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

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.

\(\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right] \)
Let A = \(\left( \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right) \)
adj A = \(\left( \begin{matrix} 2 & -4 \\ -6 & -3 \end{matrix} \right) \)
[Interchange the elements in the leading diagonal and change the sign of the elements in off diagonal]

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

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Expanding along R₁ we get,
|A| = \(2\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| -3\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| +1\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \)
= 2(8-7) -3 (6-3) +1(21-12)
= 2(1) - 3(3) + 1(9)
Since A is a non-singular matrix, A^-1 exis
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)+(14-9) \\ +(3-4)+(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)
Now, A-1 = \(\frac { 1 }{ |A| } \)adj A
⇒ A-1 = \(\frac{1}{2} \left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

Let A = \(\left[ \begin{matrix} 3 & 2 & 5 \\ 1 & 1 & 1 \\ 3 & 3 & 6 \end{matrix} \right] \). Then A is a matrix of order 3 \(\times\) 3. So ρ(A) ≤ min {3, 3} = 3. 
The highest order of minors of A is 3. There is only one third order minor of A.
It is \(\left| \begin{matrix} 3 & 2 & 5 \\ 1 & 1 & 1 \\ 3 & 3 & 6 \end{matrix} \right| \) = 3(6 - 6) - 2(6 - 6) + 5(3 - 3) = 0. So, ρ(A) < 3.
Next consider the second - order minors of A.
We find that the second order minor \(\left| \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right| \) = 3 - 2 ≠ 0. So  ρ(A) = 2.

\(\left[ \begin{matrix} 0 \\ -1 \\ 4 \end{matrix}\begin{matrix} 3 \\ 0 \\ 2 \end{matrix}\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}\begin{matrix} 6 \\ 5 \\ 0 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 4 & 2 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 0 & 0 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 2 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 8 & 20 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 2 }{ 3 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ \frac { 22 }{ 3 } & 16 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 3R_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 22 & 48 \end{matrix} \right] \)

Given A =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] =\left[ \begin{matrix} -3+10 & -9+4 \\ -7+25 & -21+10 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 7 & -5 \\ 18 & -11 \end{matrix} \right] \)
|AB| = -77+90 = 13 ≠ 0 ⇒ (AB)^-1 exists
|A| = 15-14 = 1 ≠ 0 ⇒ A^-1 exists
|B| = -2+15 = 13 ≠ 0 ⇒ B^-1 exists
(AB)^-1 = \(\frac { 1 }{ |AB| } adj(AB)=\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ...............(1)
B^-1 = \(\frac { 1 }{ |B| } adj(B)=\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \)
A^-1 = \(\frac { 1 }{ |A| } \)(adj A)
= \(\frac { 1 }{ 1 } \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) =\left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
∴ B^-1A^-1 = \(\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} 10-21 & -4+9 \\ -25+7 & 10-3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ..............(2)
 From (1) or (2) it is proved that
(AB)^-1 = B^-1 A^-1

Given A =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \)
A² = \(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 25-3 & 15-6 \\ -5+2 & -3+4 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22 & 9 \\ -2 & 1 \end{matrix} \right] \)
∴ A²- 3A - 7I₂
=\(\left[ \begin{matrix} 22 & 9 \\ -3 & 1 \end{matrix} \right] -3\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -7\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22-15-7 & 9-9+0 \\ -3+3+0 & 1+6-7 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)= O₂
Hence proved.
∴ A²-3A-7I₂ = O₂
Postmultiplying by A^-1 we get,
A²-A^-1-3AA^-1-7I₂A^-1 = 0.A^-1
⇒ A(AA^-1)-3(AA^-1)-7(A^-1) = 0
[∵ I₂A^-1 = A^-1 and | (0)A^-1= 0]
⇒ AI-3I-7A-1 = 0  [∵ AA^-1= 1]
⇒ AI-3I = 7A^-1
⇒ A^-1 = \(\frac { 1 }{ 7 } \)[A - 3I] [∴ AI = A]
⇒ A^-1 = \(\frac { 1 }{ 7 } =\left[ \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \right] \)
⇒ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 5-3 & 3-0 \\ -1-0 & -2-3 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \).

For the given A, We get |A| = (2)(7) - (9)(1) = 14 - 9 = 5. So, A^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -9 \\ -1 & 2 \end{matrix} \right] =\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 9 }{ 5 } \\ -\frac { 1 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] \).
Then, (A^-1)^T = \(\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 1 }{ 5 } \\ -\frac { 9 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).           ....(1)
For the given A, We get A^T = \(\left[ \begin{matrix} 2 & 1 \\ 9 & 7 \end{matrix} \right] \). So |A^T| = (2)(7) - (1)(9) = 5
Then, (A^T)^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).             ...(2)
From (1) and (2), we get (A^-1)^T = (A^T)^-1. Thus, we have verified the given property.

Let A = \(\left[ \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right] \). Then |A| = \(\left| \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right| \) = 2(7) + (-12) + 3(-1) = -1 ≠ 0.
Therefore, A⁻¹ exists. Now, we get
adj A = \({ \left[ \begin{matrix} +\left| \begin{matrix} 3 & 1 \\ 2 & 3 \end{matrix} \right| & -\left| \begin{matrix} -5 & 1 \\ -3 & 3 \end{matrix} \right| & +\left| \begin{matrix} -5 & 3 \\ -3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} -1 & 3 \\ 2 & 3 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ -3 & 3 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ -3 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} -1 & 3 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ -5 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right| \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 7 & 12 & -1 \\ 9 & 15 & -1 \\ -10 & -17 & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] \).
Hence, A^-1 = \(\frac { 1 }{ \left| A \right| } \)(adj A) = \(\frac { 1 }{ \left( -1 \right) } \left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] =\left[ \begin{matrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{matrix} \right] \).

2x + 3y − z = 0, x − y − 2z = 0, 3x + y + 3z = 0
Reducing the augmented matrix to row - echelon form we get
[A|0]=\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 2 & 3 & -1 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 4 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 0 & \frac { 33 }{ 5 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|0] = 3
So, \(\rho \)(A) = \(\rho \)(A|0]) = 3 = Number of unknowns Hence, the system is consistent with unique solutions.
Thus, the system has trivial solution only.
x = 0, y = 0, z = 0

Let the amount of 50% acid be x and the amount of 25% acid be y litre
By the given data, x + y = 10 ..............(1)
and \(x\left( \frac { 50 }{ 100 } \right) +y\left( \frac { 25 }{ 100 } \right) =10\left( \frac { 40 }{ 100 } \right) \)
⇒ 50x + 25y = 400 ⇒ 2x + y = 16 ...............(2)
The matrix from of the equation is \(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \end{matrix} \right] ,B=\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ X = A^-1N |A| = \(\left| \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right| \) = 1 - 2 = -1
⇒ X = \(\frac { 1 }{ |A| } \)adj A.B
⇒ X= \(-1\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
= -\(\left[ \begin{matrix} 10-16 \\ -20+16 \end{matrix} \right] \)
⇒ X = -\(\left[ \begin{matrix} -6 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 6 \\ 4 \end{matrix} \right] \)
Thus, the amount of 50% acid is 6 litre and the amount of 25% acid is 4 litre = 10 litres of 40% acid solution.

Let the prices per unit for the commodities A, B and C be Rs. x, Rs. y and Rs. z.
By the given data,
2x - 4y + 5z = 15000
3x + y - 2z = 1000
-x + 3y + z = 4000
The matrix form of the system of equations is
\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 12 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and B =\(\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| +4\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| +5\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \)
= 2 (1 + 6) + 4 (3 - 2) + 5 (9 + 1)
= 2 (7) + 4 (1) + 5(10) = 14 + 4 + 50 = 68.
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 5 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 5 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -1 & 3 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 5 \\ 1 & -2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 5 \\ 3 & -2 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ 3 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
= \(\left[ \begin{matrix} +(1+6) & -(3-2) & +(9+1) \\ -(-4-15) & +(2+5) & -(6-4) \\ +(8-5) & -(4-15) & +(2+12) \end{matrix} \right] \)
= \(\left[ \begin{matrix} 7 & -1 & 10 \\ 19 & 7 & -2 \\ 3 & 19 & 14 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adj=\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 105000+19000+12000 \\ -15000+7000+76000 \\ 150000-2000+56000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 136000 \\ 68000 \\ 204000 \end{matrix} \right] =\left[ \begin{matrix} 2000 \\ 1000 \\ 3000 \end{matrix} \right] \)
∴ x = 2000, y = 1000, z = 3000
Hence the prices per unit of the commodities A, B and C are Rs. 2000, Rs. 1000 and Rs. 3000 respectively.

Given A =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
|A| = 0-1\(\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \)
= -1(0-1) + 1(1-0) = 1 + 1 = 2
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} (0-1) & -(0-1) & +(1-0) \\ -(0-1) & +(0-1) & -(0-1) \\ +(1+0) & -(0-1) & +(0-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) ................(1)
Now A² =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{matrix} \right] =\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] \)
A²- 3I =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 2-3 & 1-0 & 1-0 \\ 1-0 & 2-3 & 1-0 \\ 1-0 & 1-0 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) .............(2)
From (1) and (2), it is proved that A^-1 = \(\frac{1}{2}\) [A² - 3I]

Identity matrix

Order of A is 2m + 2