If A is a singular matrix, n |A|=0
In this case, if (adj A)B≠O, n solution does not exist and system of equations is called inconsistent.

The given system have non-trivial solution if \(\begin{vmatrix}\beta & sin \alpha & cos \alpha \\1 & cos \alpha & sin \alpha \\ -1 & sin \alpha & -cos \alpha \end {vmatrix}\) = 0
On opening determinant we get β = sin 2α + cos 2 α
Therefore, -√2 ≤ β ≤ √2
Now, for β = 1,
sin 2α + cos 2 α = 1
=> (1/√2)sin 2α + (1/√2) cos 2α = (1/√2)
Or, cos(2α ? π/4) = 1/√2 = cos(2nπ ± π/4)
=> 2α = 2nπ ± π/4 + π/4

The given system of equations can be expressed in form of AX=B,
?X=A^-1 B
We know that, A^-1=\(\frac{1}{|A|}\) adj A
A^-1=\(\frac{1}{-32} \begin{bmatrix}-4&-4\\-5&3\end{bmatrix}\)=\(\begin{bmatrix}\frac{1}{8}&\frac{1}{8}\\\frac{5}{32}&-\frac{3}{32}\end{bmatrix}\)
?X=\(\begin{bmatrix}\frac{1}{8}&\frac{1}{8}\\\frac{5}{32}&-\frac{3}{32}\end{bmatrix}\begin{bmatrix}6\\4\end{bmatrix}\)=\(\begin{bmatrix}\frac{1}{8}×6+\frac{1}{8}×4\\\frac{5}{32}×6-\frac{3}{32}×4\end{bmatrix}\)=\(\begin{bmatrix}\frac{6}{8}+\frac{4}{8}\\\frac{30}{32}-\frac{12}{32}\end{bmatrix}\)=\(\begin{bmatrix}\frac{10}{8}\\\frac{18}{32}\end{bmatrix}\)=\(\begin{bmatrix}\frac{5}{4}\\\frac{9}{16}\end{bmatrix}\)

?=\(\begin{vmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{vmatrix}\)
?=(\(\sqrt{3}\)×2\(\sqrt{3}\))+\(\sqrt{2}\)
?=6+\(\sqrt{2}\).

\(\frac{bx}{y+z}\) = a  bx ? ay ? az = 0
\(\frac{cy}{z+x}\) = b  bx ? cy + bz = 0
\(\frac{az}{x+y}\) = c  cx + cy ? az = 0
\(\begin{vmatrix}b & -a & -a \\b & -c & b \\c & c & -a \end {vmatrix}\) = 0
Or, b(ca ? bc) + a(-ab ? bc) ? a(bc + c²) = 0
or, abc ? b²c ? a²b ? abc ? abc ? ac² = 0
or, a²b + b²c + c²a + abc = 0 which is required eliminate.

We have,\(\begin{vmatrix}1 & ab & (\frac{1}{a} + \frac{1}{b}) \\1 & bc & (\frac{1}{b} + \frac{1}{c}) \\1 & ca & (\frac{1}{c} + \frac{1}{a})\end {vmatrix}\)
= (1/abc)\(\begin{vmatrix}1 & ab & \frac{b + a}{ab} * abc \\1 & bc & \frac{b + c}{bc} * abc \\1 & ca & \frac{c + a}{ac} * abc \end {vmatrix}\)
= (1/abc)\(\begin{vmatrix}1 & ab & bc + ac \\1 & bc & ac + ab \\1 & ca & ab + bc \end {vmatrix}\)
Operating, C₃ = C₃ + C₂
= (1/abc)\(\begin{vmatrix}1 & ab & ab + bc + ac \\1 & bc & ab + bc + ac \\1 & ca & ab + bc + ac \end {vmatrix}\)
= ((ab + bc + ac)/abc)\(\begin{vmatrix}1 & ab & 1 \\1 & bc & 1 \\1 & ca & 1 \end {vmatrix}\)
= 0

Give that, A=\(\begin{bmatrix}8&5\\4&1\end{bmatrix}\)
adj A=\(\begin{bmatrix}1&-5\\-4&8\end{bmatrix}\)
|A|=8×1-(-5)×(-4)=8-20=-12
A^-1=\(\frac{1}{|A|}\) adj A=\(\frac{1}{-12} \begin{bmatrix}1&-5\\-4&8\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{12}&\frac{5}{12}\\\frac{1}{3}&-\frac{2}{3}\end{bmatrix}\).

Given that vertices are (3,2), (1,2), (5,k)
Therefore, area of triangle with vertices (3,2), (1,2), (5,k) is given by
Δ=\(\frac{1}{2}\) \(\begin{Vmatrix}3&2&1\\1&2&1\\5&k&1\end{Vmatrix}\)=0
Applying R₁→R₁-R₂, we get
\(\frac{1}{2}\begin{Vmatrix}2&0&0\\1&2&1\\5&k&1\end{Vmatrix}\)=0
Expanding along R₁, we get
\(\frac{1}{2}\) {2(2-k)-0+0}=0
2-k=0
k=2.

Δ=\(\begin{vmatrix}1&0&1\\0&0&1\\1&0&1\end{vmatrix}\)
Δ=1\(\begin{vmatrix}0&1\\0&1\end{vmatrix}\)-0\(\begin{vmatrix}0&1\\1&1\end{vmatrix}\)+1\(\begin{vmatrix}0&0\\1&0\end{vmatrix}\)
Δ=1(0-0)-0(0-1)+1(0-0)
Δ=0-0+0=0.

The given statement is true. The given system of equations can be expressed in form AX=B, where
A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\), B=\(\begin{bmatrix}5\\3\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\end{bmatrix}\)
|A|=2(1)-5(3)=2-15=-13≠0
?A is non-singular matrix.
?A^-1 exists. Hence, given system of equations is consistent.