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)^T= A^T-B^T
L.H.S = \((A-B)=\left[ \begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix} \right] -\left[ \begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix} \right] =\left[ \begin{matrix} -3 & 2 \\ 0 & -2 \end{matrix} \right] ...(2)\)
\({ (A-B) }^{ T }=\left[ \begin{matrix} -3 & 0 \\ 2 & -2 \end{matrix} \right] ...(1)\)
\({ A }^{ T }=\left[ \begin{matrix} 1 & 1 \\ 2 & 3 \end{matrix} \right] ,{ B }^{ T }=\left[ \begin{matrix} 4 & 1 \\ 0 & 5 \end{matrix} \right] \)
\(=\left[\begin{array}{rr} -3 & 0 \\ 2 & -2 \end{array}\right]\)
From (1) and (2)
(A - B)^T= A^T-B^T
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.

\(A=\left[\begin{array}{cc} \cos \theta & 0 \\ 0 & \cos \theta \end{array}\right]\),
\(B=\left[\begin{array}{cc} \sin \theta & 0 \\ 0 & \sin \theta \end{array}\right]\)
A² = A.A
\(=\left[\begin{array}{cc} \cos \theta & 0 \\ 0 & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & 0 \\ 0 & \cos \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \cos ^{2} \theta+0 & 0+0 \\ 0+0 & 0+\cos ^{2} \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \cos ^{2} \theta & 0 \\ 0 & \cos ^{2} \theta \end{array}\right]\)
B² = B.B
\(=\left[\begin{array}{cc} \sin \theta & 0 \\ 0 & \sin \theta \end{array}\right]\left[\begin{array}{cc} \sin \theta & 0 \\ 0 & \sin \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \sin ^{2} \theta+0 & 0+0 \\ 0+0 & \sin ^{2} \theta \end{array}\right]\)
\(\text { Now } A^{2}+B^{2}=\left[\begin{array}{cc} \cos ^{2} \theta & 0 \\ 0 & \cos ^{2} \theta \end{array}\right]+\left[\begin{array}{cc} \sin ^{2} \theta & 0 \\ 0 & \sin ^{2} \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \cos ^{2} \theta+\sin ^{2} \theta & 0+0 \\ 0+0 & \cos ^{2} \theta+\sin ^{2} \theta \end{array}\right]\)
\(=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I\)
Hence Proved.

A = \(\left[ \begin{matrix} 1 & 3 \\ 5 & -1 \end{matrix} \right] \)
B = \(\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{matrix} \right] \)
C = \(\left[ \begin{matrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{matrix} \right] \) 
\(B+C=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right]+\left[\begin{array}{rrr} 1 & 3 & 2 \\ -4 & 1 & 3 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 2 & 2 & 4 \\ -1 & 6 & 5 \end{array}\right]\)
\(A(B+C)=\left[\begin{array}{cc} 1 & 3 \\ 5 & -1 \end{array}\right]\left[\begin{array}{rrr} 2 & 2 & 4 \\ -1 & 6 & 5 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 2-3 & 2+18 & 4+15 \\ 10+1 & 10-6 & 20-5 \end{array}\right]\)
\(=\left[\begin{array}{rrr} -1 & 20 & 19 \\ 11 & 4 & 15 \end{array}\right]\)
\(A B=\left[\begin{array}{cc} 1 & 3 \\ 5 & -1 \end{array}\right]\left[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 1+9 & -1+15 & 2+6 \\ 5-3 & -5-5 & 10-2 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 10 & 14 & 8 \\ 2 & -10 & 8 \end{array}\right]\)
\(A C=\left[\begin{array}{cc} 1 & 3 \\ 5 & -1 \end{array}\right]\left[\begin{array}{rrr} 1 & 3 & 2 \\ -4 & 1 & 3 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 1-12 & 3+3 & 2+9 \\ 5+4 & 15-1 & 10-3 \end{array}\right]\)
\(=\left[\begin{array}{rrr} -11 & 6 & 11 \\ 9 & 14 & 7 \end{array}\right]\)
\(A B+A C=\left[\begin{array}{ccc} 10 & 14 & 8 \\ 2 & -10 & 8 \end{array}\right]+\left[\begin{array}{ccc} -11 & 6 & 11 \\ 9 & 14 & 7 \end{array}\right]\)
\(=\left[\begin{array}{rrr} -1 & 20 & 19 \\ 11 & 4 & 15 \end{array}\right]\)
From (1), (2)
A (B + C) = AB + AC , Hence verified

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.

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.

\(\left(\begin{matrix} x-3 & 3x-z \\ x+y+7 & x+y+z \end{matrix} \right) =\left( \begin{matrix} 1 & 0 \\ 1 & 6 \end{matrix} \right) \)
x-3 =1 ⇒ x = 4
3x-z = 0
3(4)-z = 0
-z=-12 ⇒ z = 12
x + y + 7 = 1
x + y = -6
4 + y = -6
y = -10
x = 4, y = -10, z = 12

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

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

(B+C) = \(\left[ \begin{matrix} 2 & 3 & 4 \\ 1 & 9 & 2 \\ -7 & 1 & -1 \end{matrix} \right] \)+\(\left[ \begin{matrix} 8 & 3 & 4 \\ 1 & -2 & 3 \\ 2 & 4 & -1 \end{matrix} \right] \)=\(\left[ \begin{matrix} 10 & 6 & 8 \\ 2 & 7 & 5 \\ -5 & 5 & -2 \end{matrix} \right] \)
\(A+(B+C)=\left[ \begin{matrix} 4 & 3 & 1 \\ 2 & 3 & -8 \\ 1 & 0 & -4 \end{matrix} \right] +\left[ \begin{matrix} 10 & 6 & 8 \\ 2 & 7 & 5 \\ -5 & 5 & -2 \end{matrix} \right] =\left[ \begin{matrix} 14 & 9 & 9 \\ 4 & 10 & -3 \\ -4 & 5 & -6 \end{matrix} \right] .... (1)\)
R.H.S = (A+B)+C
\((A+B)=\left[ \begin{matrix} 4 & 3 & 1 \\ 2 & 3 & -8 \\ 1 & 0 & -4 \end{matrix} \right] +\left[ \begin{matrix} 2 & 3 & 4 \\ 1 & 9 & 2 \\ -7 & 1 & -1 \end{matrix} \right] =\left[ \begin{matrix} 6 & 6 & 5 \\ 3 & 12 & -6 \\ -6 & 1 & -5 \end{matrix} \right] \quad \quad \quad (1)\)
\((A+B)+C=\left[ \begin{matrix} 6 & 6 & 5 \\ 3 & 12 & -6 \\ -6 & 1 & -5 \end{matrix} \right] +\left[ \begin{matrix} 8 & 3 & 4 \\ 1 & -2 & 3 \\ 2 & 4 & -1 \end{matrix} \right] =\left[ \begin{matrix} 14 & 9 & 9 \\ 4 & 10 & -3 \\ -4 & 5 & -6 \end{matrix} \right] ...(2)\)
(1) = (2) ⇒ L.H.S. = R.H.S Hence verified.

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_ij = |i - 2j|
a₁₁ = |1-2 x 1| = |1-2| = |-1| = 1
a₁₂ = |1-2 x 2| = |1-4| = |-3| = 3
a₁₃ = |1-2 x 3| = |1-6| = |-5| = 5
a₂₁ = |2-1 x 1| = |2-2| = 0
a₂₂ = |2-2 x 2| = |-2| = 2
a₂₃ = |2-2 x 3| = |-4| = 4
a₃₁= |3-2 x 1| = |1| = 1
a₃₂ = |3-2 x 2| = |-1| = 1
a₃₃ = |3-2 x 3| = |-3| = 3
\(\therefore \left[ \begin{matrix} 1 & 3 & 5 \\ 0 & 2 & 4 \\ 1 & 1 & 6 \end{matrix} \right] \) is the required 3x3 matrix.

Given that a matrix has 18 elements
1 x 18, 2 x 9, 3 x 6, 6 x 3, 9 x 2, 18 x 1
If a matrix has 6 elements, then the possible orders
1 x 6, 2 x 3, 3 x 2, 6 x 1

\(A=\left[ \begin{matrix} 3 & 0 \\ 4 & 5 \end{matrix} \right] ,B=\left[ \begin{matrix} 6 & 3 \\ 8 & 5 \end{matrix} \right] ,C=\left[ \begin{matrix} 3 & 6 \\ 1 & 1 \end{matrix} \right] \)
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
\(\therefore D=\left[ \begin{matrix} 122 & 71 \\ -58 & -34 \end{matrix} \right] \)

\(A=\left[ \begin{matrix} p & 0 \\ 0 & 2 \end{matrix} \right] ,B=\left[ \begin{matrix} 0 & -q \\ 1 & 0 \end{matrix} \right] ,C=\left[ \begin{matrix} 2 & -2 \\ 2 & 2 \end{matrix} \right] \)
\(BA={ C }^{ 2 }\Rightarrow \left[ \begin{matrix} 0 & -q \\ 1 & 0 \end{matrix} \right] \left[ \begin{matrix} p & 0 \\ 0 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2 & -2 \\ 2 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 & -2 \\ 2 & 2 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 0 & -2q \\ p & 0 \end{matrix} \right] =\left[ \begin{matrix} (4-4) & (-4-4) \\ (4+4) & (-4+4) \end{matrix} \right] \)
\({ -2q=-8\\ q=4 }{ | }{ p=8\\ q=4 }\)

L.H.S = \(\cos { \theta \left( \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right) }+ \sin { \theta \left( \begin{matrix} x & -\cos { \theta } \\ \cos { \theta } & x \end{matrix} \right) } \)
\(=\left[ \begin{matrix} { cos }^{ 2 }\theta & cos\sin { \theta } \\ -\sin { \theta cos\theta } & { cos }^{ 2 }\theta \end{matrix} \right] +\left[ \begin{matrix} xsin\theta & -sin\theta \cos { \theta } \\ sin\theta \cos { \theta } & xsin\theta \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { cos }^{ 2 }\theta +xsin\theta & 0 \\ 0 & { cos }^{ 2 }+xsin\theta \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
∴ cos²θ + x sin θ = 1
x sin θ = 1-cos²θ
\(x=\frac { { sin }^{ 2 }\theta }{ sin\quad \theta } =sin\theta \)

We observe that A is a 2 x 3 matrix and B is a 3 x 3 matrix, hence AB is defined and it will be of the order 2 × 3..
Given A = \({ \left[ \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 5 \end{matrix} \right] }_{ 2\times 3 }\), B = \({ \left[ \begin{matrix} 8 & 3 & 1 \\ 2 & 4 & 1 \\ 5 & 3 & 1 \end{matrix} \right] }_{ 2x3 }\)
AB = \({ \left[ \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 5 \end{matrix} \right] }\times \left[ \begin{matrix} 8 & 3 & 1 \\ 2 & 4 & 1 \\ 5 & 3 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8+4+0 & 3+8+0 & 1+2+0 \\ 24+2+25 & 9+4+15 & 3+1+5 \end{matrix} \right] =\left[ \begin{matrix} 12 & 11 & 3 \\ 51 & 28 & 9 \end{matrix} \right] \)

L.H.S=A+B=\(\left[ \begin{matrix} 1 & 9 \\ 3 & 4 \\ 8 & -3 \end{matrix} \right] \)+\(\left[ \begin{matrix} 5 & 7 \\ 3 & 3 \\ 1 & 0 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6 & 16 \\ 6 & 7 \\ 9 & -3 \end{matrix} \right] \) ...(1)
R.H.S=B+A =\(\left[ \begin{matrix} 5 & 7 \\ 3 & 3 \\ 1 & 0 \end{matrix} \right] \) +\(\left[ \begin{matrix} 1 & 9 \\ 3 & 4 \\ 8 & -3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6 & 16 \\ 6 & 7 \\ 9 & -3 \end{matrix} \right] \) ...(2)
(1)+(2) ⇒ L.H.S=R.H.S. Hence verified.

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.

Since A, B are of the same order 3 x 3, subtraction of 4A and 3B is defined.
4A - 3B = \(4\left[ \begin{matrix} 5 & 4 & -2 \\ \frac { 1 }{ 2 } & \frac { 3 }{ 4 } & \sqrt { 2 } \\ 1 & 9 & 4 \end{matrix} \right] -3\left[ \begin{matrix} -7 & 4 & -3 \\ \frac { 1 }{ 4 } & \frac { 7 }{ 2 } & 3 \\ 5 & -6 & 9 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 20 & 16 & -8 \\ 2 & 3 & 4\sqrt { 2 } \\ 4 & 36 & 16 \end{matrix} \right] +\left[ \begin{matrix} 21 & -12 & 9 \\ -\frac { 3 }{ 4 } & -\frac { 21 }{ 2 } & -9 \\ -15 & 18 & -27 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 41 & 4 & 1 \\ \frac { 5 }{ 4 } & -\frac { 15 }{ 2 } & 4\sqrt { 2 } -9 \\ -11 & 54 & -11 \end{matrix} \right] \)

A + B = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right] +\left[ \begin{matrix} 1 & 7 & 0 \\ 1 & 3 & 1 \\ 2 & 4 & 0 \end{matrix} \right] =\left[ \begin{matrix} 1+1 & 2+7 & 3+0 \\ 4+1 & 5+3 & 6+1 \\ 7+2 & 8+4 & 9+0 \end{matrix} \right] =\left[ \begin{matrix} 2 & 9 & 3 \\ 5 & 8 & 7 \\ 9 & 12 & 9 \end{matrix} \right] \)

If A=\(\left[ \begin{matrix} 5 & 2 & 2 \\ -\sqrt { 17 } & 0.7 & \frac { 5 }{ 2 } \\ 8 & 3 & 1 \end{matrix} \right] \), A^T=\(\left[ \begin{matrix} 5 & -\sqrt { 17 } & 8 \\ 2 & 0.7 & 3 \\ 2 & \frac { 5 }{ 2 } & 1 \end{matrix} \right] \)
(A^T)^T =\(\left[ \begin{matrix} 5 & 2 & 2 \\ -\sqrt { 17 } & 0.7 & \frac { 5 }{ 2 } \\ 8 & 3 & 1 \end{matrix} \right] \)= A
∴ verified

If A = \(\left[ \begin{matrix} \sqrt { 7 } & -3 \\ -\sqrt { 5 } & 2 \\ \sqrt { 3 } & -5 \end{matrix} \right] \)
-A\(\left[ \begin{matrix}- \sqrt { 7 } &3 \\ \sqrt { 5 } & -2 \\- \sqrt { 3 } & 5 \end{matrix} \right] \)
Transpose od -A = (-A)^T = \(\left[ \begin{matrix} -\sqrt { 7 } & +\sqrt { 5 } & -\sqrt { 3 } \\ +3 & -2 &+ 5 \end{matrix} \right] \)

If A =\(\left[ \begin{matrix} 5 & 1 & 3 \\ 4 & -7 & 8 \\ 3 & 9 & 2 \end{matrix} \right] \)
Transpose of A = A^T =\(\left[ \begin{matrix} 5 & 1 & 3 \\ 4 & -7 & 8 \\ 3 & 9 & 2 \end{matrix} \right] \)

16

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

We know that a matrix of order m x n has mn elements. Thus to find all possible orders of a matrix with 16 elements, we will find all ordered pairs of natural numbers whose product is 16.
Such ordered pairs are (1, 16), (16, 1), (4,4), (8,2), (2,8)
Hence possible orders are 1 x 16, 16 x 1, 4 x 4, 2 x 8, 8 x 2