Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

Suppose A is symmetric. Then, A^T = A and so, by theorem (vi), we get
adj(A^T) = (adj A)^T ⇒ adj A = (adj A)^T ⇒ adj A is symmetric.

Let A be a non-singular matrix of order 2m+1, where m = 0, 1, 2,... Then, we get |A| ≠ 0 and, by property (ii), we have |adj A| = |A|^(2m+1) − 1 = |A|^2m.
Since |A|^2m is always positive, we get that |adj A| is positive.

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 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).

Let A = \(\left[ \begin{matrix} 4 & 3 \\ -3 & -1 \\ 6 & 7 \end{matrix}\begin{matrix} 1 & -2 \\ -2 & 4 \\ -1 & 2 \end{matrix} \right] \). Then A is a matrix of order 3 \(\times\) 4. So ρ(A)  ≤ min {3, 4} = 3.
The highest order of minors of A is 3. We search for a non-zero third-order minor of A. But we find that all of them vanish. In fact, we have
\(\left| \begin{matrix} 4 & 3 & 1 \\ -3 & -1 & -2 \\ 6 & 7 & -1 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 3 & -2 \\ -3 & -1 & 4 \\ 6 & 7 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 1 & -2 \\ -3 & -2 & 4 \\ 6 & -1 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 3 & 1 & -2 \\ -1 & -2 & 4 \\ 7 & -1 & 2 \end{matrix} \right| \) = 0.
So, ρ(A) < 3. Next, we search for a non-zero second-order minor of A.
We find that \(\left| \begin{matrix} 4 & 3 \\ -3 & -1 \end{matrix} \right| \) = -4 + 9 = 5 ≠ 0. So, ρ(A) = 2.

\(\frac { 3 }{ x } \) + 2y = 12, \(\frac { 2 }{ x } \) + 3y = 13
Let \(\frac { 1 }{ x } \)
∴ z+2y = 12, 2z+3z = 13
∴ Δ = \(\left| \begin{matrix} 3 & 2 \\ 2 & 3 \end{matrix} \right| \)= 9 - 4 = 5
Δ₁ = \(\left| \begin{matrix} 12 & 2 \\ 13 & 3 \end{matrix} \right| \)= 36 - 26 = 10
Δ₂ = \(\left| \begin{matrix} 3 & 12 \\ 2 & 13 \end{matrix} \right| \)= 39 - 26 = 10
∴ z = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 10 }{ 5 } =2\Rightarrow \frac { 1 }{ x } =2\Rightarrow x=\frac { 1 }{ 2 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 15 }{ 5 } \) = 3
∴ Solution set {\(\frac{1}{2}\), 3}

2x-y = 8, 3x+2y+2 = -2
The matrix form of the system is
\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ AX = B where A =\(\\ \left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)
B =\(\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ X = A^-1N
Now, |A| =\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)= 4 + 3 = 7
∴ A^-1= \(\frac { 1 }{ |A| } \)adj A
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 16-2 \\ -24-4 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 14 \\ -28 \end{matrix} \right] =\left[ \begin{matrix} \frac { 14 }{ 7 } \\ \frac { -28 }{ 7 } \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \)
∴ x = 2, y = -4
Hence, the solution set is {2, -4}

\(\left[ \begin{matrix} 3 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -8 \\ \begin{matrix} -5 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 5 \\ \begin{matrix} 1 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ -2 \end{matrix} \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} 3 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -8 \\ \begin{matrix} -5 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 5 \\ \begin{matrix} 1 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ -2 \end{matrix} \end{matrix} \right] \)
A = \(\overset { { { R } }_{ 3 }\leftrightarrow { { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 2 \\ 3 \end{matrix}\begin{matrix} 2 \\ -5 \\ -8 \end{matrix}\begin{matrix} 3 \\ 1 \\ 5 \end{matrix}\begin{matrix} -2 \\ 4 \\ 2 \end{matrix} \right] \)
\(\begin{matrix} { { R } }_{ 2 }\rightarrow { { R } }_{ 2 }+2{ { R } }_{ 1 } \\ \longrightarrow \\ { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }+2{ { R } }_{ 1 } \end{matrix}\left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 2 \\ -1 \\ -2 \end{matrix}\begin{matrix} 3 \\ 7 \\ 14 \end{matrix}\begin{matrix} -2 \\ 0 \\ -4 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 2 \\ -1 \\ -1 \end{matrix}\begin{matrix} 3 \\ 7 \\ 7 \end{matrix}\begin{matrix} -2 \\ 0 \\ -2 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 2 \\ -1 \\ 0 \end{matrix}\begin{matrix} 3 \\ 7 \\ 0 \end{matrix}\begin{matrix} -2 \\ 0 \\ -2 \end{matrix} \right] \)
The last equivalent matrix is in row-echelon form. It has three non-zero rows.
∴ \(\rho \)(A) = 3

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 1 \\ 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \right] \overset { { { R } }_{ 2 }\rightarrow { { R } }_{ 2 }3{ { R } }_{ 1 }\\ { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-{ { R } }_{ 1 } }{ \underset { { { R } }_{ 4 }\rightarrow { { R } }_{ 4 }-{ { R } }_{ 1 } }{ \longrightarrow } } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} -4 \\ -3 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} -1 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 1 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 4 }\rightarrow { { R } }_{ 4 }-3{ { R } }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 1 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow 7{ { R } }_{ 3 }-{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 4 }\rightarrow 2{ { R } }_{ 4 }-{ { R } }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix} \right] \)
The last equivalent matrix is in row echelon form It has three non-zero rows.
∴ \(\rho \)(A) = 3

\(\frac { 3 }{ x } -\frac { 4 }{ y } -\frac { 2 }{ z } \) - 0, \(\frac { 1 }{ x } +\frac { 2 }{ y } +\frac { 1 }{ z } \) - 2 = 0, \(\frac { 2 }{ x } -\frac { 5 }{ y } -\frac { 4 }{ z } \) + 1 = 0
Put \(\frac { 1 }{ x } =u,\frac { 1 }{ y } =v,\frac { 1 }{ z } =w\)
We get 3u - 4v - 2w = 1, u + 2v + w = 2, 2u - 5v - 4w = -1
∴ \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| =3\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(- 8 + 5) + 4'(- 4 - 2) - 2(- 5 - 4)
= 3(- 3) + 4(- 6) - 2(- 9)
= - 9 - 24 + 18 = -15
Δ₁ = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ -1 & -5 \end{matrix} \right| \)
= 1(- 8 + 5) + 4(- 8 + 11 -2(-10 + 2)
= 1(- 3) + 4(-7) - 2(- 8)
= - 3 - 28 + 16= -15
Δ₂ = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| \)
= 3(- 8 + 1) - 1(- 4 - 2) - 2(- 1 - 4)
= 3(-7) - 1(- 6) - 2(- 5)
= -21 + 6 + 10 = -5
Δ₃ = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 2 \\ -5 & -1 \end{matrix} \right| +4\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 - 4)+ 1(-5 - 4)
= 3(8) + 4(- 5) + 1(- 9)
= 24 - 20 - 9 = - 5
∴ \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -15 }{ -15 } =1\Rightarrow \frac { 1 }{ x } =1\Rightarrow \)x = 1
v = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \Rightarrow \)y = 3
w = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 3 } \Rightarrow \)z = 3
∴ Solution set is {1, 3, 3}

3x + 3y − z = 11, 2x − y + 2z = 9, 4x + 3y + 2z = 25.
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-2-6)-3(4-8)-1(6+4)·
= 3(- 8) - 3(- 4) - 1(10)
= - 24 + 12 - 10 = - 22
Δ₂ = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= \(11\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -1\left| \begin{matrix} 9 & -1 \\ 25 & 3 \end{matrix} \right| \)
= 11(-2-6)-3(18-50)-1(27+25)
= 11(-8)-3(-32)-1(52)
= -88+96-52 = -44
Δ₂ = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -11\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| 2\begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| \)
= 3(18 - 50) -11(4 - 8) - 1(50- 36)
= 3(- 32) - 11(- 4) - 1(14)
= -96+44-14 = - 66
Δ₃ = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
\(3\left| \begin{matrix} -1 & 9 \\ 3 & 25 \end{matrix} \right| -3\left| \begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| -11\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-25-27)-3(50-36)+ 11(6+4)
= 3(- 52) - 3(14) + 11(10)
= -156 - 42 + 110= - 88
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -44 }{ -22 } \) = 2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -66 }{ -22 } \) = 3
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { -88 }{ -22 } \) = 4
∴ Solution set is {2, 3, 4}

x+y+z-2 = 0, 6x-4y+5z-31= 0, 5x+2y+2z = 13
The matrix form of the system is
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 13 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right| =1\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| -1\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \)
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| & +\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ -4 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 6 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 6 & -4 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(-8-10) & -(12-25) & +(12+20) \\ -(2-2) & +(2-5) & -(2-5) \\ +(5+4) & -(5-6) & +(-4-6) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 13 & 32 \\ 0 & -3 & 3 \\ 9 & 1 & -10 \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ X = A^-1B
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -36 & +0 & +117 \\ 26 & -93 & +13 \\ 64 & +93 & -130 \end{matrix} \right] =\frac { 1 }{ 27 } \left[ \begin{matrix} 81 \\ -54 \\ 27 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ 1 \end{matrix} \right] \)
∴ x = 3, y = -2, z = 1
∴ Solution set is {3, -2, 1}

2x + 3y - z = 9, x + y + z = 9, 3x - y - z = -1
The matrix form of the system is
\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ X = A^-1N
|A| = \(\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \)
= 2(-1+1)-3(-1-3)-1(-1-3)
= 0-3(-4) = 12+4 = 16
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & -1 \\ -1 & -1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ 3 & -1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & -1 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(-1+1) & -(-1-3) & +(-1-3) \\ -(-3-1) & +(2+3) & -(-2-9) \\ +(3+1) & -(2+1) & +(2-3) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 0 & 4 & -4 \\ 4 & 1 & 11 \\ 4 & -3 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
= \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0+36-4 \\ 36+9+3 \\ -36+99+1 \end{matrix} \right] =\frac { 1 }{ 16 } \left[ \begin{matrix} 32 \\ 48 \\ 64 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \right] \)
∴ x = 2, y = 3, z = 4
∴ Solution set is {2, 3, 4}