -1 & -2 \end{array}\right]\)
\(=\left[\begin{array}{rr} 2-2 & 4-4 \\ -1+1 & -2+2 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=O=\mathrm{RHS}\)
Hence proved

Given \(\left[\begin{array}{cc}x & x-y \\ 2 x+y & 7\end{array}\right]=\left[\begin{array}{ll}3 & 1 \\ 8 & 7\end{array}\right]\)
On equating the corresponding elements, we get
x = 3 and x - y = 1 ⇒ y = x - 1 = 3 -1 = 2

Given matrix equation is \( {\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]\left[\begin{array}{cc} 1 & -3 \\ -2 & 4 \end{array}\right]}  =\left[\begin{array}{ll} -4 & 6 \\ -9 & x \end{array}\right] \)
\(\Rightarrow {\left[\begin{array}{cc} 2-6 & -6+12 \\ 5-14 & -15+28 \end{array}\right]} =\left[\begin{array}{ll} -4 & 6 \\ -9 & x \end{array}\right]\)
\(\Rightarrow  {\left[\begin{array}{cc} -4 & 6 \\ -9 & 13 \end{array}\right]} =\left[\begin{array}{ll} -4 & 6 \\ -9 & x \end{array}\right]\)
On equating the corresponding elements, we get
x = 13

Given \(\left[\begin{array}{ll}2 x & 4\end{array}\right]\left[\begin{array}{c}x \\ -8\end{array}\right]=O\) ⇒ [2x² - 32] = 0
On equating the corresponding elements, we get
⇒ 2x² - 32 = 0
⇒ 2x² - 32 ⇒ x² - 16
⇒ x = 士4
∴ Positive value of x is 4.

Given, \(\left[\begin{array}{cc}x \cdot y & 4 \\ z+6 & x+y\end{array}\right]=\left[\begin{array}{ll}8 & w \\ 0 & 6\end{array}\right]\)
On equating the corresponding elements, we get 
x . y  = 8 ..(i) 
z+6 =0 \(\Rightarrow z=-6\) ..(ii)
and x+y  =6 ...(iii)
Now, on adding Eqs. (ii) and (iii), we get \( x+y+z=6+(-6)=0\)

We have, \(A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\)
Now, \(A^2=A \times A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\)
\(\Rightarrow  A^2=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \)
\(\Rightarrow  A^3=A^2 \cdot A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right] \)
\(\Rightarrow A^3=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right]\)

k = 2

2

Given matrix equation is
\( \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]=O \)
\(\Rightarrow \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]\right)=O \) 
[by associative law of multiplication]
\(\Rightarrow \left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{c}x+16 \\ -3 x\end{array}\right]=0\)
\(\Rightarrow [2 x(x+16)-9 x]=[O]\)
\(\Rightarrow\) 2 x^{2} + 32 x - 9x = 0
\(\Rightarrow\)\(2 x^{2}+23 x=0\)
\(\Rightarrow x(2 x+23)=0\)
\(\Rightarrow\) x = 0 and x = -23 / 2

Given, \(2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right] \)
\(\Rightarrow \cdot\left[\begin{array}{cc}6 & 8 \\ 10 & 2 x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right] \)
\(\Rightarrow \left[\begin{array}{cc}7 & 8+y \\ 10 & 2 x+1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right]\)
On comparing the corresponding elements, we get
8+y = 0 and 2x+1 = 5
\(\Rightarrow\)  y = -8 and x = \frac{5-1}{2} = 2
∴  x - y = 2-(-8) = 10

a₁₁ = a₂₂ = 0,
a₁₂ = a₂₁ = 2
\(\Rightarrow A=\left[ \begin{matrix} 0 & 2 \\ 2 & 0 \end{matrix} \right] \)

x + y = 7
Alternate Method :
Given  \(\left[ \begin{matrix} 2x+1 & 2y \\ 0 & { y }^{ 2 }+1 \end{matrix} \right] =\left[ \begin{matrix} x+3 & 10 \\ 0 & 26 \end{matrix} \right] \)
By equating, 2x + 1 = x + 3  ⇒  x = 2
2y = 10
⇒  y = 5
∴  x + y = 2 + 5 = 7

\(A=\left[ \begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix} \right]\)
Alternate Method :
Given \(3A-B=\left[ \begin{matrix} 5 & 0 \\ 1 & 1 \end{matrix} \right] , B=\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \)
\(\Rightarrow 3A-\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] =\left[ \begin{matrix} 5 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(\Rightarrow 3A=\left[ \begin{matrix} 5 & 0 \\ 1 & 1 \end{matrix} \right] +\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \)
\(\Rightarrow 3A=\left[ \begin{matrix} 9 & 3 \\ 3 & 6 \end{matrix} \right] \)
\(\Rightarrow A=\left[ \begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix} \right]\)

x + y + z = 10
Alternate Method :
\(\left[ \begin{matrix} x-y & 2y \\ 2y+z & x+y \end{matrix} \right] =\left[ \begin{matrix} 1 & 4 \\ 9 & 5 \end{matrix} \right]\)
By equating,   x - y = 1 ...(i)
and  2y = 4 ⇒  y = 2 .. (ii)
Put y = 2 in (i) ⇒  x - 2 = 1 ⇒  x = 3
2y + z = 9
⇒  2(2) + z = 9
⇒  z = 9 - 4 ⇒  z = 5
∴∴   x + y + z = 3 + 2 + 5 = 10

Given
\(\left[ \begin{matrix} x & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 \\ -2 & 0 \end{matrix} \right] =0 \)
By using matrix multiplication, we get [ x - 2 0] = [0  0] 
On equating the corresponding elements, we get
\(\Rightarrow   \left[ \begin{matrix} x-2 & 0 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \end{matrix} \right]\)
By equating, x - 2 = 0 ⇒  x = 2

x - y = 8
Alternate Method :
\(2\left[ \begin{matrix} 3 & 4 \\ 5 & x \end{matrix} \right] =\left[ \begin{matrix} 1 & y \\ 10 & 5 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 6 & 8 \\ 10 & 2x \end{matrix} \right] =\left[ \begin{matrix} 1 & y \\ 0 & 5 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 7 & 8+y \\ 10 & 2x+5 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right]\)
By equating,
8 + y = 0  ⇒  y = - 8
and 2x + 5 = 5 ⇒  x = 0
∴  x - y = 0 - (- 8)
= 8

\(x=\pm 4\)
Alternate Method :
\(\left( \begin{matrix} 2x & 4 \end{matrix} \right) \left( \begin{matrix} x \\ -8 \end{matrix} \right) =0 \Rightarrow \left( { 2x }^{ 2 }-32 \right) =0\)
By equating,
2x² - 32 = 0
\(\Rightarrow   2x^2 = 32 \)
\(\Rightarrow   x^2 = 16 \)
\(\Rightarrow   x=\pm 4 \Rightarrow  x = 4\)

x + y + z = 0
Alternate method :
\(\left[ \begin{matrix} xy & 4 \\ z+6 & x+y \end{matrix} \right] =\left[ \begin{matrix} 8 & w \\ 0 & 6 \end{matrix} \right] \)
By equating z + 6 = 0 and x + y = 6
⇒  z = - 6, x + y = 6
x + y + z = 6 - 6
⇒  x + y + z = 0

\(\left| AdjA \right| =64\left| A \right| ^{ 3-1 }\Rightarrow \left| A \right| ^{ 2 }=64\Rightarrow \left| A \right| =\pm 8\)

Given \( \left| A \right| =2\Rightarrow \left| A' \right| =2\)
Also \(\left| AA' \right| =\left| A \right| \left| A' \right| =2\times 2=4.\)

Given, matrix equation can be rewritten as \(\left[ \begin{matrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{matrix} \right] =A+\left[ \begin{matrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{matrix} \right] \)
\(\Rightarrow A=\left[ \begin{matrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{matrix} \right] -\left[ \begin{matrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{matrix} \right] =\left[ \begin{matrix} 9-1 & -1-2 & 4+1 \\ -2-0 & 1-4 & 3-9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 8 & -3 & 5 \\ -2 & -3 & -6 \end{matrix} \right]\)

\({ A }^{ T }-{ B }^{ T }=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} 3+1 & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{matrix} \right] =\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{matrix} \right] \)

\(\begin{bmatrix} 3x\quad + & 4 \\ 2x\quad + & x \end{bmatrix}=\left[ \begin{matrix} 19 \\ 15 \end{matrix} \right] \Rightarrow 3x=15\Rightarrow x=5\)