\( A_{r} =r^{2}-(r-1)^{2}=2 r-1 \)
\(\Sigma A_{r} =\Sigma(2 r-1)=2 \Sigma r-\Sigma(1)\)
\( =2 \frac{r(r+1)}{2}-1=r^{2}+1-1=r^{2} \)
\(\sum_{r=1}^{2013} A_{r} =(2013)^{2} \Rightarrow \sqrt{\left(\sum_{r=1}^{2013} A_{r}\right)}=2013\)

\( |\mathrm{A}|=\cos ^{2} \alpha+\sin ^{2} \alpha=1 \)
\( \mid \operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))) | \)
\( \operatorname{adj}(\operatorname{adj} \mathrm{A})=|\mathrm{A}|^{\mathrm{n}-2} \mathrm{~A} \)
\( =|\mathrm{A}|^{3-2} \mathrm{~A}=|\mathrm{A}| \cdot \mathrm{A} \text {. }\)
adj (adj A) = A
\( \therefore|(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))|=|\operatorname{adj}(\operatorname{adj} \mathrm{A})| \)
\( =|\mathrm{A}|^{(\mathrm{n}-1)^{2}} \)
\( =|\mathrm{A}|^{(3-1)^{2}} \)
\( =|\mathrm{A}|^{4}=1^{4}=1\)

A² - 5A +7I = 0
\( \mathrm{A}^{2}=5 \mathrm{~A}-7 \mathrm{I} \)
\( \mathrm{A}^{3}=\mathrm{A} \cdot \mathrm{A}^{2} \)
=  A(5 A - 7I)
= 5 A² -7 A
= 5(5 A - 7I) - 7A
= 25 A - 35 I - 7 A
= 18A - 35I
\( \mathrm{A}^{4}=\mathrm{A} \cdot \mathrm{A}^{3} \)
= A (18A - 35I)
= 18A² - 35 A
= 18(5A -7I) - 35A
= 90 A -126 I - 35A
= 55 A - 126 I
\( \mathrm{A}^{5} =\mathrm{A} \cdot \mathrm{A}^{4} \)
\( =\mathrm{A}(55 \mathrm{~A}-126 \mathrm{I}) \)
\( =55 \mathrm{~A}^{2}-126 \mathrm{~A}\)
= 55 (5A - 7I) - 126A
= 275 A - 385 I - 126A
= 149 A - 385 I
Given A⁵ = aA + bI
149 A - 385 I = aA + bI
a = 149
b = -385
2a - 3b = 2(149) -3(-385)
= 298 + 1155 = 1453

|A| = 0 -2 (-2 + 3) +3 (-2 + 3)
= -2 + 3 = 1 \(\neq\) 0. A^-1 exists
\(\operatorname{adj} A=\left[\begin{array}{ccc} +(-6+6) & -(-2+3) & +(-2+3) \\ -(4-6) & -(0-3) & -(0-2) \\ +(-6+9) & -(0+3) & +(0+2) \end{array}\right]\)
\(=\left[\begin{array}{lll} 0 & 1 & 1 \\ 2 & 3 & 2 \\ 3 & 3 & 2 \end{array}\right]^{T}=\left[\begin{array}{ccc} 0 & 2 & 3 \\ -1 & -3 & -3 \\ 1 & 2 & 2 \end{array}\right]\)
\(A^{-1}=\frac{1}{|A|} \text { adj } A=\left|\begin{array}{ccc} 0 & 2 & 3 \\ -1 & -3 & -3 \\ 1 & 2 & 2 \end{array}\right|\)

\(A B=\left[\begin{array}{cc} 1 & -3 \\ -2 & 7 \end{array}\right]\left[\begin{array}{ll} 7 & 3 \\ 2 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
and BA = \(\left[\begin{array}{cc} 7 & 3 \\ 2 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -3 \\ -2 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
Hence AB = BA = I
\(\therefore\) B = A^-1

Given rank of \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \) is 2
⇒ The value of the third order determinant is zero
⇒ \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \)=0
⇒ λ\(\left| \begin{matrix} \lambda & -1 \\ 0 & \lambda \end{matrix} \right| +1\left| \begin{matrix} 0 & -1 \\ -1 & \lambda \end{matrix} \right| \)+0 = 0
⇒ λ(λ² - 0) + 1(0 - 1) = 0
⇒ λ³ - 1 = 0 ⇒ λ³ = 1 ⇒ λ = 1
∴ λ = 1

Augmented matrix [A|B] =\(\left[ \begin{matrix} 1 & 1 & 3 \\ 2 & 2 & 6 \\ 2 & 1 & 1 \end{matrix}|\begin{matrix} 4 \\ 7 \\ 10 \end{matrix} \right] \)
[A|B] \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -5 \end{matrix}|\begin{matrix} 4 \\ -1 \\ 2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 4 \\ 2 \\ -1 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 [only 2 two-non zero rows]
And \(\rho\) ([A|B]) = 3 [There are 3 non-zero rows]
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
∴ The system is inconsistent.

The matrix from of the system of equations is
\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 7 \\ 14 \end{matrix} \right] \) ⇒ AX = B where
A =\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 7 \\ 14 \end{matrix} \right] \)
Transforming augmented matrix to row-echelon form we get
[A|B] =\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & -3 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ 0 \end{matrix} \right] \)
Here \(\rho\) (A) = \(\rho\) [A|B] = 1
∴ \(\rho\) (A) = \(\rho\) [A|B] = 1 < the number of unknowns, the system is consistent and has one parameter family of solutions
∴ Put y = t, where t \(\in \) R
Writing the row-echelon form to equations we get
 2x - 3y = 7
∴ 2x - 3t = 7
⇒  2x = 7 + 3t
⇒ x = \(\frac { 1 }{ 2 } \)(7 + 3t)
∴ Solution set is {\(\frac { 1 }{ 2 } \)(7+3t), t} where t \(\in \) R.

|A| =\(\left[ \begin{matrix} -2 & -3 \\ 5 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & 3 \\ -5 & 2 \end{matrix} \right] \)
(A^-1)^T = \(\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & 2 \end{matrix} \right] \)...(1)
A^T =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \)
|A^T| =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ (A^T)^-1 = \(\frac { 1 }{ |A^{ T }| } adj(A^{ T })=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & -2 \end{matrix} \right] \)...(2)
From (1) and (2), (A^-1)^T = (A^T)^-1

Let A =\(\left[ \begin{matrix} 4 \\ -2 \\ 1 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 0 \\ -1 \\ 8 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
A\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ -2 \\ 4 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 8 \\ -1 \\ 0 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 4 \\ 11 \\ -12 \end{matrix}\begin{matrix} 8 \\ 15 \\ -32 \end{matrix}\begin{matrix} 7 \\ 19 \\ -25 \end{matrix} \right] \)
A is in row - echelon form and it has 3 non-zero rows.
∴ \(\rho\) (A) = 3

Let A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & h-2 & 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} & \begin{matrix} 0 \\ 0 \end{matrix} & \begin{matrix} h+2 \\ 3 \end{matrix} \end{matrix} \right] \)
The rank of A will be less than 3 if every minor of order 3 vanishes
∴ \(\left| \begin{matrix} 1 & 0 & 0 \\ 0 & h-2 & 0 \\ 0 & 0 & 3 \end{matrix} \right| \) = 0
⇒ 1\(\left| \begin{matrix} h-2 & 0 \\ 0 & 3 \end{matrix} \right| \) + 0 + 0 =0 ⇒ 3(h-2) = 0
⇒ h-2 = 0 ⇒ h = 2

AB =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] =\left[ \begin{matrix} 8+3 & 10+4 \\ 20+9 & 25+12 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
|AB| =\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
= 407 - 406 = 1 ≠ 0
(AB)^-1 = \(\frac { 1 }{ |AB| } \) adj(AB)
= \(\frac { 1 }{ 1 } \left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \) ....(1)
|A| =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \) = 6 - 5 =1
|B| =\(\left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] \) = 16 - 15 = 1
B^-1 = \(\frac { 1 }{ |B| } adjB=\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \)
A^-1 = \(\frac { 1 }{ |A| } adjA=\left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
∴ B^-1A^-1 =\(\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
=\(\\ \left[ \begin{matrix} 12+25 & -4-10 \\ -9-20 & 3+8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \)....(2)
From (1) and (2), (AB)^-1 = B^-1A^-1

Δ = \(\left| \begin{matrix} 3 & a \\ 2 & a \end{matrix} \right| \) = 3a - 2a = a
Δ₁ =\(\left| \begin{matrix} 4 & a \\ 2 & a \end{matrix} \right| \) = 4a - 2a = 2a
Δ₂ = \(\left| \begin{matrix} 3 & 4 \\ 2 & 2 \end{matrix} \right| \)= 6 - 8 = -2

\(\therefore x=\frac{\Delta_{1}}{\Delta}=\frac{2 \not a}{\not a}=2=y y=\frac{\Delta_{2}}{\Delta}=\frac{-2}{a}\)
∴ Solution set is {2,\(\frac { -2 }{ a } \)}

\(\Delta = \left| \begin{matrix} t & 3 & -1 \\ 1 & 2 & 1 \\ -t & 1 & 2 \end{matrix} \right| =t\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ -t & 2 \end{matrix} \right| -\left| \begin{matrix} 1 & 2 \\ -t & 1 \end{matrix} \right| \)
= t(4 - 1) -3 (2 + t) -1(1 + 2t)
= 3t - 6i - 3t - 1 - 2t = - 7 - 2t
The system will fail to have unique solution if
Δ = 0 ⇒ -7-2t = 0 ⇒ -2t = 7 ⇒ t = \(\frac { -7 }{ 2 } \)
∴ t = \(\frac { -7 }{ 2 } \).

Δ = \(\left| \begin{matrix} 2 & 3 \\ 1 & 6 \end{matrix} \right| \) = 12 - 3 = 9 ≠ 0
Δ₁ = \(\left| \begin{matrix} 10 & 3 \\ 4 & 6 \end{matrix} \right| \) = 60 - 12 = 48
Δ₂ = \(\left| \begin{matrix} 2 & 10 \\ 1 & 4 \end{matrix} \right| \) = 8 - 10 = -2
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 48 }{ 9 } =\frac { 16 }{ 3 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -2 }{ 9 } \)
∴ Solution set is \(\left\{ \frac { 16 }{ 3 } ,\frac { -2 }{ 9 } \right\} \).