\(A=\left[\begin{array}{rr} 2 & -2 \\ -3 & 4 \end{array}\right]\)
\(A^{2}=A \cdot A\)
\(=\left[\begin{array}{rr} 2 & -2 \\ -3 & 4 \end{array}\right]\left[\begin{array}{rr} 2 & -2 \\ -3 & 4 \end{array}\right] \)
\(=\left[\begin{array}{rr} 4+6 & -4-8 \\ -6-12 & 6+16 \end{array}\right] \)
\(=\left[\begin{array}{rr} 10 & -12 \\ -18 & 22 \end{array}\right] \)
\(\therefore \ -A^{2}+6 A=\left[\begin{array}{rr} -10 & 12 \\ 18 & -22 \end{array}\right]+6\left[\begin{array}{rr} 2 & -2 \\ -3 & 4 \end{array}\right]\)
\(=\left[\begin{array}{rr} -10 & 12 \\ 18 & -22 \end{array}\right]+\left[\begin{array}{rr} 12 & -12 \\ -18 & 24 \end{array}\right]\)
\(=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\)

\(5 \mathrm{~A} -3 \mathrm{~B}+2 \mathrm{C} \)
\(=5\left[\begin{array}{cc} 1 & -2 \\ 3 & 0 \end{array}\right]-3\left[\begin{array}{rr} -1 & 4 \\ 2 & 3 \end{array}\right]+2\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right] \)
\(=\left[\begin{array}{cc} 5 & -10 \\ 15 & 0 \end{array}\right]-\left[\begin{array}{rr} -3 & 12 \\ 6 & 9 \end{array}\right]+\left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right] \)
\(=\left[\begin{array}{cc} 8 & -20 \\ 7 & -9 \end{array}\right] \)

\(\left[ \begin{matrix} x+y+z \\ x+z \\ y+z \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 5 \\ 7 \end{matrix} \right] \)
⇒ x + y + z = 9 ...(1)
⇒ x + z = 5..(2)
⇒ y + z = 7 ...(3)

Sub y = 4 in (3)
4 + z = 7
z = 3
Sub z = 3 in (2)
x + 3 = 5
x = 2
x = 2, y = 4, x = 3

A=\(\left( \begin{matrix} 5 & -4 \\ 6 & -5 \end{matrix} \right) \)
\({ A }^{ 2 }=\begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix}\begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix}\)
\(=\begin{pmatrix} (25-24) & (-20+20) \\ (30-30) & (-24+25) \end{pmatrix}\)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) =I\)
A² = I
Hence it is proved

A = \(\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \) 
\({ A }^{ T }=\left( \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right) \)
\(A.{ A }^{ T }=\left( \begin{matrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix} \right) \left( \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right) \)
\(=\left( \begin{matrix} { cos }^{ 2 }\theta +s{ in }^{ 2 }\theta & -cos\theta sin\theta +cos\theta sin\theta \\ -sin\theta \quad cos\theta +cos\theta sin\theta & { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta \end{matrix} \right) \)
\(\\ =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=I\)
AA^T  = 1
Hence proved.

\(A=\left[\begin{array}{ll} 2 & 5 \\ 4 & 3 \end{array}\right], B=\left[\begin{array}{cc} 1 & -3 \\ 2 & 5 \end{array}\right]\)
\(A B=\left[\begin{array}{ll} 2 & 5 \\ 4 & 3 \end{array}\right]\left[\begin{array}{cc} 1 & -3 \\ 2 & 5 \end{array}\right]\)
\(=\left[\begin{array}{cc} 2+10 & -6+25 \\ 4+6 & -12+15 \end{array}\right]=\left[\begin{array}{cc} 12 & 19 \\ 10 & 3 \end{array}\right]\)
\(\mathrm{BA}=\left[\begin{array}{cc} 1 & -3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 2 & 5 \\ 4 & 3 \end{array}\right]\)
\(=\left[\begin{array}{cc} 2-12 & 5-9 \\ 4+20 & 10+15 \end{array}\right]=\left[\begin{array}{cc} -10 & -4 \\ 24 & 25 \end{array}\right]\)
\(\therefore A B \neq B A\)

Given A has 'a' rows and 'a + 3'columns
B has 'b' rows and '17- b' columns
Since AB and BA exist
For AB, No. of columns in A = No. of rows in B
a + 3 = b
For BA, No. of columns in B = No of rows in A
17 - b = a
17- (a + 3) = a
17- a - 3 = a
2a = 14
a = 7
b = a + 3 = 7 + 3 = 10
a = 7, b = 10.

First, we add the two matrices on both left, right hand sides to get
\(\left[ \begin{matrix} d+3 & 8+a \\ 3b-2 & a-4 \end{matrix} \right] =\left[ \begin{matrix} 2 & 2a+1 \\ b-5 & 4c \end{matrix} \right] \)
Equating the corresponding elements of the two matrices, we have
d + 3 = 2 gives d = –1
8 + a = 2a + 1 gives a = 7
3b - 2 = b - 5 gives b = \(\frac {-3}{2}\)
Substituting a = 7 in a - 4 = 4c gives c = \(\frac {3}{4}\)
Therefore, a = 7, b = \(\frac {-3}{2}\), c = \(\frac {3}{4}\), d = -1.

The given matrices are equal. Thus all corresponding elements are equal.
Therefore, a - b = 1 …(1)
2a + c = 5 …(2)
2a - b = 0 …(3)
3c + d = 2 …(4)
(3) gives 2a - b = 0
2a = b …(5)
Put 2a = b in equation (1), a - 2a = 1 gives a = −1
Put a = −1 in equation (5), 2(-1) = b gives b = −2
Put a = −1 in equation (2), 2(-1) + c = 5 gives c = 7
Put c = 7 in equation (4), 3(7) + d = 2 gives d = −19
Therefore, a = −1, b = −2, c = 7, d = −19

The general 3 x 3 matrix is given by A = \(\left( \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right) \)  a_ij = i²j²
a₁₁ = 1² x 1² = 1 x 1 = 1; a₁₂ = 1² x 2² = 1 x 4 = 4; a₁₃ = 1² x 3² = 1 x 9 = 9
a₂₁ = 2² x 1² = 2 x 1 = 2; a₂₂ = 2² x 2² = 4 x 4 = 16; a₂₃ = 2² x 3² = 4 x 9 = 36
a₃₁ = 3² x 1² = 3 x 1 = 3; a₃₂ = 3² x 2² = 9 x 4 = 36; a₃₃ = 3² x 3² = 9 x 9 = 81
Hence the required matrix is A = \(\left( \begin{matrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{matrix} \right) \)

LHS = (AB)^T 
AB = \({ \left[ \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{matrix} \right] }_{ 2\times 3 }\times { \left[ \begin{matrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{matrix} \right] }_{ 3\times 2 }\)
= \(\left[ \begin{matrix} 2-2+0 & -1+8+2 \\ 4+1+0 & -2-4+2 \end{matrix} \right] =\left[ \begin{matrix} 0 & 9 \\ 5 & -4 \end{matrix} \right] \)
(AB)^T = \({ \left[ \begin{matrix} 0 & 9 \\ 5 & -4 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 0 & 5 \\ 9 & -4 \end{matrix} \right] \) ....(1)
RHS = (B^TA^T)
B^T = \(\left[ \begin{matrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{matrix} \right] \), A^T = \(\left[ \begin{matrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{matrix} \right] \)
B^TA^T = \({ \left[ \begin{matrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{matrix} \right] }_{ 2\times 3 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{matrix} \right] }_{ 3\times 2 }\)
= \(\left[ \begin{matrix} 2-2+0 & 4+1+0 \\ -1+8+2 & -2-4+2 \end{matrix} \right] \)
B^TA^T = \(\left[ \begin{matrix} 0 & 5 \\ 9 & -4 \end{matrix} \right] \)...(2)}
From (1) and (2), (AB)^T = B^TA^T.
Hence proved.

LHS = A(B + C)
B + C = \(\left[ \begin{matrix} 1 & 2 \\ -4 & 2 \end{matrix} \right] +\left[ \begin{matrix} -7 & 6 \\ 3 & 2 \end{matrix} \right] =\left[ \begin{matrix} -6 & 8 \\ -1 & 4 \end{matrix} \right] \)
A(B + C) = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} -6 & 8 \\ -1 & 4 \end{matrix} \right] =\left[ \begin{matrix} -6-1 & 8+4 \\ 6-3 & -8+12 \end{matrix} \right] =\left[ \begin{matrix} -7 & 12 \\ 3 & 4 \end{matrix} \right] \) ......(1)
RHS = AB + AC
AB = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 2 \\ -4 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1-4 & 2+2 \\ -1-12 & -2+6 \end{matrix} \right] =\left[ \begin{matrix} -3 & 4 \\ -13 & 4 \end{matrix} \right] \)
AC = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} -7 & 6 \\ 3 & 2 \end{matrix} \right] =\left[ \begin{matrix} -7+3 & 6+2 \\ 7+9 & -6+6 \end{matrix} \right] =\left[ \begin{matrix} -4 & 8 \\ 16 & 0 \end{matrix} \right] \)
Therefore, AB + AC = \(\left[ \begin{matrix} -3 & 4 \\ -13 & 4 \end{matrix} \right] +\left[ \begin{matrix} -4 & 8 \\ 16 & 0 \end{matrix} \right] =\left[ \begin{matrix} -7 & 12 \\ 3 & 4 \end{matrix} \right] \) ....(2)
From (1) and (2), A(B + C) = AB + AC. Hence proved.

LHS (AB)C
AB = \({ \left[ \begin{matrix} 1 & -1 & 2 \end{matrix} \right] }_{ 1\times 3 }{ \left[ \begin{matrix} 1 & -1 \\ 2 & 1 \\ 1 & 3 \end{matrix} \right] }_{ 3\times 2 }=\left[ \begin{matrix} 1-2+2 & -1-1+6 \end{matrix} \right] =\left[ \begin{matrix} 1 & 4 \end{matrix} \right] \)
(AB)C = \({ \left[ \begin{matrix} 1 & 4 \end{matrix} \right] }_{ 1\times 2 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right] }_{ 2\times 2 }=\left[ \begin{matrix} 1+8 & 2-4 \end{matrix} \right] =\left[ \begin{matrix} 9 & -2 \end{matrix} \right] \) ....(1)
RHS = A(BC)
BC = \({ \left[ \begin{matrix} 1 & -1 \\ 2 & 1 \\ 1 & 3 \end{matrix} \right] }_{ 3\times 2 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right] }_{ 2\times 2 }=\left[ \begin{matrix} 1-2 & 2+1 \\ 2+2 & 4-1 \\ 1+6 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 3 \\ 4 & 3 \\ 7 & -1 \end{matrix} \right] \)
A(BC) = \({ \left[ \begin{matrix} 1 & -1 & 2 \end{matrix} \right] }_{ 1\times 3 }{ \left[ \begin{matrix} -1 & 3 \\ 4 & 3 \\ 7 & -1 \end{matrix} \right] }_{ 3\times 2 }\)
A(BC) = \(\left[ \begin{matrix} -1-4+14 & 3-3-2 \end{matrix} \right] =\left[ \begin{matrix} 9 & -2 \end{matrix} \right] \) ....(2)
From (1) and (2), (AB)C = A(BC).

\({ \left[ \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right] }_{ 2\times 2 }{ \left[ \begin{matrix} x \\ y \end{matrix} \right] }_{ 2\times 1 }=\left[ \begin{matrix} 4 \\ 5 \end{matrix} \right] \)
By matrix multiplication \(\left[ \begin{matrix} 2x+y \\ x+2y \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 5 \end{matrix} \right] \)
Rewriting 2x + y = 4  ....(1)
x + 2y = 5  ....(2)

Substituting y = 2 in (1), 2x + 2 = 4 gives x = 1
Therefore, x = 1, y = 2.

x²-4x = 5
y²-2y = 8
y²-2y-8 = 0
(y-4) (y+2) = 0
y = 4, -2
x² - 4x - 5 = 0
(x - 5) (x + 1) = 0
x = 5, -1
x = -15, y = 4, -2

\(\left[ \begin{matrix} 2{ x }^{ 2 } & 2x \\ 3x & { x }^{ 2 } \end{matrix} \right] +2\left[ \begin{matrix} 8 & 5x \\ 4 & 4x \end{matrix} \right] =\left[ \begin{matrix} { 2x }^{ 2 }+16 & 24 \\ 10 & 6x \end{matrix} \right] \)
12x=48 ⇒ x=4

4x - 2y = 4   ...(1) 
-3x + 3y = 6  ...(2)

y  =   6                                          
Sub y = 6 in (1) ⇒ 4x - 2(6) = 4
4x  = 16
x = 4
x = 4, y = 6

X + Y = \(\left[ \begin{matrix} 7 & 0 \\ 3 & 5 \end{matrix} \right] \)    ...(1)
X - Y= \(\left[ \begin{matrix} 3 & 0 \\ 0 & 4 \end{matrix} \right] \)   ...(2)
______________
\((1)+(2)\Rightarrow 2x=\left[ \begin{matrix} 10 & 0 \\ 3 & 9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 \\ \frac { 3 }{ 2 } & \frac { 9 }{ 2 } \end{matrix} \right] \)
\((1)-(2)\Rightarrow X+Y=\left[ \begin{matrix} 7 & 0 \\ 3 & 5 \end{matrix} \right] \)

\(2Y=\left[ \begin{matrix} 4 & 0 \\ 3 & 1 \end{matrix} \right] \Rightarrow Y=\frac { 1 }{ 2 } \left[ \begin{matrix} 4 & 0 \\ 3 & 1 \end{matrix} \right] \)
\(\therefore Y=\left[ \begin{matrix} 2 & 0 \\ \frac { 3 }{ 2 } & \frac { 1 }{ 2 } \end{matrix} \right] \)
\(X=\left[ \begin{matrix} 5 & 0 \\ \frac { 3 }{ 2 } & \frac { 9 }{ 2 } \end{matrix} \right] , Y=\left[ \begin {matrix} 2 & 0 \\ \frac { 3 }{ 2 } & \frac { 1 }{ 2 } \end{matrix} \right] \)

3A + 2B - C = \(3\left[ \begin{matrix} 1 & 8 & 3 \\ 3 & 5 & 0 \\ 8 & 7 & 6 \end{matrix} \right] +2\left[ \begin{matrix} 8 & -6 & -4 \\ 2 & 11 & -3 \\ 0 & 1 & 5 \end{matrix} \right] -\left[ \begin{matrix} 5 & 3 & 0 \\ -1 & -7 & 2 \\ 1 & 4 & 3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 3 & 24 & 9 \\ 9 & 15 & 0 \\ 24 & 21 & 18 \end{matrix} \right] +\left[ \begin{matrix} 16 & -12 & -8 \\ 4 & 22 & -6 \\ 0 & 2 & 10 \end{matrix} \right] +\left[ \begin{matrix} -5 & -3 & 0 \\ 1 & 7 & -2 \\ -1 & -4 & -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 14 & 9 & 1 \\ 14 & 44 & -8 \\ 23 & 19 & 25 \end{matrix} \right] \)