Expanding along first row, we get
?=8x+1(3x+5)-(2x-2)(x²-1)
=(24x²+43x+5)-(2x³-2x²-2x+2)
=-2x³+26x²+45x+3.

Δ=\(\begin{vmatrix}-a&b&c\\-2a+4x&2b-4y&2c+4z\\x&-y&z\end{vmatrix}\)
Using properties of determinants, given determinant can be expressed as a sum of two determinants.
Δ=\(\begin{vmatrix}-a&b&c\\-2a&2b&2c\\x&-y&z\end{vmatrix}\)+\(\begin{vmatrix}-a&b&c\\4x&-4y&4z\\x&-y&z\end{vmatrix}\)
Δ=2\(\begin{vmatrix}-a&b&c\\-a&b&c\\x&-y&z\end{vmatrix}\)+4\(\begin{vmatrix}-a&b&c\\x&-y&z\\x&-y&z\end{vmatrix}\)
Since two rows are similar in each of determinants, determinant is 0.

Here, f(x) = f?(x)
=> f(x) is purely real

We have, \(\begin{vmatrix}sin^2 a & sina \,cosa & cos^2 a\\sin^2 b & sinb\, cosb & cos^2 b\\sin^2 c & sinc\, cosc & cos^2 c \end {vmatrix}\)
Now, multiplying by 2 in both numerator and denominator of column 2 and C₁ = C₁ + C₃ we get,
1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & 2sina\, cosa & cos^2 a\\sin^2 b + cos^2 b & 2sinb\, cosb & cos^2 b\\sin^2 c + cos^2 c & 2sinc\, cosc & cos^2 c \end {vmatrix}\)
= 1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & sin2a & cos^2 a\\sin^2 b + cos^2 b & sin2b & cos^2 b\\sin^2 c + cos^2 c & sin2c & cos^2 c \end {vmatrix}\)
= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2 a\\1 & sin2b & cos^2 b\\1 & sin2c & cos^2 c \end {vmatrix}\)
Solving furr,
= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2?a \\0 & sin2b-sin2a & cos^2? b-cos^2 ?a \\0 & sin2c-sin2a & cos^2 ?c-cos^2 ?a \end {vmatrix}\)
= 1/2 [(sin2b – sin2a)(cos²?c – cos²?a) ? (cos² b – cos²a)(sin2c – sin2a)]
Now, since, [cos² ?A + cos² B = sin(A + B) * sin(B ? A)]
So, 1/2 [2 cos(a + b) sin(b ? a) * sin(c + a)sin(a ? c) ? sin(a + b)sin(a ? b) * 2 cos(a + c)sin(c ? a)]
= sin(a ? b)sin(c ? a)[sin(c + a)cos(a + b) ? cos(c + a) sin(a + b)]
= sin(a ? b) sin(c ? a) sin(c + a ? a ? b)
= -sin(a ? b) sin(b ? c) sin(c ? a)

Given that, A=\(\begin{bmatrix}1&0\\9&4\end{bmatrix}\)
We know that, A(adj A)=(adj A)A=|A|I
?|A|=4-0=4
?A(adj A)=|A|I=\(\begin{bmatrix}4&0\\0&4\end{bmatrix}\).

Solving given system of equation by Cramer?s rule, we get,
x = D₁/D, y = D₂/D, z = D₃/D where,
D = \(\begin{vmatrix}2 & -3 & 4 \\3 & -4 & 5 \\4 & -5 & 6 \end {vmatrix}\)
D = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 5 \\4 & 5 & 6 \end {vmatrix}\)
Now, performing, C₃ = C₃ ? C₂ and C₂ = C₂ ? C₁ we get,
D = –\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D = 0
Similarly,
D₁ = \(\begin{vmatrix}7 & -3 & 4 \\8 & -4 & 5 \\9 & -5 & 6 \end {vmatrix}\)
Now, performing, C₁ = C₁ ? C₃
D₁ = –\(\begin{vmatrix}3 & -3 & 4 \\3 & -4 & 5 \\3 & -5 & 6 \end {vmatrix}\)
Now, performing, C₃ = C₃ ? C₂
D₁ = -3\(\begin{vmatrix}1 & -3 & 1 \\1 & -4 & 1 \\1 & -5 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D₁ = 0
D₂ = \(\begin{vmatrix}2 & 7 & 4 \\3 & 9 & 5 \\4 & 8 & 6 \end {vmatrix}\)
Now, performing,
D₂ = –\(\begin{vmatrix}2 & 3 & 2 \\3 & 3 & 2 \\4 & 3 & 2 \end {vmatrix}\)
Now, performing, C₂ = C₂ ? C₃ and C₃ = C₃ ? C₁
D₂ = 6\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D₂ = 0
D₃ = \(\begin{vmatrix}2 & -3 & 7 \\3 & -4 & 9 \\4 & -5 & 6 \end {vmatrix}\)
D₃ = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 4 \\4 & 5 & 4 \end {vmatrix}\)
Now, performing, C₂ = C₂ ? C₂ and C₃ = C₃ ? C₂
D₃ = -4\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D₃ = 0
Since, D = D₁ = D₂ = D₃ = 0, thus, it has infinitely many solutions.

Δ=\(\begin{vmatrix}sin? \,y&0&sin? \,y\\cos? \,y&1&cos? \,x\\sin? \,y&1&sin? \,y \end{vmatrix}\)
Δ=sin? y \(\begin{vmatrix}1&cos? \,x\\1&sin \,?y \end{vmatrix}\)-0\(\begin{vmatrix}cos ?\,y&cos? \,x \\sin? \,y&sin \,y \end{vmatrix}\)+sin? y \(\begin{vmatrix}cos ?\,y&1\\sin \,?y&1\end{vmatrix}\)
Δ=sin ?y (sin? y-cos? x)-0+sin ?y (cos? y-sin ?y)
Δ=sin²?y-sin ?y cos? x+sin? y cos ?y-sin²?y=sin ?y (cos? y-cos? x)

The given system of equations can be expressed in form of AX=B,
?X=A^-1 B
A=\(\begin{bmatrix}3&-2\\2&2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\end{bmatrix}\), B=\(\begin{bmatrix}3\\4\end{bmatrix}\)
We know that, A^-1=\(\frac{1}{|A|}\) adj A
A^-1=\(\frac{1}{10} \begin{bmatrix}2&2\\-2&3\end{bmatrix}\)=\(\begin{bmatrix}\frac{1}{5}&\frac{1}{5}\\-\frac{1}{5}&\frac{3}{10}\end{bmatrix}\)
?X=A^-1 B=\(\begin{bmatrix}\frac{1}{5}&\frac{1}{5}\\-\frac{1}{5}&\frac{3}{10}\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}\)=\(\begin{bmatrix}\frac{1}{5}�3+\frac{1}{5}�4\\-\frac{1}{5}�3+\frac{3}{10}�4\end{bmatrix}\)=\(\begin{bmatrix}\frac{7}{5}\\\frac{3}{5}\end{bmatrix}\)

The area of triangle with vertices (2,3), (4,1), (5,0) is given by
Δ=\(\frac{1}{2}\begin{Vmatrix}2&3&1\\4&1&1\\5&0&1\end{Vmatrix}\)
Applying R₂→R₂-R₃
Δ=\(\frac{1}{2}\begin{Vmatrix}2&3&1\\-1&1&0\\5&0&1\end{Vmatrix}\)
Expanding along R₂, we get
Δ=\(\frac{1}{2}\) {-(-1)(3-0)+1(2-5)}
Δ=\(\frac{1}{2}\) (0-0)=0.

Here, in L.H.S we have,
\(\begin{vmatrix}2 & 5 & 8 \\1 & 4 & 7 \\6 & 4 & \alpha \end {vmatrix}\)
So, for trivial roots above value is = 0
=>\(\begin{vmatrix}2 & 5 & 8 \\1 & 4 & 7 \\6 & 4 & \alpha \end {vmatrix}\) = 0
Solving it furr we get α = 12