Matrices and Determinants

Given system of equation

$ \begin{aligned} x+y-z & =6 \\ 3 x+2 y-z & =5 \\ 2 x-y-2 z & =-3 \end{aligned} $

We can represent this as

$ A \cdot X=B $

Where

$ \begin{aligned} & A=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 3 & 2 & -1 \\ 2 & -1 & -2 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 6 \\ 5 \\ -3 \end{array}\right] \\ & \text { Now, }[A / B]=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 3 & 2 & -1 \\ 2 & -1 & -2 \end{array}\right]\left[\begin{array}{c} 6 \\ 5 \\ -3 \end{array}\right] \end{aligned} $

$ \begin{aligned} & R_2 \rightarrow R_2-3 R_1=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -1 & 2 \\ 2 & -1 & -2 \end{array}\right]\left[\begin{array}{c} 6 \\ -13 \\ -3 \end{array}\right] \\ & R_3 \rightarrow R_3-2 R_1=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -1 & 2 \\ 0 & -3 & 0 \end{array}\right]\left[\begin{array}{c} 6 \\ -13 \\ -15 \end{array}\right] \\ & \begin{aligned} R_2 \rightarrow-R_2 & =\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -2 \\ 0 & -3 & 0 \end{array}\right]\left[\begin{array}{c} 6 \\ 13 \\ -15 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & R_3 \rightarrow R_3+3 R_2=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & -6 \end{array}\right]\left[\begin{array}{c} 6 \\ 13 \\ 24 \end{array}\right] \\ & R_3 \rightarrow \frac{R_3}{-6} \\ &=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 6 \\ 13 \\ -4 \end{array}\right] \end{aligned} $

$ \begin{aligned} R_2 \rightarrow R_2+2 R_3 & =\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 6 \\ 5 \\ -4 \end{array}\right] \end{aligned} $

Now, we get

$ \begin{aligned} x+y-z & =6 \\ y & =5 \\ z & =-4 \\ x+5+4 & =6 \\ x+9 & =6 \\ x & =-3 \end{aligned} $

Thus, $x=\alpha=-3$

$ \begin{aligned} & y=\beta=5 \\ & z=\gamma=-4 \end{aligned} $

then, $\alpha+\beta=-3+5=2$

We have, $\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$

$ \begin{aligned} & C_2 \rightarrow C_2-C_1 \text { and } C_3 \rightarrow C_3-C_1 \\ & =\left|\begin{array}{ccc} 1 & 0 & 0 \\ a^2 & b^2-a^2 & c^2-a^2 \\ a^3 & b^3-a^3 & c^3-a^3 \end{array}\right| \end{aligned} $

$\begin{aligned} & =(b-a)(c-a) \\ & \left|\begin{array}{ccc}1 & 0 & 0 \\ a^2 & (b+a) & (c+a) \\ a^3 & \left(b^2+a^2+a b\right) & \left(c^2+a^2+a c\right)\end{array}\right| \\ & =(b-a)(c-a) \\ & \left|\begin{array}{cc}b+a & c+a \\ b^2+a^2+a b & c^2+a^2+a c\end{array}\right| \\ & =\left[(b-a)(c-a)\left[b c^2+a^2 b+a b c+a c^2+a^3\right.\right. \\ & \left.+a^2 c-b^2 c-a^3-a^2 b-a^2 c-a b c-a b^2\right] \\ & =(b-a)(c-a)\left[b c^2+a c^2-b^2 c-a b^2\right] \\ & =(b-a)(c-a)[b c(c-b) \\ & +a(c-b)(c+b)] \\ & =(a-b)(b-c)(c-a)(a b+b c+c a)\end{aligned}$

To solve the given equation $\alpha A^2 + \beta A = 2I$ where $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$, we first need to compute $A^2$:

$ A^2 = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix} $

Calculating the matrix multiplication:

$ A^2 = \begin{bmatrix} (1)(1) + (2)(-2) & (1)(2) + (2)(-5) \\ (-2)(1) + (-5)(-2) & (-2)(2) + (-5)(-5) \end{bmatrix} = \begin{bmatrix} -3 & -8 \\ 8 & 21 \end{bmatrix} $

Next, substitute $A^2$ into the equation $\alpha A^2 + \beta A = 2I$:

$ \alpha \begin{bmatrix} -3 & -8 \\ 8 & 21 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $

Break this down into:

$ \begin{bmatrix} -3\alpha & -8\alpha \\ 8\alpha & 21\alpha \end{bmatrix} + \begin{bmatrix} \beta & 2\beta \\ -2\beta & -5\beta \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $

Equate the corresponding elements from both sides:

$-3\alpha + \beta = 2$

$-8\alpha + 2\beta = 0$

$8\alpha - 2\beta = 0$

$21\alpha - 5\beta = 2$

From the second equation $-8\alpha + 2\beta = 0$, we solve for $\beta$:

$ 2\beta = 8\alpha \Rightarrow \beta = 4\alpha $

Substitute $\beta = 4\alpha$ into the first equation:

$ -3\alpha + 4\alpha = 2 \Rightarrow \alpha = 2 $

Using $\beta = 4\alpha$, we find:

$ \beta = 4(2) = 8 $

Finally, calculate $\alpha + \beta$:

$ \alpha + \beta = 2 + 8 = 10 $

To find the value of $ a $ that makes the system of equations have no solution, we need to ensure that the determinant of the coefficient matrix is zero. This will indicate that the equations are linearly dependent, leading to no unique solutions.

Given the system of equations:

$ \begin{align*} x + 2y + 3z &= 6, \\ x + 3y + 5z &= 9, \\ 2x + 5y + az &= 12, \end{align*} $

we consider the coefficient matrix:

$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{bmatrix} $

Calculate the determinant of this matrix:

$ D = \begin{vmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{vmatrix} $

The determinant is calculated as follows:

$ \begin{align*} D &= 1 \cdot (3a - 25) - 2 \cdot (a - 10) + 3 \cdot (5 - 6) \\ &= 3a - 25 - 2a + 20 - 3 \\ &= 3a - 2a - 25 + 20 - 3 \\ &= a - 8. \end{align*} $

Setting the determinant $ D = 0 $ gives:

$ a - 8 = 0, $

which simplifies to:

$ a = 8. $

Therefore, the system of equations has no solution when $ a = 8 $.

We have.

$ \left|\begin{array}{ccc} 1-x & -2 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0 $

$ R_1 \rightarrow R_1+R_2+R_y \text { we get } $

$ \Rightarrow\left|\begin{array}{ccc} -x & -x & -x \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0 $

$ \Rightarrow-x\left|\begin{array}{ccc} 1 & 1 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0 $

$C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$, we get

$ \begin{aligned} & \Rightarrow-x\left|\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 6-x & 0 \\ 1 & -3 & -x \end{array}\right|=0 \\ & \Rightarrow \quad(-x)(-x)(6-x)=0 \\ & \quad x^{\prime}-6 x^2=0 \end{aligned} $

Sum of product of zero

$ =\frac{\text { Coefficient of } x}{\text { Cocflicient of } x^3}=0 $

Hence, $\alpha \beta+\beta y+p^2=0$

We are given that the determinant of a 3rd order matrix $ A $ is $ K $. Hence, $|A| = K$.

For a matrix $ A $, the determinant properties used here are:

The determinant of the transpose of a matrix is the same as the determinant of the matrix itself. So, $|A^T| = |A| = K$.

The determinant of the product of a matrix with its transpose is $|AA^T| = |A||A^T| = K \times K = K^2$.

Now, we are asked to find the sum of the determinants of matrices $ A^4 $ and $ (A - A^4) $.

Since the determinant of a product is the product of the determinants and given $|A| = K$, we have:

$|A^4| = |A|^4 = K^4$.

To find $|A - A^4|$, note that:

By matrix properties, $|(A - A^4)| = \text{simplify or evaluate separately}$.

Given the expresion $|(AA^T) + (A - A^T)|$ in the context, we simplify:

$|A A^T| + |(A - A^4)| = K^2 + |A| - |A^T| = K^2 + K - K = K^2$.

Thus the result for the sum of these determinants aligns with the property calculations stated, giving us the simplified determinant result of $ K^2 $.

We have,

$ \begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5 \end{array}\right| \\ &=1(-5-2)-1(10+2)+1(2-1) \\ &=-7-12+1 \\ &=-18 \neq 0 \\ & \Delta_1=\left|\begin{array}{ccc} 5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5 \end{array}\right| \\ &=5(-5-2)-1(20-22)+1(4+11) \\ &=-35+2+15 \\ &=-18 \\ & \Delta_3=\left|\begin{array}{ccc} 1 & 1 & 5 \\ 2 & -1 & 4 \\ -1 & 1 & 11 \end{array}\right| \\ &=1(-11-4)-1(22+4)+5(2-1) \\ &=-15-26+5=-36 \\ & \therefore \quad \alpha=\frac{\Delta_1}{\Delta}=\frac{-18}{-18}=1, \\ & \therefore \quad \beta=\frac{\Delta_3}{\Delta}=\frac{-36}{-18}=2 \end{aligned} $

Hence, $\alpha^2+\beta^2=1^2+2^2=5$.

Here,

$\begin{aligned} & A=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right] \\ & A^T=\left[\begin{array}{ccc} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{array}\right] \\ & A A^T=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{array}\right] \\ & A A^T=\left[\begin{array}{ccc} 9+9+16 & 6+9+16 & 0+3+4 \\ 6+9+16 & 4+9+16 & 0+3+4 \\ 0+3+4 & 0+3+4 & 0+1+1 \end{array}\right] \\ & A A^T=\left[\begin{array}{ccc} 34 & 31 & 7 \\ 31 & 29 & 7 \\ 7 & 7 & 2 \end{array}\right] \\ & \left(A A^T\right)^T=\left[\begin{array}{ccc} 34 & 31 & 7 \\ 31 & 29 & 7 \\ 7 & 7 & 2 \end{array}\right] \\ \end{aligned}$

So, $A A^T$ is a symmetric matrix.

The given system of equation is $5 x-y+2 z=3, x+y+z=6$ and $2 x+y-z=-5$ In matrix form, if may be represented as where

$A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 5 & -1 & 2 \\ 2 & 1 & -1 \end{array}\right], x=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } D=\left[\begin{array}{c} 6 \\ 3 \\ -5 \end{array}\right]$

Thus, $A$ is a non-singular matrix, so $A^{-1}$ exist.

$\begin{aligned} & C_{11}=(-1)^{1+1}\left|\begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array}\right|=-1 \\ & C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 5 & 2 \\ 2 & -1 \end{array}\right|=9 \\ & C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 5 & -1 \\ 2 & 1 \end{array}\right|=7 \\ & C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right|=2 \end{aligned}$

$\begin{aligned} & C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right|=-3 \\ & C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=1 \\ & C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array}\right|=3 \\ & C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 5 & 2 \end{array}\right|=3 \\ & C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 5 & -1 \end{array}\right|=-6 \end{aligned}$

$\begin{aligned} \text { Then, } \operatorname{adj}(A) & =\left[\begin{array}{ccc} -1 & 9 & 7 \\ 2 & -3 & 1 \\ 3 & 3 & -6 \end{array}\right]^T=\left[\begin{array}{ccc} -1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6 \end{array}\right] \\ (\operatorname{adj} A) \cdot D & =\left[\begin{array}{ccc} -1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6 \end{array}\right]\left[\begin{array}{c} 6 \\ 3 \\ -5 \end{array}\right] \\ & =\left[\begin{array}{c} -6+6-15 \\ 54-9-15 \\ 42+3+30 \end{array}\right]=\left[\begin{array}{c} -15 \\ 30 \\ 75 \end{array}\right] \end{aligned}$

$\begin{aligned} & A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right], B=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow A^2=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \\ & \Rightarrow A^2=\left[\begin{array}{ll} 1+0 & 0+0 \\ 2+2 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 4 & 1 \end{array}\right] \\ & \Rightarrow A^3=\left[\begin{array}{ll} 1 & 0 \\ 4 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \quad\left[A^2 \cdot A=A^3\right]\\ & \Rightarrow A^3=\left[\begin{array}{ll} 1+0 & 0+0 \\ 4+2 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 6 & 1 \end{array}\right] \\ & \text { Similarly, } A^6=\left[\begin{array}{ll} 1 & 0 \\ 12 & 1 \end{array}\right] \quad \left[A^n=A_{21}=n \times 2\right]\\ & \text { Now, } B=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow B^2=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow B^2=\left[\begin{array}{ll} 1+0 & 3+3 \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow B^3=\left[\begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 9 \\ 0 & 1 \end{array}\right] \end{aligned}$

Similarly, $B^6=\left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right]\quad \left[B^n=B_{12}=n \times 3\right]$

$\begin{aligned} &\begin{aligned} & \therefore \quad\left(A^6+B^6\right)=\left[\begin{array}{cc} 1 & 0 \\ 12 & 1 \end{array}\right]+\left[\begin{array}{cc} 1 & 18 \\ 0 & 1 \end{array}\right] \\ & A^6+B^6=\left[\begin{array}{cc} 2 & 18 \\ 12 & 2 \end{array}\right] \\ & \end{aligned}\\ &\operatorname{Det}\left(A^6+B^6\right)=4-216=-212 \end{aligned}$

$\text { Here, } G(x)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$

$G(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right]$

So, $x+y=0 \Rightarrow y=(-x)$

Now, $G(-x)=\left[\begin{array}{ccc}\cos (-x) & -\sin (-x) & 0 \\ \sin (-x) & \cos (-x) & 0 \\ 0 & 0 & 1\end{array}\right]$

$\begin{aligned} G(x) \cdot G(y) & =\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}$

Given, $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 1\end{array}\right]$

$\begin{aligned} \left(A^T\right)^2 & +(12 A)^T \\ & =\left[\begin{array}{cc} 2 & -4 \\ -3 & 1 \end{array}\right]^2+\left[\begin{array}{cc} 24 & -36 \\ -48 & 12 \end{array}\right]^T \\ & =\left[\begin{array}{cc} 16 & -12 \\ -9 & 13 \end{array}\right]+\left[\begin{array}{cc} 24 & -48 \\ -36 & 12 \end{array}\right] \\ & =\left[\begin{array}{cc} 16+24 & -12-48 \\ -9-36 & 13+12 \end{array}\right]=\left[\begin{array}{cc} 40 & -60 \\ -45 & 25 \end{array}\right] \end{aligned}$

Given, $a, b$ and $c$ are respectively 5th, 8th, and 13th terms of arithmetic progression.

$\begin{aligned} & \therefore a=P+(5-1) d \\ & \Rightarrow a=P+4 d \\ & b=P+(8-1) d \\ & \Rightarrow \quad b=P+7 d \\ & c=P+(13-1) d \\ & \Rightarrow c=P+12 d \\ & \therefore \quad\left|\begin{array}{ccc} a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1 \end{array}\right|=\left|\begin{array}{ccc} P+4 d & 5 & 1 \\ P+7 d & 8 & 1 \\ P+12 d & 13 & 1 \end{array}\right| \\ & =\left|\begin{array}{ccc} P & 5 & 1 \\ P & 8 & 1 \\ P & 13 & 1 \end{array}\right|+\left|\begin{array}{ccc} 4 d & 5 & 1 \\ 7 d & 8 & 1 \\ 12 d & 13 & 1 \end{array}\right| \\ \end{aligned}$

$\begin{aligned} & =P\left|\begin{array}{ccc} 1 & 5 & 1 \\ 1 & 8 & 1 \\ 1 & 13 & 1 \end{array}\right|+d\left|\begin{array}{ccc} 4+1 & 5 & 1 \\ 7+1 & 8 & 1 \\ 12+1 & 13 & 1 \end{array}\right| \\ & =P \cdot 0+d\left|\begin{array}{ccc} 5 & 5 & 1 \\ 8 & 8 & 1 \\ 13 & 13 & 1 \end{array}\right|=P \cdot 0+d \cdot 0=0 \end{aligned}$

Given that,

$\begin{aligned} A & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ b+c a+b & 0 & 1 \end{array}\right] \end{aligned}$

$\begin{aligned} & \text { Since, } A^2=I \text { (given) } \\ & \Rightarrow b+c a+b=0 \\ & \Rightarrow b=\frac{-c a}{2} \end{aligned}$

Given,

$A=\left[\begin{array}{ccc} -2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1 \end{array}\right]$

Given, $\operatorname{det} A=\left|\begin{array}{ccc}-2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1\end{array}\right|=0$

$\begin{array}{ll} \Rightarrow & -2(-1-3)+x(2+x)+1(3 x-2)=0 \\ \Rightarrow & 8+2 x+x^2+3 x-2=0 \\ \Rightarrow & x^2+5 x+6=0 \Rightarrow(x+2)(x+3)=0 \\ \Rightarrow & x=-2 \text { and }-3 \\ \therefore & l=-2 \text { and } m=-3 \\ & l^3-m^3=(-2)^3-(-3)^3=-8+27=19 \end{array}$

$\begin{aligned} & \text { Given, } a_i^2+b_i^2+c_i^2=1 \\ & \Rightarrow \quad a_1^2+b_1^2+c_1^2=1 \\ & \qquad a_2^2+b_2^2+c_2^2=1 \\ & \qquad a_3^2+b_3^2+c_3^2=1 \\ \end{aligned}$

Given, $a_i a_j+b_i b_j+c_i c_j=0\quad [i, j=1,2,3, i \neq j]$

$\begin{aligned} \text { Let } \mathbf{p} & =a_1 \mathbf{i}+b_1 \mathbf{j}+c_1 \mathbf{k} \\ \mathbf{q} & =a_2 \mathbf{i}+b_2 \mathbf{j}+c_2 \mathbf{k} \\ \mathbf{r} & =a_3 \mathbf{i}+b_3 \mathbf{j}+c_3 \mathbf{k} \\ |\mathbf{p}| & =|\mathbf{q}|=|\mathbf{r}|=1 \\ \mathbf{p} \cdot \mathbf{q} & =\left(a_1 \mathbf{i}+b_1 \mathbf{j}+c_1 \mathbf{k}\right) \cdot\left(a_2 \mathbf{i}+b_2 \mathbf{j}+c_3 \mathbf{k}\right) \\ & =a_1 a_2+b_1 b_2+c_1 c_2=0 \\ \Rightarrow \mathbf{p} & \perp \mathbf{q} \end{aligned}$

Similarly, $\mathbf{q} \perp \mathbf{r}, \mathbf{r} \perp \mathbf{p}$

$\begin{aligned} & \text { Now, } A=\left|\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right|=\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=\left[\begin{array}{lll} \mathbf{p} & \mathbf{q} & \mathbf{r} \end{array}\right]=1 \\ & \Rightarrow \operatorname{det}\left(A A^\tau\right)=1 \end{aligned}$

$\begin{aligned} & \text { } \begin{aligned} & A=\frac{1}{7}\left[\begin{array}{ccc} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{array}\right] \\ &|A|=\frac{1}{7^3}\left[\begin{array}{ccc} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{array}\right] \quad\left[\because \operatorname{det}(k A)=k^n \operatorname{det} A\right] \\ &=\frac{1}{7^3}[3(-9-12)+2(-18+4)+6(-36-6)] \\ &=\frac{1}{7^3}[-63-28-252] \\ &=\frac{-343}{7^3}=-1 \neq-0 \end{aligned} \end{aligned}$

$\operatorname{Adj}(A)=\frac{1}{7^2}\left[\begin{array}{ccc}-21 & 14 & -42 \\ 42 & 21 & -14 \\ 14 & -42 & -21\end{array}\right]$

$\left[\because \operatorname{adj}(k A)=k^{n-1} \operatorname{adj} A\right]$

$=\frac{1}{7^2}\left[\begin{array}{ccc} -21 & 42 & 14 \\ 14 & 21 & -42 \\ -42 & -14 & -21 \end{array}\right]$

Now, $A^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\frac{-1}{7^2}\left[\begin{array}{ccc}-21 & 42 & 14 \\ 14 & 21 & -42 \\ -42 & -14 & -21\end{array}\right]$

$\begin{aligned} & =\frac{1}{7}\left[\begin{array}{ccc} 3 & -6 & -2 \\ -2 & -3 & 6 \\ 6 & 2 & 3 \end{array}\right] \\ A^{-1} & =A^{T} \end{aligned}$

$A=\left[\begin{array}{cc}\alpha^2 & 5 \\ 5 & -\alpha\end{array}\right]$ and $\left|A^{10}\right|=1024$

$\begin{aligned} & |A|=-\alpha^3-25 \\ & \because \quad\left|A^{10}\right|=1024 \\ & \Rightarrow \quad|A|^{10}=2^{10} \\ & \Rightarrow \quad|A|= \pm 2 \\ \end{aligned}$

Taking positive value, $-\alpha^3-25=2 \Rightarrow-\alpha^3=27$

$\begin{aligned} & \Rightarrow \alpha^3=-27 \\ & \Rightarrow \alpha=-3 \end{aligned}$

Taking negative value, $\alpha^3-25=-2 \Rightarrow-\alpha^3=23$

$\Rightarrow \alpha=(-23)^{\frac{1}{3}}$

$A=\left[\begin{array}{ccc} 5 & \sin ^2 \theta & \cos ^2 \theta \\ -\sin ^2 \theta & -5 & 1 \\ \cos ^2 \theta & 1 & 5 \end{array}\right]$

$\begin{aligned} & |A|=5(-25-1)-\sin ^2 \theta\left(-5 \sin ^2 \theta-\cos ^2 \theta\right) \\ & +\cos ^2 \theta\left(-\sin ^2 \theta+5 \cos ^2 \theta\right) \\ & =-130-\sin ^2 \theta\left[-5\left(1-\cos ^2 \theta\right)-\cos ^2 \theta\right] \\ & +\cos ^2 \theta\left[-\sin ^2 \theta+5\left(1-\sin ^2 \theta\right)\right] \\ & =-130-\sin ^2 \theta\left(-5+4 \cos ^2 \theta\right)+\cos ^2 \theta\left(-6 \sin ^2 \theta+5\right) \\ & =-130+5 \sin ^2 \theta-4 \sin ^2 \theta \cos ^2 \theta \\ & -6 \sin ^2 \theta \cos ^2 \theta+5 \cos ^2 \theta \\ & =-130+5-10 \sin ^2 \theta \cos ^2 \theta \\ & |A|=-125-10 \sin ^2 \theta \cos ^2 \theta \\ & \therefore \quad|A|_{\max }=-125 \\ & \end{aligned}$

$\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$

$\begin{aligned} \text { Let } x^2+1 & =k \\ \Rightarrow \quad x^2 & =k-1 \end{aligned}$

Substituting $x^2=k-1$ in LHS, we get

$\begin{aligned} & \frac{(k-1)^2+24(k-1)+28}{k^3}=\frac{k^2+22 k+5}{k^3} \\ &=\frac{1}{k}+\frac{22}{k^2}+\frac{5}{k^3} \\ &=\frac{1}{\left(x^2+1\right)}+\frac{22}{\left(x^2+1\right)^2}+\frac{5}{\left(x^2+1\right)^3} \end{aligned}$

On comparing with RHS, we get

$\begin{aligned} & A x+B=1, C x+D=22 \text { and } E x+F=5 \\ & \text { At } x=1, \\ & A+B=1, C+D=22 \text { and } E+F=5 \\ & \therefore A+B+C+D+E+F=1+22+5=28 \end{aligned}$

$\operatorname{det} A=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=k$ (given)

$\begin{aligned} \text { Then, } \operatorname{det} B & =\left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2+2 a_1 & b_2+2 b_1 & c_2+2 c_1 \\ a_3 & b_3 & c_3 \end{array}\right| \\ & =\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|+\left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ 2 a_1 & 2 b_1 & 2 c_1 \\ a_3 & b_3 & c_3 \end{array}\right| \\ & =k+0 \\ \Rightarrow \operatorname{det} B & =k \end{aligned}$

$\begin{aligned} A & =\left[\begin{array}{llll}\sqrt{2020} & \sqrt{2022} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092}\end{array}\right] \\ & =\left[\begin{array}{cccc}\sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{2} \cdot \sqrt{2020} & \sqrt{2} \sqrt{2021} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{3} \sqrt{2020} & \sqrt{3} \sqrt{2021} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{4} \sqrt{2020} & \sqrt{4} \sqrt{2021} & \sqrt{8088} & \sqrt{8092}\end{array}\right] \\ R_2 & \rightarrow R_2-\sqrt{2} R_1, R_3 \rightarrow R_3-\sqrt{3} R_1, R_4 \rightarrow R_4-\sqrt{4} R_1 \\ A & \sim\left[\begin{array}{cccc}\sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]\end{aligned}$

$\because$ A has one non-zero row

$\therefore$ Rank (A) = 1

Option (a) is correct.

The given equation is:

$ \left| \begin{array}{lll} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{array} \right| = 0 $

We can rewrite the third column as (1) + (the cube term), so this is the same as:

$ \left| \begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array} \right| + \left| \begin{array}{lll} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{array} \right| = 0 $

Let’s call the first determinant as $A$ and the second as $B$.

Step 1: Find the value of $|A|$

Let: $ A = \left| \begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array} \right| $

Subtract first row from second and third, $ R_2 \rightarrow R_2 - R_1,\quad R_3 \rightarrow R_3 - R_1 $ so, $ \left| \begin{array}{ccc} x & x^2 & 1 \\ y-x & y^2 - x^2 & 0 \\ z-x & z^2 - x^2 & 0 \end{array} \right| $

Now, since last column for $R_2$ and $R_3$ is 0, expand along third column:

$|A| = (y-x)(z^2 - x^2) - (z-x)(y^2 - x^2)$

$= (y-x)(z-x)(z+x) - (z-x)(y-x)(y+x)$

$= (y-x)(z-x)(z+x - y-x)$

$= (y-x)(z-x)(z-y)$

Step 2: Find the value of $|B|$

$ B = \left| \begin{array}{ccc} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{array} \right| $

Factor $x, y, z$ from each row: $ = x y z \left| \begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array} \right| $

Now, do the same row operation as before: $ R_2 \rightarrow R_2 - R_1,\quad R_3 \rightarrow R_3 - R_1 $

So determinant becomes: $ x y z \left| \begin{array}{ccc} 1 & x & x^2 \\ 0 & y-x & y^2 - x^2 \\ 0 & z-x & z^2 - x^2 \end{array} \right| $

Expand this along the first column:

$|B| = x y z \left\{ (y-x)(z^2 - x^2) - (z-x)(y^2 - x^2) \right\}$

$= x y z (y-x)(z-x)(z-y)$

Step 3: Use the original equation

We have: $ |A| + |B| = 0 $ So, $ (y-x)(z-x)(z-y) + x y z (y-x)(z-x)(z-y) = 0 $

Factor out $(y-x)(z-x)(z-y)$: $ (y-x)(z-x)(z-y)(1 + x y z) = 0 $

Because $x, y, z$ are all different, $(y-x)(z-x)(z-y) \neq 0$. So, $ 1 + x y z = 0 $

Therefore, $ x y z = -1 $

Let $n=2$ (say) and $A=\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$, Then $\operatorname{adj} A=\left[\begin{array}{cc}c & -b \\ 0 & a\end{array}\right]$ which is a upper triangular matrix. Then, for any $n \times n$ matrix $A$, $\operatorname{adj} A$ is also upper triangular matrix.

$f(x)=\left|\begin{array}{ccc}x & x^2 & x^3 \\ 1 & 2 x & 3 x^2 \\ 0 & 2 & 6 x\end{array}\right|$

Expand along column 1

$\begin{aligned} f(x) & =x\left(12 x^2-6 x^2\right)-\left(6 x^3-2 x^3\right) \\ & =x\left(6 x^2\right)-4 x^3=6 x^3-4 x^3=2 x^3 \\ \therefore \quad f(x) & =2 x^3 \Rightarrow f^{\prime}(x)=6 x^2 \\ f^{\prime \prime}(x) & =12 x \end{aligned}$

Then, $f^{\prime \prime}(x): f^{\prime}(x)=12 x: 6 x^2=2: x$

Trace of A = Sum of diagonal elements of A

Given, $A=\left[\begin{array}{ccc}1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9\end{array}\right]$

$=1+7+9=17$

$\begin{aligned} & \text { }-x+y \cos C+z \cos B=0 \\ & x \cos C-y+z \cos A=0 \\ & x \cos B+y \cos A-Z=0 \\ & \Delta=\left|\begin{array}{ccc}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1\end{array}\right| \\ & =-1\left(1-\cos ^2 A\right)+\cos C(\cos A \cos B+\cos C) \\ & +\cos B(\cos A \cos C+\cos B) \\ & \end{aligned}$

$\begin{aligned} & =-1+\cos ^2 A+2 \cos A \cos B \cos C \\ & \qquad+\cos ^2 C+\cos ^2 B \\ & =-\sin ^2 A+\cos ^2 B+\cos ^2 C+2 \cos A \cos B \cos C \\ & \text { Clearly, } \Delta>0 . \\ & \Rightarrow \text { System of equations has a non-zero solution. }\end{aligned}$

$\begin{aligned} \text { Let } A & =\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right] \\ A^{-1} & =\frac{\operatorname{adj}(A)}{|A|} \\ & =\left(\frac{1}{\sec ^2 \theta}\right)\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right] \end{aligned}$

So, $\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$

$\begin{aligned} &=\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc} 1-\tan ^2 \theta & -2 \tan \theta \\ 2 \tan \theta & -\tan ^2 \theta+1 \end{array}\right] \\ & {\left[\begin{array}{ll} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] } \\ & \Rightarrow \quad a=\cos 2 \theta, b=\sin 2 \theta \end{aligned}$

$\begin{aligned} & {\left[\begin{array}{ccc}a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b\end{array}\right] \quad\left[R_1 \rightarrow R_1+R_3\right]} \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 1 & 1 \\ a-b & b-c & c-a \\ b+c & c+a & a+b\end{array}\right] \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 1 & 1 \\ a+c & b+a & b+c \\ b+c & c+a & a+b\end{array}\right] \quad {\left[R_2 \rightarrow R_2+R_3\right]} \\ & \end{aligned}$

$\begin{aligned} C_2 & \rightarrow C_2-C_1 \text { and } C_3 \rightarrow C_3-C_1 \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 0 & 0 \\ a+c & b-c & b-a \\ b+c & a-b & a-c\end{array}\right] \\ & =(a+b+c)[(b-c)(a-c)-(b-a)(a-b)] \\ & =(a+b+c)\left(a b-b c-c a+c^2-b a+b^2+a^2-a b\right) \\ & =(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\ & =a^3+b^3+c^3-3 a b c\end{aligned}$

$\begin{aligned} & \text { }\left|\begin{array}{ccc} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{array}\right| \\ & =(b+c)[(c+a)(a+b)-b c]-a\left(a b+b^2-b c\right) \\ & =(b+c)\left(a c+b c+a^2+a b-b c\right) \quad+a\left(b c-c^2-a c\right) \\ & -a^2 b-a b^2+a b c+a b c-a c^2-a^2 c \\ & =a b c+a^2 b+a b^2+a c^2+a^2 c+a b c -a^2 b-a b^2+a b c+a b c-a c^2-a^2 c\\ & =4 a b c \end{aligned}$

Option (c) is correct.

${D^T} = ABC$

and $S = ABCD$

${S^2} = (ABCD)(ABCD) = ({D^T}\,.\,D)({D^T}D)$

$ = {({D^T}D)^2}$ ..... (i)

$\because$ $S = ABCD = {D^T}D$

${S^T} = {({D^T}D)^T} = {D^T}D$

$\therefore$ From Eq. (i), we get

${S^2} = {({S^T})^2}$

$\therefore$ ${S^2} = {({S^T})^2} = {({S^2})^T}$

$\begin{aligned} A & =\left[\begin{array}{lll} a^2 & 15 & 31 \\ 12 & b^2 & 41 \\ 35 & 61 & c^2 \end{array}\right], \\ B & =\left[\begin{array}{ccc} 2 a & 3 & 5 \\ 2 & 2 b & 8 \\ 1 & 4 & 2 c-3 \end{array}\right] \end{aligned}$

Given, trace of $A=\operatorname{trace}$ of $B$

$\begin{aligned} & \Rightarrow a^2+b^2+c^2=2 a+2 b+2 c-3 \\ & \Rightarrow\left(a^2-2 a+1\right)+\left(b^2-2 b+1\right)+\left(c^2-2 c+1\right)=0 \\ & \Rightarrow(a-1)^2+(b-1)^2+(c-1)^2=0 \\ & \Rightarrow a=1, b=1, c=1 \end{aligned}$

Then product of diagonal elements of $B$

$=(2 a)(2 b)(2 c-3)=(2)(2)(2-3)=-4$

$\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$

$ +\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{array}\right|=0$

$\begin{aligned} & \because\left.|A|=\mid A^T\right] \\ & \Rightarrow\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right| \\ &+(-1)^n\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ a & -b & c \end{array}\right|=0 \end{aligned}$

$\begin{aligned} & R_2 \leftrightarrow R_3 \text { and } R_3 \leftrightarrow R_1 \text { in 2nd determinant } \\ & \Rightarrow\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right| \\ & +(-1)^n\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right|=0 \\ \end{aligned}$

Both determinants are equal and their sum is zero when $n$ is an odd integer.

$\begin{array}{lr}\text { }\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^2\end{array}\right|=0 \\ \Rightarrow 1\left(6 x^2-12 x\right)+3\left(x^2-4\right)+1(3 x-6)=0 \\ \Rightarrow 6 x^2-12 x+3 x^2-12+3 x-6=0 \\ \Rightarrow 9 x^2-9 x-18=0 \\ \Rightarrow x^2-x-2=0\end{array}$

$\begin{aligned} & \text { } A=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] \\ & \text { Then, } A^T=\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right] \\ & \because \quad A A^T=20 I \\ & \Rightarrow \quad\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]=20\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{array}\right]=\left[\begin{array}{cc} 20 & 0 \\ 0 & 20 \end{array}\right] \\ & \therefore \quad a^2+b^2=20 \\ & \Rightarrow \quad(a+b)^2-2 a b=20 \\ & \Rightarrow \quad(a+b)^2=20+2 \times \frac{5}{2} \quad\left[\because a b=\frac{5}{2}\right] \\ & \text { or } \quad(a+b)^2=25 \\ & or\quad a+b= \pm 5 \\ \end{aligned}$

or $\quad x^2 \pm 5 x+\frac{5}{2}=0$ or $2 x^2 \pm 10 x+5=0$

$\begin{aligned} & \text { (c) } A=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array}\right] \\ & \therefore|A|=1(1+3)+1(2+3)+1(2-1)=10 \quad ... (\mathrm{i})\\ & \text { and } 10 B=\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{array}\right] \\ & \therefore|10 B|=4(2 \alpha)-2(-15-\alpha)+2(10) \\ & \Rightarrow 10^3|B|=10 \alpha+50 \quad\left[\because|k A|=k^n|A|\right] \ldots \mathrm{(ii)}\\ & \because B=A^{-1}\\ & \Rightarrow|B|=\left|A^{-1}\right|=|A|^{-1} \\ & \Rightarrow|B|=\frac{1}{|A|}=\frac{1}{10} \quad [\text { [rom Eq. (i)} \end{aligned}$

[from Eq. (i)] From Eq. (ii), we get

$1000~.~\frac{1}{10}=10\alpha+50$

or $\alpha=\frac{50}{10}=5$

Let $A=\left[\begin{array}{ccc}4 & 2 & 1-x \\ 5 & k & 1 \\ 6 & 3 & 1+x\end{array}\right]$

Given that, rank of matrix $A$ is 1 .

$\Rightarrow$ All the minors of order 2 are zero.

$\therefore {M_{33}} = 0$

$ \Rightarrow \left| {\matrix{ 4 & 2 \cr 5 & k \cr } } \right| = 0$

or $4 k-10=0$ or $k=\frac{5}{2}$

and $M_{11}=\left|\begin{array}{cc}k & 1 \\ 3 & 1+x\end{array}\right|=0$

or $\left|\begin{array}{cc}\frac{5}{2} & 1 \\ 3 & 1+x\end{array}\right|=0$

or $\frac{5}{2}+\frac{5}{2} x-3=0$

or $\frac{5}{2} x =\frac{1}{2}$

or $x =\frac{1}{5}$

$a_1, a_2, \ldots, a_9$, are in GP.

Then, $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\ldots \frac{a_9}{a_8}=r$ [common ratio]

Now, $\left|\begin{array}{lll}\log a_1 & \log a_2 & \log a_3 \\ \log a_4 & \log a_5 & \log a_6 \\ \log a_7 & \log a_8 & \log a_9\end{array}\right|$

$C_3 \rightarrow C_3-C_2, C_2 \rightarrow C_2-C_1 $

$\left| {\matrix{ {\log {a_1}} & {\log {{{a_2}} \over {{a_1}}}} & {\log {{{a_3}} \over {{a_2}}}} \cr {\log {a_4}} & {\log {{{a_5}} \over {{a_4}}}} & {\log {{{a_6}} \over {{a_5}}}} \cr {\log {a_7}} & {\log {{{a_8}} \over {{a_7}}}} & {\log {{{a_9}} \over {{a_8}}}} \cr } } \right|$

or $\left| {\matrix{ {\log {a_1}} & {\log r} & {\log r} \cr {\log {a_4}} & {\log r} & {\log r} \cr {\log {a_7}} & {\log r} & {\log r} \cr } } \right|=0$

[$\because$ C$_2$ and C$_3$ are identical]

$\begin{aligned} & \text { } \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{a} \cdot \mathbf{a}=4+1+9=14 \\ & \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}=2+3-3=2\end{aligned}$

$\begin{aligned} & \mathbf{a} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=6-1-6=-1 \\ & \text { b. } \mathbf{b}=1+9+1=11 \\ & \mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{b}=3-3+2=2 \\ & \mathbf{c} \cdot \mathbf{c}=9+1+4=14 \\ & \text { Now, }\left|\begin{array}{lll}\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c}\end{array}\right|=\left|\begin{array}{rrr}14 & 2 & -1 \\ 2 & 11 & 2 \\ -1 & 2 & 14\end{array}\right| \\ & =14(154-4)-2(28+2)-1(4+11) \\ & =2100-60-15 \\ & =2025 \\ & \end{aligned}$