Determinant of 2x2 Matrix

Statement I

Put $$a=\cos\alpha,\;b=\cos\beta,\;c=\cos\gamma$$.
The first determinant becomes

$$\Delta_1=\begin{vmatrix}1&a&b\\a&1&c\\b&c&1\end{vmatrix}$$

For any symmetric matrix of this type the standard expansion gives
$$\Delta_1=1+2abc-a^{2}-b^{2}-c^{2}\;.\;-(1)$$

The second determinant is

$$\Delta_2=\begin{vmatrix}0&a&b\\a&0&c\\b&c&0\end{vmatrix}$$

Expanding along the first row:

$$\Delta_2=0\cdot\Bigl|\begin{smallmatrix}0&c\\c&0\end{smallmatrix}\Bigr| -a\Bigl|\begin{smallmatrix}a&c\\b&0\end{smallmatrix}\Bigr| +b\Bigl|\begin{smallmatrix}a&0\\b&c\end{smallmatrix}\Bigr| = -a(-bc)+b(ac)=2abc\;.\;-(2)$$

Given $$\Delta_1=\Delta_2$$, substitute $$(1)$$ and $$(2)$$:

$$1+2abc-a^{2}-b^{2}-c^{2}=2abc \;\;\Longrightarrow\;\;1-a^{2}-b^{2}-c^{2}=0$$

Hence $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=a^{2}+b^{2}+c^{2}=1$$, not $$\frac{3}{2}$$.
So Statement I is false.

Statement II

Let

$$D(x)=\begin{vmatrix}x^{2}+x & x+1 & x-2\\ 2x^{2}+3x-1 & 3x & 3x-3\\ x^{2}+2x+3 & 2x-1 & 2x-1\end{vmatrix}$$

Use the cofactor rule $$D=A_1E_1-B_1E_2+C_1E_3$$ with
$$(A_1,B_1,C_1)=(x^{2}+x,\;x+1,\;x-2).$$

$$E_1=\begin{vmatrix}0&3x-3\\2x-1&0\end{vmatrix}=3(2x-1)=6x-3$$
$$\;\;\Longrightarrow\;A_1E_1=(x^{2}+x)(6x-3)=6x^{3}+3x^{2}-3x$$

$$E_2=\begin{vmatrix}2x^{2}+3x-1&3x-3\\x^{2}+2x+3&2x-1\end{vmatrix} =x^{3}+x^{2}-8x+10$$
$$\;\;\Longrightarrow\;B_1E_2=(x+1)(x^{3}+x^{2}-8x+10) =x^{4}+2x^{3}-7x^{2}+2x+10$$

$$E_3=\begin{vmatrix}2x^{2}+3x-1&3x\\x^{2}+2x+3&2x-1\end{vmatrix} =x^{3}-2x^{2}-14x+1$$
$$\;\;\Longrightarrow\;C_1E_3=(x-2)(x^{3}-2x^{2}-14x+1) =x^{4}-4x^{3}-10x^{2}+29x-2$$

Therefore

$$D(x)=\bigl(6x^{3}+3x^{2}-3x\bigr) -\bigl(x^{4}+2x^{3}-7x^{2}+2x+10\bigr) +\bigl(x^{4}-4x^{3}-10x^{2}+29x-2\bigr)$$

Simplifying term by term:

$$D(x)=24x-12=12(2x-1)$$

Thus $$p=24,\;q=-12$$ and
$$p^{2}=24^{2}=576,\quad 196q^{2}=196\cdot144=28224\neq576.$$

Hence Statement II is also false.

Both statements are false  →  Option B.

First, compute $$A^T$$, the transpose of A:

$$A^T = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix}$$.

Now, compute $$A^T A$$:

$$A^T A = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix} \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I$$.

Thus, $$A^T A = 3I$$.

 $$C = ABA^T$$.

 $$C^2 = (ABA^T)(ABA^T) = AB(A^T A)BA^T = AB(3I)BA^T = 3AB^2A^T$$, since scalar multiplication commutes.

 compute $$B^2$$:

$$B^2 = B \cdot B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} (1)(1) + (0)(1) & (1)(0) + (0)(1) \\ (1)(1) + (1)(1) & (1)(0) + (1)(1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$$.

So, $$C^2 = 3A B^2 A^T = 3A \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} A^T$$.

Now, $$X = A^T C^2 A = A^T \left(3A B^2 A^T\right) A = 3 A^T A B^2 A^T A$$.

Substitute $$A^T A = 3I$$:

$$X = 3 (3I) B^2 (3I) = 3 \cdot 3 \cdot 3 \cdot I B^2 I = 27 B^2$$, since the identity matrix commutes.

$$X = 27 \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 27 & 0 \\ 54 & 27 \end{pmatrix}$$.

compute $$\det X$$:

For a matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, $$\det = ad - bc$$.

So, $$\det X = (27)(27) - (0)(54) = 729 - 0 = 729$$.

Given matrix $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$$ and the condition $$|2A|^3 = 2^{21}$$, with $$\alpha, \beta \in \mathbb{Z}$$.

First, recall that for an $$n \times n$$ matrix, $$|kA| = k^n |A|$$. Here, $$A$$ is a 3x3 matrix, so $$n=3$$. Thus, $$|2A| = 2^3 |A| = 8 |A|$$.

Substitute into the given condition: $$(8 |A|)^3 = 2^{21}$$. Since $$8 = 2^3$$, this becomes: $$(2^3 |A|)^3 = 2^{21}$$. Simplifying the exponents: $$2^{9} |A|^3 = 2^{21}$$.

Divide both sides by $$2^9$$: $$|A|^3 = 2^{12}$$. Taking the cube root: $$|A| = 2^{4} = 16$$. Therefore, the determinant of $$A$$ is 16.

Now, compute the determinant of $$A$$. Expanding along the first row: $$|A| = 1 \cdot \begin{vmatrix} \alpha & \beta \\ \beta & \alpha \end{vmatrix} - 0 \cdot (\ldots) + 0 \cdot (\ldots) = \alpha^2 - \beta^2$$.

So, $$\alpha^2 - \beta^2 = 16$$. Factorizing: $$(\alpha - \beta)(\alpha + \beta) = 16$$. Since $$\alpha$$ and $$\beta$$ are integers, both $$\alpha - \beta$$ and $$\alpha + \beta$$ are integers and must be even (as their product is even and both must have the same parity).

Set $$x = \alpha - \beta$$ and $$y = \alpha + \beta$$, so $$x \cdot y = 16$$, and both $$x$$ and $$y$$ are even integers. The even factor pairs of 16 are: $$(2,8)$$, $$(4,4)$$, $$(8,2)$$, $$(-2,-8)$$, $$(-4,-4)$$, $$(-8,-2)$$.

Solve for $$\alpha$$ and $$\beta$$ in each pair using $$\alpha = \frac{x + y}{2}$$ and $$\beta = \frac{y - x}{2}$$:

• For $$(2,8)$$: $$\alpha = \frac{2+8}{2} = 5$$, $$\beta = \frac{8-2}{2} = 3$$
• For $$(4,4)$$: $$\alpha = \frac{4+4}{2} = 4$$, $$\beta = \frac{4-4}{2} = 0$$
• For $$(8,2)$$: $$\alpha = \frac{8+2}{2} = 5$$, $$\beta = \frac{2-8}{2} = -3$$
• For $$(-2,-8)$$: $$\alpha = \frac{-2-8}{2} = -5$$, $$\beta = \frac{-8-(-2)}{2} = -3$$
• For $$(-4,-4)$$: $$\alpha = \frac{-4-4}{2} = -4$$, $$\beta = \frac{-4-(-4)}{2} = 0$$
• For $$(-8,-2)$$: $$\alpha = \frac{-8-2}{2} = -5$$, $$\beta = \frac{-2-(-8)}{2} = 3$$

The possible values of $$\alpha$$ are $$5, 4, -5, -4$$. Comparing with the options: A. $$3$$, B. $$5$$, C. $$17$$, D. $$9$$, the value $$\alpha = 5$$ is present (option B).

Verification for $$\alpha = 5$$, $$\beta = 3$$: $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & 3 \\ 0 & 3 & 5 \end{bmatrix}$$, $$|A| = 5^2 - 3^2 = 25 - 9 = 16$$. Then $$|2A| = 8 \times 16 = 128$$, and $$|2A|^3 = 128^3 = (2^7)^3 = 2^{21}$$, which satisfies the condition.

Checking other options:

• For $$\alpha = 3$$: $$\alpha^2 - \beta^2 = 9 - \beta^2 = 16 \implies \beta^2 = -7$$, not real.
• For $$\alpha = 17$$: $$289 - \beta^2 = 16 \implies \beta^2 = 273$$, not a perfect square.
• For $$\alpha = 9$$: $$81 - \beta^2 = 16 \implies \beta^2 = 65$$, not a perfect square.

Thus, the only valid value from the options is $$\alpha = 5$$.