\(A^{2}=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{ll} 5 & -8 \\ 2 & -3 \end{array}\right]\)

\( \mathrm{A}^{2}=\mathrm{A} \times \mathrm{A} \)
\( =\left[\begin{array}{ccc} 1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right] \)
\(=\left[\begin{array}{ccc} 1+10-2 & 2+4-1 & 1+12-3 \\ 5+10-12 & 10+4-6 & 5+12-18 \\ -2-5+6 & -4-2+3 & -2-6+9 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 9 & 5 & 10 \\ 3 & 8 & -1 \\ -1 & -3 & 1 \end{array}\right]\)
\( \mathrm{A}^{3}=\mathrm{A^2} \times \mathrm{A} \)
\(=\left[\begin{array}{ccc} 9 & 5 & 10 \\ 3 & 8 & -1 \\ -1 & -3 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 9+25-20 & 18+10-10 & 9+30-30 \\ 3+40+2 & 6+16+1 & 3+48+3 \\ -1-15-2 & -2-6-1 & -1-18-3 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 14 & 18 & 9 \\ 45 & 23 & 54 \\ -18 & -9 & -22 \end{array}\right]\)

adj A = \(\left[\begin{array}{ccc} +(1-4) & -(2+4) & +(-4-2) \\ -(-2-4) & +(-1+4) & -(2+4) \\ +(4+2) & -(2+4) & +(-1+4) \end{array}\right]^{T}\)
\(=\left[\begin{array}{ccc} -3 & -6 & -6 \\ 6 & 3 & -6 \\ -6 & -6 & 3 \end{array}\right]^{T}=3\left[\begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right]^{T}\)
adj A = 3A^T

|A| = 8 + 2(a - 6) = 2a - 4
cofactor of a₁₂ (= 1) is 8 in |A|
In A^-1 now A₁₂ = \(\frac{8}{|A|}=-4 \Rightarrow \frac{8}{2 a-4}=-4\)
\(\frac{2}{2 a-4}=-1\)
\( \Rightarrow 2=-2 a+4 \Rightarrow 2 a=2 \Rightarrow a=1 \ \ \therefore|A|=-2 \)
\(\text {In } A^{-1}, A_{23}=c =\frac{\text { cofactor } a\ \text {in }|A|}{|A|} \)
\(c =\frac{2}{-2}=-1 \)

Let A = \(\left[\begin{array}{ccc} 1 & 1 & -2 \\ 3 & -1 & 1 \\ 3 & 3 & -6 \end{array}\right]\)
\(|A|=1(6-3)-1(-18-3)-2(9+3)\)
Since |A| = 3 + 21 - 24 = 0
Hence the solution of the given linear equations does not exists.

Δ = \(\left| \begin{matrix} 6 & -7 \\ 9 & -5 \end{matrix} \right| \) = -30 + 63 = 33
Δ₁ = \(\left| \begin{matrix} 16 & -7 \\ 35 & -5 \end{matrix} \right| \) = -80 + 245 = 165
Δ₂ = \(\left| \begin{matrix} 6 & 16 \\ 9 & 35 \end{matrix} \right| \) = 210 - 144 = 66
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 165 }{ 33 } \) = 5
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 66 }{ 33 } \) = 2
∴ Solution set is { 5, 2}

The equations can be written in matrix form as
\(\left( \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 4 \end{matrix} \right) \) ⇒ AX = B where
A =\(\left( \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right) ,X=\left( \begin{matrix} x \\ y \end{matrix} \right) ,B=\left( \begin{matrix} 3 \\ 4 \end{matrix} \right) \)
∴ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right| \) = 2+5 =7 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 7 } \left[ \begin{matrix} 1 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ X = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 1 & 1 \\ -5 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 4 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 3+4 \\ -15+8 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 7 \\ -7 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ -1 \end{matrix} \right] \)
∴ Solution set is {1, -1}

Matrix form of the given system of equations is
\(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \)
Homogeneous system of equations has trivial solution only if |A| ≠ 0.
∴ \(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \) ≠ 0
Expanding along R₁,
\(1\left| \begin{matrix} -3 & -3 \\ 1 & k \end{matrix} \right| -2\left| \begin{matrix} 1 & -3 \\ 2 & k \end{matrix} \right| +2\left| \begin{matrix} 1 & -3 \\ 2 & 1 \end{matrix} \right| \neq 0\)
⇒ 1 (-3k + 3) - 2 (k + 6) + 2 (1 + 6) ≠ 0
⇒ -3k + 3 -2k - 12 + 14 ≠ 0
⇒ -5k + 5 ≠ 0
⇒ -5k ≠ -5 ⇒ k ≠ \(\frac { -5 }{ -5 } \) =1
⇒ k ≠ 1

The matrix form of the system is
\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where
A=\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
|A| =\(\left| \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right| =3\left| \begin{matrix} 2 & 12 \\ 1 & 7 \end{matrix} \right| -1\left| \begin{matrix} 3 & 12 \\ 2 & 7 \end{matrix} \right| +9\left| \begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix} \right| \)
= 3 (14 - 12) - 1 (21 -24) + 9 (3 -4)
= 3 (2) -1 (-3) + 9 (-1)
= 6 + 3 - 9 = 9 - 9 = 0
Since |A| = 0, the homogeneous system of equations have non-trivial solutions also.

A =\(\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 0 \end{matrix}\begin{matrix} 1 \\ 8 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ -1 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 12 \end{matrix}\begin{matrix} 1 \\ 6 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 4 \end{matrix}\begin{matrix} -3 \\ 5 \end{matrix}\begin{matrix} 12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }+(-1){ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 4 \end{matrix}\begin{matrix} 13 \\ 5 \end{matrix}\begin{matrix} -12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \end{matrix}\begin{matrix} 13 \\ -47 \end{matrix}\begin{matrix} -12 \\ 42 \end{matrix}\begin{matrix} -6 \\ 25 \end{matrix} \right] \)
The equivalent row-echelon matrix hats two non zero rows.
∴ \(\rho\) (A) = 2

Let A =\(\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{matrix} \right] \)
Now |A| = \(\left| \begin{matrix} 6 & -5 \\ -2 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & -5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & 6 \\ 5 & -2 \end{matrix} \right| \)
= 3 (12 - 10) + 1 (-30 + 25) + 1 (30 - 30)
= 3(2) + 1 (-5) + 0 = 6 - 5 = 1 ≠ 0
∴ Rank of A is 3.

Agumented matrix
[A|B] \(\left[ \begin{matrix} 2 & 5 \\ 6 & 15 \end{matrix}|\begin{matrix} 7 \\ 13 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 5 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ -8 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 and \(\rho\)([A|B]) = 3
∴ \(\rho\) (a) ≠ \(\rho\) ([AIB])
Hence the system is inconsistent.

Given A³ = I ⇒ IA³I = |I|
⇒ |A.A.A| = 1
⇒ |A|. |A|·|A| = 1
⇒ |A|³ = 1
∴ |A| ≠ 0
Hence, A is non-singular.

Given A³ = 1
Pre multiply by A^-1 we get,
A^-1. A³ = A^-1. I
⇒ (A^-1. A) A² = A^-1 [∵ A^-1 I = A^-1]
⇒ I. A² = A^-1  I∵  A^-1. A = I]
⇒ A² = A^-1  [∵ I. A² = A²]
∴ A^-1 = A²

Given A (adj A) =\(\left[ \begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix} \right] =10\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)...(1)
We know A (adj A) = (adj A) A = |A|. I₂ ...(2)
Comparing (1) and (2), we get |A| = 10