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 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

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

Let A = \(\left[ \begin{matrix} 6 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -9 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \). Then A is a matrix of order 4 × 3 and ρ(A) ≤ 3.
The last two rows are zero rows. There are several second order minors.
We find that there is a second order minor, for example, \(\left| \begin{matrix} 6 & 0 \\ 0 & 2 \end{matrix} \right| \) = (6)(2) = 12 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third and fourth rows are zero rows.

Let A = \(\left[ \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3
The third order minor |A| = \(\left| \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right| \) = (2)(3)(1) = 6 ≠ 0. So, ρ(A) = 3.
Note that there are three non-zero rows.

We first find adj A. By definition, we get adj A = \({ \left[ \begin{matrix} +{ M }_{ 11 } & -{ M }_{ 12 } \\ -{ M }_{ 21 } & +{ M }_{ 22 } \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} d & -c \\ -b & a \end{matrix} \right] }^{ T }=\left[ \begin{matrix} d & -c \\ -c & a \end{matrix} \right] \).
Since A is non-singular, |A| = ad - bc ≠ 0.
As \({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } \) adj A, we get A^-1 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).

\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 4
∴ \(\rho \)(A) ≤ min(3, 4) = 3 
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 8 & 1 & 0 \end{matrix} \right| \) = 0+0-8(4-4) = 0
[Expanded along C₁]
Also, \(\left| \begin{matrix} 0 & 2 & 1 \\ 0 & 4 & 3 \\ 8 & 0 & 2 \end{matrix} \right| =0+0-8\left| \begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix} \right| \)
[Expanded along C₁]
= -8(6-4) = -8(2) = -16 ≠ 0
∴ \(\rho \)(A) = 3

We compute |adj A| = \(\left| \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right| \) = 9.
So, we get
A^-1 = \(\pm \frac { 1 }{ \sqrt { \left| adjA \right| } } \) adj(A) = \(\frac { 1 }{ \sqrt { 9 } } \left[ \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] =\pm \frac { 1 }{ 3 } \left[ \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] \).