We find that |A| = \(\left| \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & 4 \\ 2 & -4 & 3 \end{matrix} \right| \) = 8(21 - 16) + 6(-18 + 8) + 2(24 - 14) = 40 - 60 + 20 = 0
By the definition of adjoint, we get
adj A = \({ \left[ \begin{matrix} \left( 21-16 \right) & -\left( -18+8 \right) & \left( 24-14 \right) \\ -\left( -18+8 \right) & \left( 24-4 \right) & -\left( 32+12 \right) \\ \left( 24-14 \right) & -\left( -32+12 \right) & \left( 56-36 \right) \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
So, we get
A(adj A) = \(\left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & 80-120+40 & 80-120+40 \\ -30+70-40 & -60+140-80 & -60+140-80 \\ 10-40+30 & 20-80+60 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃,
Similarly, we get
(adj A)A = \(\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & -30+70-40 & 10-40+30 \\ 80-120+40 & -60+140-80 & 20-80+60 \\ 80-120+40 & -60+140-80 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃.
Hence, A(adj A) = (adj A)A = |A|I₃.

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.

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

AB =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] =\left[ \begin{matrix} 8+3 & 10+4 \\ 20+9 & 25+12 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
|AB| =\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
= 407 - 406 = 1 ≠ 0
(AB)^-1 = \(\frac { 1 }{ |AB| } \) adj(AB)
= \(\frac { 1 }{ 1 } \left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \) ....(1)
|A| =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \) = 6 - 5 =1
|B| =\(\left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] \) = 16 - 15 = 1
B^-1 = \(\frac { 1 }{ |B| } adjB=\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \)
A^-1 = \(\frac { 1 }{ |A| } adjA=\left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
∴ B^-1A^-1 =\(\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
=\(\\ \left[ \begin{matrix} 12+25 & -4-10 \\ -9-20 & 3+8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \)....(2)
From (1) and (2), (AB)^-1 = B^-1A^-1

Δ = \(\left| \begin{matrix} 3 & a \\ 2 & a \end{matrix} \right| \) = 3a - 2a = a
Δ₁ =\(\left| \begin{matrix} 4 & a \\ 2 & a \end{matrix} \right| \) = 4a - 2a = 2a
Δ₂ = \(\left| \begin{matrix} 3 & 4 \\ 2 & 2 \end{matrix} \right| \)= 6 - 8 = -2

\(\therefore x=\frac{\Delta_{1}}{\Delta}=\frac{2 \not a}{\not a}=2=y y=\frac{\Delta_{2}}{\Delta}=\frac{-2}{a}\)
∴ Solution set is {2,\(\frac { -2 }{ a } \)}

Let A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 0 & 5 \end{matrix} \right] \). Applying elementary row operations, we get
A \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & -6 & -4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & 0 & 0 \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ(A) = 2.

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.

First, we find |adj (A)| = \(\left| \begin{matrix} 7 & 7 & -7 \\ -1 & 11 & 7 \\ 11 & 5 & 7 \end{matrix} \right| \) =  7(77 - 35) - 7(-7 - 77) - 7(-5 - 121) = 1764 > 0
So, we get
A = \(\pm \frac { 1 }{ \sqrt { \left| adjA \right| } } \) adj(adj A) = \(\pm \frac { 1 }{ \sqrt { 1764 } } { \left[ \begin{matrix} +\left( 77-35 \right) & -\left( -7-77 \right) & +\left( -5-121 \right) \\ -\left( 49+35 \right) & +\left( 49+77 \right) & -\left( 35-77 \right) \\ +\left( 49+77 \right) & -\left( 49-7 \right) & +\left( 77+7 \right) \end{matrix} \right] }^{ T }\)
= \(\pm \frac { 1 }{ 42 } { \left[ \begin{matrix} 42 & 84 & -126 \\ -84 & 126 & 42 \\ 126 & -42 & 84 \end{matrix} \right] }^{ T }=\pm \left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ -3 & 1 & 2 \end{matrix} \right] \).

A is orthogonal

May be 4