x = 3,
y = 2

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

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.

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

\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₃] =\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 6 & -2 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-6{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & -1 \\ 0 & 4 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -6 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 0 & 1 & 0 \\ -1 & -4 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\\ \overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+R_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -2 & -3 & 1 \\ -1 & 1 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -2 & -3 & 1 \\ -3 & - & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
So, We get A^-1=\(\left[ \begin{matrix} -2 & -3 & 1 \\ -3 & -3 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \).

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 4 & -2 & 6 \\ 1 & 1 & -3 \\ 15 & -3 & 9 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 8 \\ -1 \\ 21 \end{matrix} \right] \)
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 4 & -2 & 6 \\ 1 & 1 & -3 \\ 15 & -3 & 9 \end{matrix}|\begin{matrix} 8 \\ -1 \\ 21 \end{matrix} \right] \) \(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 4 & -2 & 6 \\ 15 & -3 & 9 \end{matrix}|\begin{matrix} -1 \\ 8 \\ 21 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-15{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & -6 & 18 \\ 0 & -18 & 54 \end{matrix}|\begin{matrix} -1 \\ 12 \\ 36 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -6 \right) , \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -18 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 1 & -3 \end{matrix}|\begin{matrix} -1 \\ -2 \\ -2 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -1 \\ -2 \\ 0 \end{matrix} \right] \).
So, ρ(A)= ρ([A | B]) = 2 < 3. From the echelon form, we get the equivalent equations
x + y − 3z = −1, y − 3z = −2 , 0 = 0.
The equivalent system has two non-trivial equations and three unknowns. So, one of the unknowns should be fixed at our choice in order to get two equations for the other two unknowns. We fix z arbitrarily as a real number t , and we get y = 3t − 2, x = −1−(3t − 2) + 3t = 1. So, the solution is (x = 1, y = 3t − 2, z = t), where t is real. The above solution set is a one-parameter family of solutions. Here, the given system is consistent and has infinitely many solutions which form a one parameter family of solutions.

The path y = ax² + bx + c passes through the points (10, 8), (20, 16), (40, 22). So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16, 1600a + 40b + c = 22. To apply Cramer’s rule, we find
Δ = \(\left| \begin{matrix} 100 & 10 & 1 \\ 400 & 20 & 1 \\ 1600 & 40 & 1 \end{matrix} \right| =1000\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{matrix} \right| \) = 1000 [-2 + 12 - 6] = -6000,
Δ₁ = \(\left| \begin{matrix} 8 & 10 & 1 \\ 16 & 20 & 1 \\ 22 & 40 & 1 \end{matrix} \right| =20\left| \begin{matrix} 4 & 1 & 1 \\ 8 & 2 & 1 \\ 11 & 4 & 1 \end{matrix} \right| \) = 20[-8 + 3 + 10] = 100,
Δ₂ = \(\left| \begin{matrix} 100 & 8 & 1 \\ 400 & 16 & 1 \\ 1600 & 22 & 1 \end{matrix} \right| =200\left| \begin{matrix} 1 & 4 & 1 \\ 4 & 8 & 1 \\ 16 & 11 & 1 \end{matrix} \right| \) = 200[-3 + 48 - 84] = -7800,
Δ₃ = \(\left| \begin{matrix} 100 & 10 & 8 \\ 400 & 20 & 16 \\ 1600 & 40 & 22 \end{matrix} \right| =2000\left| \begin{matrix} 1 & 1 & 4 \\ 4 & 2 & 8 \\ 16 & 4 & 11 \end{matrix} \right| \) = 2000[-10 + 84 - 64] = 20000.
By Cramer’s rule, we get a = \(\frac { { \Delta }_{ 1 } }{ \Delta } =-\frac { 1 }{ 60 } \), b = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { 7800 }{ 6000 } =\frac { 78 }{ 60 } =\frac { 13 }{ 10 } \), c = \(\frac { { \Delta }_{ 3 } }{ \Delta } =\frac { 20000 }{ 6000 } =-\frac { 20 }{ 6 } =-\frac { 10 }{ 3 } \).
So, the equation of the path is y = \(\frac { 1 }{ 60 } { x }^{ 2 }+\frac { 13 }{ 10 } x-\frac { 10 }{ 3 } \).
When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the team won the match.

Given F (\(\alpha\)) = \(\left[ \begin{matrix} \cos { \alpha } & 0 & \sin { \alpha } \\ 0 & 1 & 0 \\ -\sin { \alpha } & 0 & \cos { \alpha } \end{matrix} \right] \)
Expanding along R₁ we get,
|F(\(\alpha\))| = cos \(\alpha\) \(\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| -0+sin\alpha \left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \)
= cos \(\alpha\) (cos - 0) + sin \(\alpha\) (0 + sin \(\alpha\))
= cos² + sin² \(\alpha\) = 1 ≠ 0
Since F (\(\alpha\)) is a non-singular matrix, [F(\(\alpha\))]^-1 exists
Now, adj (F(\(\alpha\))) = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ sin\alpha & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & sin\alpha \\ 0 & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & 0 \\ sin\alpha & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & sin\alpha \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & sin\alpha \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & 0 \\ 0 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(cos\alpha -0) & -(0) & +(0+sin\alpha ) \\ -(0) & +(cos^{ 2 }\alpha +sin^{ 2 }\alpha & -(0) \\ +(0-sin\alpha ) & -(0) & +(cos-0) \end{matrix} \right] ^{ T }\)
\(\left[ \begin{matrix} cos\alpha & 0 & +sin\alpha \\ 0 & 1 & 0 \\ -sin\alpha & 0 & cos\alpha \end{matrix} \right] =\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
∴ F(\(\alpha\))^-1 = \(\frac { 1 }{ |F(\alpha )| } \) adj (F(\(\alpha\)))
[F(\(\alpha\))]^-1 = \(\frac { 1 }{ 1 } \left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(1)
Now, F(-\(\alpha\))=\(\left[ \begin{matrix} cos(-\alpha ) & 0 & sin(-\alpha ) \\ 0 & 1 & 0 \\ -s9n(-\alpha ) & 0 & cos(-\alpha ) \end{matrix} \right] \)
=\(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(2)
[∵ cos \(\alpha\) is an even function, cos (-\(\alpha\)) = cos \(\alpha\) and sin \(\alpha\) is an odd function, sin (-\(\alpha\)) = -sin\(\alpha\)]
From (1) and (2)
[F(\(\alpha\))]^-1 = F (-\(\alpha\))

|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

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}

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
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

Given adj A =\(\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
Now adj(adj A) =\(\left[ \begin{matrix} +\left| \begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 2 \\ -1 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 0 \\ -1 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & 1 \\ 2 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(2-0) & -(0) & +(0+2) \\ -(0) & +(1+1) & -(0) \\ +(0+2) & -(0) & +(2-0) \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 2 & 0 & 2 \\ 0 & 2 & 0 \\ -2 & 0 & 2 \end{matrix} \right] ^{ T }\)
adj(adj A) =\(\left[ \begin{matrix} 2 & 0 & -2 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{matrix} \right] \)

Given A =\(\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
adj A =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of the elements in the off diagonal]
|A| = 24 - 20 = 4
∴ A(adj A) =\(\\ \left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) ....(1)
(adj A)(A) =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] =\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \)...(2)
|A|I₂ = 4\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) .....(3)
From (1), (2) and (3), it is proved that
A (adj A) = (adj A) A = |A|I₂