We have to show that AB = BA
\(LHS\quad AB=\begin{pmatrix} 2 & -2\sqrt { 2 } \\ \sqrt { 2 } & 2 \end{pmatrix}\times \begin{pmatrix} 2 & 2\sqrt { 2 } \\ -\sqrt { 2 } & 2 \end{pmatrix}\)
\(=\left( \begin{matrix} 4+4 & 4\sqrt { 2 } -4\sqrt { 2 } \\ 2\sqrt { 2 } -2\sqrt { 2 } & 4+4 \end{matrix} \right) \)
\(=\left( \begin{matrix} 8 & 0 \\ 0 & 8 \end{matrix} \right) \)
\(RHS\quad AB=\begin{pmatrix} 2 & 2\sqrt { 2 } \\ -\sqrt { 2 } & 2 \end{pmatrix}\times \begin{pmatrix} 2 & -2\sqrt { 2 } \\ \sqrt { 2 } & 2 \end{pmatrix}\)
\(=\left( \begin{matrix} 4+4 & -4\sqrt { 2 } +4\sqrt { 2 } \\ -2\sqrt { 2 } +2\sqrt { 2 } & 4+4 \end{matrix} \right) \)
\(=\left( \begin{matrix} 8 & 0 \\ 0 & 8 \end{matrix} \right) \)
Hence LHS = RHS (i.e.) AB = BA

( A - B )C = AC - BC
L.H.S = (A - B)C
\(A-B=\left[ \begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix} \right] -\left[ \begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix} \right] =\left[ \begin{matrix} -3 & 2 \\ 0 & -2 \end{matrix} \right] \)
\((A-B)C=\left[ \begin{matrix} -3 & 2 \\ 0 & -2 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 \\ 1 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} (-6+2) & (0+4) \\ (0-2) & (0-4) \end{matrix} \right] \)
\(=\left[ \begin{matrix} -4 & 4 \\ -2 & -4 \end{matrix} \right] ..(1)\)
R.H.S = AC - BC
\(AC=\left[ \begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 \\ 1 & 2 \end{matrix} \right] =\left[ \begin{matrix} (2+2) & (0+4) \\ (2+3) & (0+6) \end{matrix} \right] \)
\(=\left[ \begin{matrix} 4 & 4 \\ 5 & 0 \end{matrix} \right] \)
\(BC=\left[ \begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 \\ 1 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} (8+0) & (0+0) \\ (2+5) & (0+10) \end{matrix} \right] \)
\(=\left[ \begin{matrix} 8 & 0 \\ 7 & 10 \end{matrix} \right] \)
\(AC-BC=\left[ \begin{matrix} 4 & 4 \\ 5 & 6 \end{matrix} \right] -\left[ \begin{matrix} 8 & 0 \\ 7 & 10 \end{matrix} \right] =\left[ \begin{matrix} -4 & 4 \\ -2 & -4 \end{matrix} \right] \)
(1) = (2) ⇒ LHS = RHS. Hence verified.

\(\\ \frac { 1 }{ 2 } A-\frac { 3 }{ 2 } B\) = \(\frac {1}{2}\)(A - 3B)
= \(\frac { 1 }{ 2 } \left( \left[ \begin{matrix} 1 & 8 & 3 \\ 3 & 5 & 0 \\ 8 & 7 & 6 \end{matrix} \right] -3\left[ \begin{matrix} 8 & -6 & -4 \\ 2 & 11 & -3 \\ 0 & 1 & 5 \end{matrix} \right] \right) \)
= \(\frac { 1 }{ 2 } \left[ \left[ \begin{matrix} 1 & 8 & 3 \\ 3 & 5 & 0 \\ 8 & 7 & 6 \end{matrix} \right] +\left[ \begin{matrix} -24 & 18 & 12 \\ -6 & -33 & 9 \\ 0 & -3 & -15 \end{matrix} \right] \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} -23 & 26 & 15 \\ -3 & -28 & 9 \\ 8 & 4 & -9 \end{matrix} \right] \)
= \(\left[ \begin{matrix} -\frac { 23 }{ 2 } & 13 & \frac { 15 }{ 2 } \\ -\frac { 3 }{ 2 } & -14 & \frac { 9 }{ 2 } \\ 4 & 2 & -\frac { 9 }{ 2 } \end{matrix} \right] \)

\(A=\left[\begin{array}{lll} 5 & 2 & 9 \\ 1 & 2 & 8 \end{array}\right], B=\left[\begin{array}{rr} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{array}\right]\)
\(A B=\left[\begin{array}{lll} 5 & 2 & 9 \\ 1 & 2 & 8 \end{array}\right]\left[\begin{array}{rr} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{array}\right]\)
\(=\left[\begin{array}{rr} 5+2+45 & 35+4-9 \\ 1+2+40 & 7+4-8 \end{array}\right]\)
\(=\left[\begin{array}{lr} 52 & 30 \\ 43 & 3 \end{array}\right]\)
\((A B)^{T}=\left[\begin{array}{rr} 52 & 43 \\ 30 & 3 \end{array}\right]\)
\(\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=\left[\begin{array}{rrr} 1 & 1 & 5 \\ 7 & 2 & -1 \end{array}\right]\left[\begin{array}{ll} 5 & 1 \\ 2 & 2 \\ 9 & 8 \end{array}\right]\)
\(=\left[\begin{array}{cc} 5+2+45 & 1+2+40 \\ 35+4-9 & 7+4-8 \end{array}\right]\)
\(=\left[\begin{array}{ll} 52 & 43 \\ 30 & 3 \end{array}\right]\)
From (1) and (2)
(AB)^T = B^T A ^T
Hence proved.

\(\mathrm{A}=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right], \mathrm{I}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
\(\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\)
\(=\left[\begin{array}{ll} a^{2}+b c & a b+b d \\ a c+c d & b c+d^{2} \end{array}\right]\)
\((a+d) A=(a+d)\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\)
\(=\left[\begin{array}{ll} a(a+d) & b(a+d) \\ c(a+d) & d(a+d) \end{array}\right]\)
\(=\left[\begin{array}{ll} a^{2}+a d & a b+b d \\ a c+c d & a d+d^{2} \end{array}\right]\)
A² - ( a + d) A
\(=\left[\begin{array}{ll} a^{2}+b c & a b+b d \\ a c+c d & b c+d^{2} \end{array}\right]-\left[\begin{array}{ll} a^{2}+a d & a b+b d \\ a c+c d & a d+d^{2} \end{array}\right]\)
\(=\left[\begin{array}{cc} b c-a d & 0 \\ 0 & b c-a d \end{array}\right]\)
Now (bc - ad) I₂ \(=(b c-a d)\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{cc} b c-a d & 0 \\ 0 & b c-a d \end{array}\right]\)
From (1) and (2)
A² - (a + d)A = (bc - ad) I₂
Hence proved.

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.

We observe that A is a 2 x 2 matrix and B is a 2 x 2 matrix, hence AB is defined and it will be of the order 2 x 2.
AB = \(\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 2 & 0 \\ 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} 4+1 & 0+3 \\ 2+3 & 0+9 \end{matrix} \right] =\left[ \begin{matrix} 5 & 3 \\ 5 & 9 \end{matrix} \right] \)
BA = \(\left[ \begin{matrix} 2 & 0 \\ 1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} 4+0 & 2+0 \\ 2+3 & 1+9 \end{matrix} \right] =\left[ \begin{matrix} 4 & 2 \\ 5 & 10 \end{matrix} \right] \)
Therefore, AB ≠ BA.

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

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

\(A=\left[ \begin{matrix} 3 & 0 \\ 4 & 5 \end{matrix} \right] ,B=\left[ \begin{matrix} 6 & 3 \\ 8 & 5 \end{matrix} \right] ,C=\left[ \begin{matrix} 3 & 6 \\ 1 & 1 \end{matrix} \right] \)
CD - AB = 0 ⇒ CD = AB
\(AB=\left[ \begin{matrix} 3 & 0 \\ 4 & 5 \end{matrix} \right] \left[ \begin{matrix} 6 & 3 \\ 8 & 5 \end{matrix} \right] =\left[ \begin{matrix} (18+0) & (9+0) \\ (24+40) & (12+25) \end{matrix} \right] \)
\(CD=\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
\(Let\quad D=\left[ \begin{matrix} x & y \\ z & w \end{matrix} \right] \)
\(\left[ \begin{matrix} 3 & 6 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x & y \\ z & w \end{matrix} \right] =\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
\(\left[ \begin{matrix} 3x+6z & 3y+6w \\ x+z & y+w \end{matrix} \right] =\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
3x + 6z = 18      ...(1)
x + z = 64    ...(2)

Sub. x=122 in(2)
122+z=64
z=64-122=-58
3y+6w=9  ....(3)
 y+w=37   ....(4)

Sub. w = -34 in (4)
y-34 = 37
y = 37 + 34 = 71
∴ Solutions: x = 122
y = 71
z = -58
w = -34
\(\therefore D=\left[ \begin{matrix} 122 & 71 \\ -58 & -34 \end{matrix} \right] \)

It is not possible to add A and B because they have different orders.

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 = \(\left( \begin{matrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{matrix} \right) \)