Matrices and Determinants

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \quad\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right| \\ & \text { Applying } C_1 \rightarrow C_1+C_2+C_3 \\ & \quad=\left|\begin{array}{ccc} 2(a+b+c) & a & b \\ 2(b+c+a) & b+c+2 a & b \\ 2(c+a+b) & a & c+a+2 b \end{array}\right| \end{aligned} \end{aligned} $

$ \begin{aligned} &=2(a+b+c)\left|\begin{array}{ccc} 1 & a & b \\ 1 & b+c+2 a & b \\ 1 & a & c+a+2 b \end{array}\right|\\ &\text { Applying } R_1 \rightarrow R_1-R_2 \text { and } R_2 \rightarrow R_2-R_3\\ &\begin{aligned} & =2(a+b+c)\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & a+b+c & -(a+b+c) \\ 1 & a & c+a+2 b \end{array}\right| \\ & =2(a+b+c)^3\left|\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & a & c+a+2 b \end{array}\right| \\ & =2(a+b+c)^3[1(1-0)]=2(a+b+c)^3 \end{aligned} \end{aligned} $

To clarify:

When we have a matrix $ B $ of order 3 with a determinant $ |B| = 6 $, the formula to find the determinant of its adjugate matrix, $|\operatorname{adj}(B)|$, is given by:

$ |\operatorname{adj}(B)| = |B|^{n-1} $

where $ n $ is the order of the matrix $ B $.

In this case, since $ n = 3 $, we have:

$ |\operatorname{adj}(B)| = 6^{3-1} = 6^2 = 36 $

This confirms that the assertion is true.

However, the reason provided, $ |\operatorname{adj}(B)| = |B|^n $, is incorrect because it states that the determinant of the adjugate matrix is raised to the power of $ n $ rather than $ n-1 $. The correct relationship involves raising the determinant of $ B $ to the power of $ n-1 $.

Thus, the assertion is true, but the given reason is false.

Given the matrix $ A = \left[\begin{array}{lll} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{array}\right] $, which is singular, we need to find the quadratic equation with roots $ k $ and $ \frac{1}{k} $.

Since $ A $ is singular, its determinant is zero:

$ |A| = \left|\begin{array}{lll} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{array}\right| = 2(5k - 2) - 3(5 - 8) + 4(1 - 4k) = 0 $

Expanding this:

$ 2(5k - 2) = 10k - 4 $

$ -3(5 - 8) = -3 \cdot 3 = -9 $

$ 4(1 - 4k) = 4 - 16k $

Putting it all together, we get:

$ 10k - 4 - 9 + 4 - 16k = 0 $

Simplifying:

$ -6k + 9 = 0 $

Solving for $ k $:

$ k = \frac{3}{2} $

Thus, the roots of the quadratic equation are $ \frac{3}{2} $ and $ \frac{2}{3} $. Using the standard form of a quadratic equation $ x^2 - (\alpha + \beta)x + \alpha \beta = 0 $ where the roots are $ \alpha = \frac{3}{2} $ and $ \beta = \frac{2}{3} $, we calculate:

$ \alpha + \beta = \frac{3}{2} + \frac{2}{3} = \frac{9}{6} + \frac{4}{6} = \frac{13}{6} $

$ \alpha \beta = \frac{3}{2} \cdot \frac{2}{3} = 1 $

Substitute these into the quadratic equation:

$ x^2 - \left(\frac{13}{6}\right)x + 1 = 0 $

To eliminate fractions, multiply through by 6:

$ 6x^2 - 13x + 6 = 0 $

This is the quadratic equation with the specified roots.

Given that $|P| = \left|\frac{A}{2}\right|$, we start by using the property of the adjoint matrix, where $|\text{adj} (A)| = |A|^{n-1}$ for an $ n \times n $ matrix. Here, since $ A $ is a $ 4 \times 4 $ matrix, $|\text{adj} (A)| = |A|^{3}$.

Substituting the given condition, we have:

$ |A|^{3} = \left(\frac{1}{2}\right)^{4} |A| $

Simplifying, this becomes:

$ |A|^{3} = \frac{1}{16} |A| $

Assuming $|A| \neq 0$, divide both sides by $|A|$:

$ |A|^{2} = \frac{1}{16} $

Taking the square root on both sides gives:

$ |A| = \pm \frac{1}{4} $

To find $|A^{-1}|$, we use the property $|A^{-1}| = \frac{1}{|A|}$:

$ |A^{-1}| = \pm 4 $

Therefore, the absolute determinant of the inverse of $ A $ is $\pm 4$.

We have,

$ \begin{array}{r} x+2 y+3 z=4 \\ 4 x+5 y+3 z=5 \\ 3 x+4 y+3 z=\lambda \end{array} $

Write the equation in matrix form $\left[A_1^1 B\right]$

$ \left[\begin{array}{lll:l} 1 & 2 & 3 & 4 \\ 4 & 5 & 3 & 5 \\ 3 & 4 & 3 & \lambda \end{array}\right] $

$R_2 \rightarrow R_2-4 R_1$ and $R_3 \rightarrow R_3-3 R_1$

$ \sim\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & -3 & -9 & -11 \\ 0 & -2 & -6 & \lambda-12 \end{array}\right] $

$ \sim\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & -3 & -9 & -11 \\ 0 & 0 & 0 & 3 \lambda-14 \end{array}\right] $

$ R_3 \leftrightarrow 3 R_3-2 R_2 $

For consistency $3 \lambda-14=0 \Rightarrow 3 \lambda=14$

Given, $3 \lambda=n+100$

Therefore, $n=14-100$

$ n=-86 $

So, the value of $n$ is -86 .

Let $\Delta=\left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|\quad \ldots \text { (i) }$

Applying, $R_1 \rightarrow R_1+R_3, R_2 \rightarrow R_2+R_3$ we get

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

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

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

$ =a b^2-a^2 b-a c^2+a^2 c-b^2 c+b c^2 $

Thus, option (a) and (b) are correct. Now, option (c),

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

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

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

$ =-1\left[\left(b^2+b-a^2-a\right)(c-a)-\left(c^2+c-a^2-a\right)(b-a)\right] $

$ =\left(c^2+c-a^2-a\right)(b-a)-\left(b^2+b-a^2-a\right)(c-a) $

$ =c^2 b-a c^2+b c-a c-a^2 b+a^3-a b+a^2-b^2 c+a b^2-b c+a b+a^2 c -a^3+a c-a^2 $

Hence, option (d) is correct required answer.

Given, $A B C D E=I$

$\Rightarrow A^{-1}(A B C D E) E^{-1}=A^{-1} I E^{-1}$

( $\because$ Pre multiply by $A^{-1}$ and post multiply by $E^{-1}$, respectively)

$ \begin{array}{ll} \Rightarrow & I B C D I=A^{-1} E^{-1} \\ \Rightarrow & B C D=A^{-1} E^{-1} \end{array} $

Now, pre multiply by $B^{-1}$ and post multiply by $D^{-1}$, we get

$ \begin{array}{l} \Rightarrow & B^{-1}(B C D) D^{-1} & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & B^{-1} B C D D^{-1} & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & I C I & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & C & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \therefore & C^{-1} & =\left(B^{-1} A^{-1} E^{-1} D^{-1}\right)^{-1} \\ & & =\left(D^{-1}\right)^{-1}\left(E^{-1}\right)^{-1}\left(A^{-1}\right)^{-1}\left(B^{-1}\right)^{-1} \\ & & \text { (using reversal law) } \end{array} $

$\begin{array} & &=D E A B \\ \therefore \quad C^{-1} & =D E A B \end{array}$

Given, $A=\left[a_{i j}\right], 1 \leq i, j \leq n$ with $n \geq 2$ and $a_{i j}=i+j$

$\therefore$ The matrix is

$A=\left[\begin{array}{ccccc}2 & 3 & 4 & \cdots & n+1 \\ 3 & 4 & 5 & \cdots & n+2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ n+1 & n+2 & n+3 & \cdots & n+n\end{array}\right]$

$ =\left[\begin{array}{cccc} 2 & 3 & \cdots & n+1 \\ 3 & 4 & \cdots & n+2 \\ 4 & 5 & \cdots & n+3 \\ \vdots & \vdots & \cdots & \vdots \\ n+1 & n+2 & \cdots & 2 n \end{array}\right] $

Use elementary row operation to subtract 1st row from the $i$ th row for $2 \leq i \leq n$, we get

$ A=\left[\begin{array}{ccccc} 2 & 3 & 4 & \cdots & n+1 \\ 1 & 1 & 1 & \cdots & 1 \\ 2 & 2 & 2 & \cdots & 2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ n-1 & n-1 & n-1 & \cdots & n-1 \end{array}\right] $

Applying, $R_3 \rightarrow R_3-2 R_2$ and $R_4 \rightarrow R_4-3 R_2$ and so on

$ A=\left[\begin{array}{ccccc} 2 & 3 & 4 & \cdots & n+1 \\ 1 & 1 & 1 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{array}\right] $

$\therefore$ Rank $A=2$

Given the matrix $ A $:

$ A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} $

We are tasked to find $ A^2 - 5A + 6I $.

Calculate $ A^2 $:

$ A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} $

Perform the matrix multiplication:

$ A^2 = \begin{bmatrix} 1\cdot1 + 0\cdot2 + 2\cdot3 & 1\cdot0 + 0\cdot1 + 2\cdot2 & 1\cdot2 + 0\cdot3 + 2\cdot4 \\ 2\cdot1 + 1\cdot2 + 3\cdot3 & 2\cdot0 + 1\cdot1 + 3\cdot2 & 2\cdot2 + 1\cdot3 + 3\cdot4 \\ 3\cdot1 + 2\cdot2 + 4\cdot3 & 3\cdot0 + 2\cdot1 + 4\cdot2 & 3\cdot2 + 2\cdot3 + 4\cdot4 \end{bmatrix} $

$ = \begin{bmatrix} 1 + 0 + 6 & 0 + 0 + 4 & 2 + 0 + 8 \\ 2 + 2 + 9 & 0 + 1 + 6 & 4 + 3 + 12 \\ 3 + 4 + 12 & 0 + 2 + 8 & 6 + 6 + 16 \end{bmatrix} $

$ = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} $

Calculate $ 5A $:

$ 5A = 5 \cdot \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} $

Calculate $ 6I $:

$ 6I = 6 \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} $

Combine the results to find $ A^2 - 5A + 6I $:

$ A^2 - 5A + 6I = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} $

$ = \begin{bmatrix} 7 - 5 + 6 & 4 - 0 + 0 & 10 - 10 + 0 \\ 13 - 10 + 0 & 7 - 5 + 6 & 19 - 15 + 0 \\ 19 - 15 + 0 & 10 - 10 + 0 & 28 - 20 + 6 \end{bmatrix} $

$ = \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} $

Thus, the resulting matrix is:

$ \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} $

We have, $\left|\begin{array}{ccc}x^2+2 x & x+2 & 1 \\ 2 x+1 & x-1 & 1 \\ x+2 & -1 & 1\end{array}\right|=0$

$ \begin{aligned} & \Rightarrow\left(x^2+2 x\right)\left|\begin{array}{cc} x-1 & 1 \\ -1 & 1 \end{array}\right|-(x+2)\left|\begin{array}{cc} 2 x+1 & 1 \\ x+2 & 1 \end{array}\right|+1 \\ & \Rightarrow\left(x^2+2 x\right)(x-1+1)-(x+2)(2 x+1-x-2) \\ & \quad+1\left(-2 x-1-x^2-x+2\right)=0 \\ & \Rightarrow x\left(x^2+2 x\right)-(x+2)(x-1)+1 \\ & \quad\left(-x^2-3 x+1\right)=0 \\ & \Rightarrow x^3+2 x^2-x^2-x+2-x^2-3 x+1=0 \\ & \Rightarrow x^3-4 x+3=0 \end{aligned} $

On putting, $x=1$, we get

$ (1)^3-4 \times 1+3=0 $

Thus, $x=1$ is a root of the given equation.

Thus, $(x-1)\left(x^2+x-3\right)=0$

Solve $x^2+x-3=0$

$ \begin{aligned} x & =\frac{-1 \pm \sqrt{1^2-4 \times 1 \times(-3)}}{2 \times 1} \\ & =\frac{-1 \pm \sqrt{1+12}}{2}=\frac{-1 \pm \sqrt{13}}{2} \end{aligned} $

Hence, positive roots are 1 and $\frac{-1+\sqrt{13}}{2}$

Sum of positive roots $=1+\left(\frac{-1+\sqrt{13}}{2}\right)$

$ =\frac{1+\sqrt{13}}{2} $

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

We know that, if $ A $ is an orthogonal matrix, then

$ A A^{T}=I $

$ \begin{array}{c} \Rightarrow \quad \frac{1}{3}\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right] \cdot \frac{1}{3}\left[\begin{array}{ccc} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{array}\right]=\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 9 & 0 & a+4+2 b \\ 0 & 9 & 2 a+2-2 b \\ a+2 b & 2 a+2-2 b & a^{2}+4+b^{2} \end{array}\right] \\ =\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] \end{array} $

$ \Rightarrow \quad a+4+2 b=0 \text { and } 2 a+2-2 b=0 $

$ \Rightarrow \quad a+2 b=-4 \quad \ldots \text { (i) } $

and $ a-b=-1 \quad \ldots \text { (ii) } $

By solving Eqs. (i) and (ii), we get

$ a=-2, b=-1 $

Let $ |A|=\left|\begin{array}{ccc}p+a & b & c \\ a & q+b & c \\ a & b & r+c\end{array}\right| $

Also, given that equations has non trivial solution $ \therefore $

$ |A|=0 $

$ \begin{aligned} & \quad\left|\begin{array}{ccc} p+a & b & c \\ a & q+b & c \\ a & b & r+c \end{array}\right|=0 \\ \Rightarrow & (p+a)[(q+b)(r+c)-b c]-b[a(r+c)-c a] +c[a b-a(q+b)]=0 \\ \Rightarrow & (p+a)[q r+q c+b r]-b[a r]+c[-a q]=0 \\ \Rightarrow & p q r+p q c+p b r+a q r=0 \end{aligned} $

Dividing whole equation by $ p q r $

$ \begin{array}{ll} \Rightarrow & \frac{p q r}{p q r}+\frac{p q c}{p q r}+\frac{p r b}{p q r}+\frac{q r a}{p q r}=0 \\ \Rightarrow & 1+\frac{c}{r}+\frac{b}{q}+\frac{a}{p}=0 \\ \Rightarrow & \frac{a}{p}+\frac{b}{q}+\frac{c}{r}=-1 \end{array} $

Given the matrix $ A = \begin{bmatrix} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{bmatrix} $ and $ A^{-1} = \frac{1}{k} \operatorname{adj}(A) $, we want to determine the value of $ k $.

We know from the properties of matrices that:

$ A^{-1} = \frac{\operatorname{adj}(A)}{|A|} $

By comparing this equation with the given relationship $ A^{-1} = \frac{1}{k} \operatorname{adj}(A) $, we can identify that:

$ k = |A| $

Therefore, we need to calculate the determinant $|A|$ of the matrix $ A $:

$ |A| = \left| \begin{array}{ccc} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array} \right| $

To find $|A|$, expand across the first row:

$ |A| = 3 \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} $

Calculate each of the 2x2 determinants:

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

$ \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} = (1 \times 1) - (0 \times -1) = 1 $

$ \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = (1 \times 1) - (0 \times 2) = 1 $

Substitute these values back into the determinant calculation:

$ |A| = 3 \times 3 + 2 \times 1 + 4 \times 1 = 9 + 2 + 4 = 15 $

Thus, the value of $ k $ is $|A| = 15$.

Given that $|A|=m-n$, and let's solve the system of linear equations to find the values of $m$ and $n$ :

$4m + n = 22$ ...... (1)

$17m + 4n = 93$ ....... (2)

We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal:

$16m + 4n = 88$ ......... (3)

Now, subtract equation (3) from equation (2):

$(17m + 4n) - (16m + 4n) = 93 - 88$

$m = 5$

Now, we can find the value of $n$ by substituting $m$ into equation (1):

$4(5) + n = 22$

$20 + n = 22$

$n = 2$

Since the order of matrix A is $m$, the order is 5. Now, let's find the determinant of $n \operatorname{adj}(\operatorname{adj}(mA))$:

We know that $\operatorname{det}(A) = m-n = 3$.

$ \begin{aligned} & \therefore \operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A))) \\\\ & =|2 \operatorname{adj}(\operatorname{adj}(5 A))| \\\\ & =2^5|5 A|^{16} \\\\ & =2^5 5^{80}|A|^{16}=2^5 \cdot 3^{16} \cdot 5^{80} \\\\ & =3^{11} 5^{80} 6^5 \end{aligned} $

So, $a+b+c=96$

$ \begin{aligned} & A=\left[\begin{array}{lll} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{array}\right] \\\\ & |A|=2 \end{aligned} $

$ \begin{aligned} \Rightarrow&1(6-1)-2(2 \alpha-1)+3(\alpha-3)=2\\\\ \Rightarrow&5-4 \alpha+2+3 \alpha-9=2\\\\ \Rightarrow&-\alpha-4=0\\\\ \Rightarrow&\alpha=-4 \end{aligned} $

Now, $\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}$

$ \Rightarrow $ $2^3|\operatorname{adj}(2 \operatorname{adj}(2 A))|$ = ${32^n}$

$ \begin{aligned} \Rightarrow & 8|\operatorname{Adj}(2 \operatorname{Adj}(2 \mathrm{~A}))| = {32^n}\\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2 \times 2^2 \operatorname{Adj}(\mathrm{A})\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2^3 \operatorname{AdjA}\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|2^6 \operatorname{Adj}(\operatorname{AdjA})\right| = {32^n} \\\\ \Rightarrow & 2^3\left(2^6\right)^3|\operatorname{Adj}(\operatorname{Adj})| = {32^n} \\\\ \Rightarrow & 2^3 \cdot 2^{18}|\mathrm{~A}|^4 = {32^n} \\\\ \Rightarrow& 2^{21} \cdot 2^4 = {32^n}\\\\ \Rightarrow& 2^{25} = {32^n}\\\\ \Rightarrow& \left(2^5\right)^5 = {32^n}\\\\ \Rightarrow& (32)^5 = {32^n} \end{aligned} $

$ \begin{array}{ll} \therefore n=5 \\\\ \Rightarrow 3 n+\alpha=15-4=11 \end{array} $

$ \begin{aligned} & 2 x+y-z=5 \\ & 2 x-5 y+\lambda z=\mu \\ & x+2 y-5 z=7 \end{aligned} $

For infinite solution $\Delta=0=\Delta_1=\Delta_2=\Delta_3$

$ \Delta=\left|\begin{array}{ccc} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{array}\right|=0 $

$ \begin{aligned} \Rightarrow& 2(25-2 \lambda)-(-10-\lambda)-(4+5)=0 \\\\ \Rightarrow& 50-4 \lambda+10+\lambda-9=0 \\\\ \Rightarrow& 51=3 \lambda \Rightarrow \lambda=17 \end{aligned} $

$ \Delta_3=\left|\begin{array}{ccc} 5 & 2 & 1 \\ \mu & 2 & -5 \\ 7 & 1 & 2 \end{array}\right|=0 $

$ \begin{aligned} \Rightarrow & 2(-35-2 \mu)-(14-\mu)+5(4+5)=0 \\\\ \Rightarrow & -70-4 \mu-14+\mu+45=0 \\\\ \Rightarrow & -3 \mu=39 \\\\ \Rightarrow & -\mu=13 \end{aligned} $

Now $(\lambda+\mu)^2+(\lambda-\mu)^2$

$ \begin{aligned} & (17+13)^2+(17-13)^2 \\\\ & 900+16 \\\\ & =916 \end{aligned} $

The given system of equations is :

1. $2x + 4y + 2az = b$

2. $x + 2y + 3z = 4$

3. $2x - 5y + 2z = 8$

We can write this in matrix form :

$ \begin{bmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} b \\ 4 \\ 8 \end{bmatrix} $

First, we need to find the determinant of the coefficient matrix, which we'll call Δ. The coefficient matrix is :

$ \begin{bmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{bmatrix} $

We find its determinant for 3 $ \times $ 3 matrices :

$ \Delta = 2(2)(2) + 4(3)(2) + 2a(1)(-5) - 2a(2)(2) - 4(1)(2) - 2(3)(-5) $

$ \Delta = 8 + 24 - 10a - 8a - 8 + 30 $

$ \Delta = -18a + 54 $

$ \Delta = -18(a - 3) $

Next, we substitute the third column of our matrix with the column of constants (b, 4, 8) and calculate the determinant Δ₃ :

$ \begin{bmatrix} 2 & 4 & b \\ 1 & 2 & 4 \\ 2 & -5 & 8 \end{bmatrix} $

We find its determinant :

$ \Delta_3 = 2(2)(8) + 4(4)(2) + b(1)(-5) - b(2)(2) - 4(1)(8) - 2(4)(-5) $

$ \Delta_3 = 32 + 32 - 5b - 4b - 32 + 40 $

$ \Delta_3 = -9b + 72 $

$ \Delta_3 = 9(8 - b) $

Now, let's analyze the options :

1. For a=3 and b=8, we have Δ = 0 and Δ₃ = 0, which indicates an infinite number of solutions.

2. For a=3 and b=6, we have Δ = 0 and Δ₃ ≠ 0, which indicates no solution.

3. For a=8 and b=8, we have Δ ≠ 0, which indicates a unique solution.

4. For a=6 and b=6, we have Δ ≠ 0, which indicates a unique solution.

Therefore, the statement that is NOT correct is Option B: "It has infinitely many solutions if a=3, b=6", because in this case the system actually has no solution.

$ B=\left[\begin{array}{lll} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{array}\right], \alpha>2 $

And $\operatorname{adj}(A)=B,|A|=2$

$ \begin{aligned} & \Rightarrow|\operatorname{adj}(A)|=|B| \\\\ & \Rightarrow 2^2=(8-3 \alpha)-3(4-3 \alpha)+\alpha(-\alpha) \\\\ & \Rightarrow \alpha^2-6 \alpha+8=0 \end{aligned} $

$ \begin{aligned} \Rightarrow & (\alpha-4)(\alpha-2)=0 \\\\ & \alpha=4,2 \text { but } \alpha>2 \text { so } \alpha=4 \end{aligned} $

Now

$ \begin{aligned} & {\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c} \alpha \\ -2 \alpha \\ \alpha \end{array}\right]=\left[\begin{array}{lll} 4-8 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right]} \\\\ & =\left[\begin{array}{lll} 12 & 12 & 8 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right] \\\\ & = {48-96+32=-16} \end{aligned} $

Sure! A symmetric matrix is a square matrix that is equal to its transpose. For a matrix to be symmetric, the element at row i and column j must be equal to the element at row j and column i. In other words, $A_{ij} = A_{ji}$.

For a 3 $ \times $ 3 symmetric matrix, it looks like this:

$ \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \\ \end{pmatrix} $

Notice that there are only 6 unique elements we need to fill because of the symmetry:

1. $a$ in the (1,1) position
2. $b$ in the (1,2) and (2,1) positions
3. $c$ in the (1,3) and (3,1) positions
4. $d$ in the (2,2) position
5. $e$ in the (2,3) and (3,2) positions
6. $f$ in the (3,3) position

Each of these unique elements can take a value from the set ${0,1,2,3,4,5,6,7,8,9}$, which has 10 elements.

We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:

$10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^{6}$

Thus, the total number of symmetric matrices of order 3 with entries from this set is $10^{6}$.

$ \begin{aligned} & \text { Let } C=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right], \mathrm{D}=\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & \mathrm{DC}=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I} \end{aligned} $

$ \begin{aligned} & \mathrm{B}=\mathrm{CAD} \\\\ & \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots(\mathrm{CAD})}_{\text {n-times }} \end{aligned} $

$ \Rightarrow \mathrm{B}^{\mathrm{n}}=\mathrm{CA}^{\mathrm{n}} \mathrm{D} $

$ \mathrm{A}^2=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right] $

$ \mathrm{A}^3=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{3}{51} \\ 0 & 1 \end{array}\right] $

$ \text { Similarly } A^{\mathrm{n}}=\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right] $

$ \begin{aligned} \mathrm{B}^{\mathrm{n}} & =\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51}+2 \\ -1 & -\frac{\mathrm{n}}{51}-1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} \frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\ -\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51} \end{array}\right] \end{aligned} $

$ \sum\limits_{n=1}^{50} B^n=\left[\begin{array}{cc} 50+\frac{50 \cdot 51}{2 \cdot 51} & \frac{50 \cdot 51}{2 \cdot 51} \\\\ \frac{-50 \cdot 51}{2 \cdot 51} & 50-\frac{50 \cdot 51}{2 \cdot 51} \end{array}\right]=\left[\begin{array}{cc} 75 & 25 \\ -25 & 25 \end{array}\right] $

$ \therefore $ Sum of the elements = 100

Given,

$ \begin{aligned} & 7 x+11 y+\alpha z=13 \\\\ & 5 x+4 y+7 z=\beta \\\\ & 175 x+194 y+57 z=361 \end{aligned} $

$ \text { For infinite solution, }\left|\begin{array}{ccc} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 175 & 194 & 57 \end{array}\right|=0 $

$ \Rightarrow\left|\begin{array}{ccc} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 0 & -81 & 57-25 \alpha \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow 81(49-5 \alpha)+(57-25 \alpha)(-27)=0 \\\\ & \Rightarrow 270 \alpha=-2430 \Rightarrow \alpha=-9 \end{aligned} $

And $\Delta_1=0$

$ \begin{aligned} & \left|\begin{array}{ccc} 13 & 11 & -9 \\ \beta & 4 & 7 \\ 361 & 194 & 57 \end{array}\right|=0 \\\\ & \Rightarrow \beta=11 \end{aligned} $

$ \therefore \alpha+\beta+2=4 $

$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$

Put $x=0$

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{array}\right|=\frac{9}{8} \times 81 \\\\ & \lambda^3=\frac{3^6}{2^3} \end{aligned} $

$ \begin{aligned} & \Rightarrow \lambda=\frac{9}{2} \\\\ & \Rightarrow \frac{\lambda}{3}=\frac{3}{2} \end{aligned} $

$ \begin{aligned} & \text { Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\\\ & \Rightarrow 4 x^2-24 x+27=0 \end{aligned} $

We have, $A^T=\alpha A+I$, where $A$ is $2 \times 2$ matrix and $\alpha \in R-\{-1,1\}$

$ \begin{aligned} \left(A^T\right)^T & =\alpha A^T+I \\\\ A & =\alpha A^T+I \\\\ A & =\alpha(\alpha A+I)+I \left[\because A^T=\alpha A+I\right]\\\\ A & =\alpha^2 A+(\alpha+1) I \\\\ A & \left(1-\alpha^2\right)=(\alpha+1) I \\\\ A & =\frac{(\alpha+1)}{(1-\alpha)(1+\alpha)} I \end{aligned} $

$ \begin{aligned} A & =\frac{1}{1-\alpha} I ..........(i)\\\\ |A| & =\frac{1}{(1-\alpha)^2} ...........(ii) \end{aligned} $

Also,

$ \begin{aligned} \left|A^2-A\right| & =4 \\\\ |A||A-I| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{1}{1-\alpha}-1\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{\alpha}{1-\alpha}\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2} \times \frac{\alpha^2}{(1-\alpha)^2} & =4 \\\\ \frac{\alpha^2}{(1-\alpha)^4} & =4 \\\\ \frac{\alpha}{(1-\alpha)^2} & = \pm 2 \\\\ 2(1-\alpha)^2 & = \pm \alpha \end{aligned} $

If $2(1-\alpha)^2=\alpha$, then $2 \alpha^2-5 \alpha+2=0$

Sum of value of $\alpha=\frac{5}{2} \quad\left[\because\right.$ Sum of zero $\left.=\frac{\text {-Coefficient of } x}{\text { Coefficient of } x^2}\right]$

If $2(1-\alpha)^2=-\alpha$, then $2 \alpha^2-3 \alpha+2=0$

$\therefore$ No real value of $\alpha$

Hence, sum of all values of $\alpha=\frac{5}{2}$

Given that

$ \begin{aligned} & A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right] \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !}\left|\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right| \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !} \times 5 ! 6 ! 7 !\left|\begin{array}{lll} 1 & 6 & 7 \times 6 \\ 1 & 7 & 8 \times 7 \\ 1 & 8 & 9 \times 8 \end{array}\right| \end{aligned} $

$ \begin{aligned} & R_3 \rightarrow R_3-R_2 \text { and } R_2 \rightarrow R_2-R_1 \\\\ & |A|=\left|\begin{array}{lll} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{array}\right|=2 \\\\ & |\operatorname{adj}(\operatorname{adj}(2 A))|=|2 A|^{(n-1)^2} \\\\ & =|2 A|^4=\left(2^3|A|\right)^4=2^{12}|A|^4=2^{16} \end{aligned} $

Given that $A$ is $3 \times 3$ matrix and $|A|=2$

$ \begin{aligned} & \text { Now, | 3adj }\left(|3 A| A^2\right) \text { | } \\\\ & =3^3\left|\operatorname{adj}\left(|3 A| A^2\right)\right| \\\\ & =3^3\left|\operatorname{adj}\left(54 A^2\right)\right| \\\\ & =3^3\left|54 A^2\right|^2 \\\\ & =3^3 \times\left(54^3\right)^2 \times|A|^4 \\\\ & =3^3 \times(54)^6 \times 2^4 ~~~~~ {[|A|=2 \text { given }]} \\\\ & =3^3 \times\left(3^3 \times 2\right)^6 \times 2^4 ~~~~~ {\left[\left(a^m\right)^n=a^{m n}\right]} \\\\ & =3^{11} \times 3^{10} \times 2^{10} ~~~~~~~ {\left[(a b)^m=a^m b^m\right]} \\\\ & =(3)^{11} \times(6)^{10} \end{aligned} $

Given system of linear equation is

$ \begin{gathered} 2 x-y+3 z=5 \\\\ 3 x+2 y-z=7 \\\\ 4 x+5 y+\alpha z=\beta \\\\ \text { Now, } \Delta=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{array}\right|=7(\alpha+5) \end{gathered} $

So, this system of equation has unique solution, if $\alpha \neq-5$

and $\Delta_1=\left|\begin{array}{ccc}5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & \alpha\end{array}\right|=17 \alpha-5 \beta+30$

and $\Delta_2=\left|\begin{array}{ccc}2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & \alpha\end{array}\right|=-11 \beta+\alpha+104$

and $\Delta_3=\left|\begin{array}{ccc}2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta\end{array}\right|=7(\beta-9)$

For infinitely many solutions,

$ \Delta=\Delta_1=\Delta_2=\Delta_3=0 $

$ \begin{aligned} & \text { So, } \Delta=0 \\\\ & \Rightarrow 7(\alpha+5)=0 \\\\ & \Rightarrow \alpha=-5 \end{aligned} $

$ \begin{aligned} & \Delta_3 =0 \\\\ & \Rightarrow 7(\beta-9) =0 \\\\ & \Rightarrow \beta =9 \end{aligned} $

If $\alpha=-5$ and $\beta=8$, then $\Delta$ equals zero but $\Delta_3$ does not, which would imply the system is inconsistent for $\alpha=-5$ and $\beta=8$.

Therefore, the option "The system is inconsistent for $\alpha=-5$ and $\beta=8$ " is correct.

$ \begin{aligned} & \mathrm{A}=\left[\begin{array}{cc} 1 & 5 \\ \lambda & 10 \end{array}\right] \\\\ & \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\\\ & \Rightarrow\left|\begin{array}{cc} 1-x & 5 \\ \lambda & 10-x \end{array}\right|=0 \\\\ & \Rightarrow(1-x)(10-x)-5 \lambda=0 \\\\ & \Rightarrow 10-11 x+x^2-5 \lambda=0 \end{aligned} $

Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$

$ \Rightarrow \alpha A^2+\beta A-I=0 $

and $A^2-11 A+(10-5 \lambda) I=0$

On solving, we get

$ \alpha=\frac{1}{5}, \beta=-\frac{11}{5} $

$ \begin{aligned} & \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\ & \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\ & =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\ & =\frac{125}{25}+9=14 \end{aligned} $

Since, the given system has a non trivial solution,

$ \text { So, } \Delta=0 $

$ \Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow 1\left(\tan ^2 \theta-\sqrt{7}\right)-1(-\tan \theta-\sqrt{7})+\sqrt{3}(-1-\tan \theta)=0 \\\\ & \Rightarrow \tan ^2 \theta-\sqrt{7}+\tan \theta+\sqrt{7}-\sqrt{3}-\sqrt{3} \tan \theta=0 \\\\ & \Rightarrow \tan \theta(\tan \theta-\sqrt{3})+1(\tan \theta-\sqrt{3})=0 \\\\ & \Rightarrow \tan \theta=\sqrt{3} \text { or } \tan \theta=-1 \\\\ & \therefore \theta=\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}, \frac{-\pi}{4}, \frac{3 \pi}{4}\right\} \\\\ & \text { So, } \frac{120}{\pi} \sum \theta=\frac{120}{\pi}\left\{\frac{4 \pi-8 \pi-3 \pi+9 \pi}{12}\right\} \\\\ & =\frac{120}{\pi}\left[\frac{2 \pi}{12}\right]=20 \end{aligned} $

We have,

$ \begin{aligned} & |\mathrm{A}|=\left|\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{array}\right|=2(4-1)-1(2-0)+0 \\\\ & =6-2=4 \\\\ & \text { So, }|2 \mathrm{~A}|=2^3|\mathrm{~A}|=8 \times 4=32 \\\\ & \text { Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|^{(n-1)^3} \\\\ & =(32)^{2^3}=32^8 \\\\ & \Rightarrow 16^n=(32)^8=2^8 \times 16^8 \\\\ & \Rightarrow 16^n=16^{2+8} \Rightarrow n=10 \end{aligned} $

Concepts :

(a) $|k \mathrm{~A}|=k^n|\mathrm{~A}|$

(b) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}$

$ \text { Here, } P=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right], A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] $

$ \text { Here, } \mathrm{PP}^{\mathrm{T}}=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right] $

$ =\left[\begin{array}{cc} \frac{3}{4}+\frac{1}{4} & -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I} $

Similarly $P^T P=1$

$ \because $ $Q=P A P^{T}$

Now, $Q^{2007}=\left(P A P^T\right)\left(P A P^T\right) \ldots 2007$ times $=P A^{2007} P^T$

$ \begin{aligned} & \text { As, } A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \\\\ & \Rightarrow A^2=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1+0 & 1+1 \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] \end{aligned} $

$ A^3=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] $
.
.
.
.
$ A^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] $

$ \begin{aligned} & \text { Hence, } \mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2007} \mathrm{P}=\mathrm{A}^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \\\\ & \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow a=1, b=2007, c=0, d=1 \\\\ & \therefore 2 a+b-3 c-4 d=2(1)+2007-3(0)-4(1) \\\\ & =2+2007-4=2005 \end{aligned} $

We have, $P^2=I-P$

$ \begin{aligned} \Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\ & =2 I-3 P \text { [Using Eq. (i)] } \end{aligned} $

$ \begin{aligned} \Rightarrow P^8 & =(2 I-3 P)^2 \\\\ & =4 I+9 P^2-12 P \\\\ & =13 I-21 P \text { [Using Eq. (i)] } \end{aligned} $

and

$ \begin{aligned} P^6 & =(I-P)(2 I-3 P) \\\\ & =2 I-3 P-2 P+3 P^2 \\\\ & =5 I-8 P \text { [Using Eq. (i)] } \end{aligned} $

$\begin{aligned} & \text { Now, } P^8+P^6=18 I-29 P \\\\ & \text { and } P^8-P^6=8 I-13 P \\\\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\\\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$

Given system of equations,

$ \begin{aligned} x+y+z & =6 ........(i)\\\\ x+2 y+\alpha z & =10 ........(ii)\\\\ x+3 y+5 z & =\beta ........(iii) \end{aligned} $

Here,

$ \begin{aligned} \Delta & =\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{array}\right| \\\\ & =1(10-3 \alpha)-1(5-\alpha)+1(3-2) \\\\ & =10-3 \alpha-5+\alpha+1 \\\\ & =6-2 \alpha \end{aligned} $

For unique solution, $\Delta \neq 0$

$ \Rightarrow 6-2 \alpha \neq 0 \Rightarrow \alpha \neq 3 $

When, $\alpha=3$

$ \begin{aligned} \Delta_1 & =\left|\begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ \beta & 3 & 5 \end{array}\right| \\\\ & =\left|\begin{array}{ccc} 0 & 1 & 0 \\ -2 & 2 & 1 \\ \beta-18 & 3 & 2 \end{array}\right|=-1(-4-\beta+18) \\\\ & =\beta-14 \end{aligned} $

and

$\begin{aligned} \Delta_2 & =\left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \beta & 5\end{array}\right| \\\\ & =1(50-3 \beta)-6(5-3)+1(\beta-10) \\\\ & =50-3 \beta-12+\beta-10 \\\\ & =28-2 \beta=2(14-\beta)\end{aligned}$

and

$ \begin{aligned} \Delta_3 & =\left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 3 & \beta \end{array}\right| \\ & =1(2 \beta-30)-1(\beta-10)+6(3-2) \\\\ & =2 \beta-30-\beta+10+6 \\\\ & =\beta-14 \end{aligned} $

Thus, at $\beta=14, \Delta_1=\Delta_2=\Delta_3=0$

$ \Rightarrow \alpha=3, \beta=14$

So, system has infinite solutions.

Given system of equations,

$ \text { and } \quad \begin{aligned} x+y+a z & =b \\ 2 x+5 y+2 z & =6 \\ x+2 y+3 z & =3 \end{aligned} $

Since, given system of equation has infinitely many solutions

$ \therefore D=0 \text { and } D_1=D_2=D_3=0 $

$ \begin{aligned} & \text { Here, } D=\left|\begin{array}{lll} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow 1(15-4)-1(6-2)+a(4-5)=0 \\\\ & \Rightarrow -a+11-4=0 \\\\ & \Rightarrow a=7 \end{aligned} $

$ \begin{aligned} & \text { and } D_1=\left|\begin{array}{ccc} b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow b(15-4)-1(18-6)+a(12-15)=0 \\\\ & \Rightarrow 11 b-12-3 a=0 \\\\ & \Rightarrow 11 b-12-21=0 ~~~~~~~(\because a=7)\\\\ & \Rightarrow 11 b-33=0 \\\\ & \Rightarrow b=3 \\\\ & \therefore 2 a+3 b=2(7)+3(3)=14+9=23 \end{aligned} $

Given, $A^2=I$

and $b=|A|$

Let $ A=\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right] $

$ \begin{aligned} \therefore \quad A^2 & =\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right]\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right] \\\\ & =\left[\begin{array}{cc} a_1^2+b_1 a_2 & a_1 b_1+b_1 b_2 \\ a_1 a_2+a_2 b_2 & b_1 a_2+b_2^2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \left(\because A^2=I\right) \end{aligned} $

By equality of matrices, we get

$ \begin{aligned} & a_1 a_2+a_2 b_2=0 \text { and } a_1 b_1+b_1 b_2=0 \\\\ &\Rightarrow a_2\left(a_1+b_2\right)=0 \text { and } b_1\left(a_1+b_2\right)=0 \\\\ &\Rightarrow a_1+b_2=0 \text { (since, } b_1, a_2 \neq 0 \text { ) }\\\\ &\Rightarrow \text { Sum of diagonal elements }=0 \\\\ &\Rightarrow a=0 \end{aligned} $

$ \begin{aligned} & \text { Now, } A^2=I \\\\ & \Rightarrow \left|A^2\right|=1 \\\\ & \Rightarrow |A|^2=1 ~~~~~~~\left(\because\left|A^n\right|=|A|^n\right)\\\\ & \Rightarrow b^2=1 ~~~~~~~~~~~(\because|A|=b)\\\\ & \therefore 3 a^2+4 b^2=3(0)+4(1)=4 \end{aligned} $

For infinite solution $\Delta=\Delta_x=\Delta_y=\Delta_z=0$

$ \Delta=\left|\begin{array}{lll} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2 $

If $\beta=1$, then all planes are overlapping

$\therefore$ Option (A) is correct.

Option (B) :

If $\alpha=2 \Rightarrow \Delta \neq 0$

$\therefore $ Unique solution exist

$\therefore$ Option (B) is incorrect.

Option (C) :

$ \begin{aligned} & \alpha=2, \beta=1 \\\\ & 2 x+y+z=1 \\\\ & x+2 y+z=1 \\\\ & x+y+2 z=1 \end{aligned} $

Adding all three equations,

$ x+y+z=\frac{3}{4} $

$\therefore$ option (C) is correct.

Option (D) :

If $\alpha=-2$ and $\beta=1$, then $\Delta=0, \Delta_x \neq 0$

$\therefore$ No solution.

$\therefore$ Option (D) is correct.

$A = {1 \over 2}\left[ {\matrix{ 1 & {\sqrt 3 } \cr { - \sqrt 3 } & 1 \cr } } \right]$

Let $\theta=\frac{\pi}{3}$

$ \begin{aligned} A^2 & =\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right] \\\\ A^3 & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 3 \theta & \sin 3 \theta \\\\ -\sin 3 \theta & \cos 3 \theta \end{array}\right] \end{aligned} $

$\begin{aligned} & \therefore \quad A^{30}=\left[\begin{array}{cc}\cos 30 \theta & \sin 30 \theta \\\\ -\sin 30 \theta & \cos 30 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\ & A^{25}=\left[\begin{array}{cc}\cos 25 \theta & \sin 25 \theta \\\\ -\sin 25 \theta & \cos 25 \theta\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]=A \\\\ & \therefore \quad A^{30}+A^{25}-A=I\end{aligned}$

$\left|\begin{array}{lll}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{array}\right|=0$

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

$\Rightarrow$ $\lambda=1,-2$

For $\lambda=1 \Rightarrow \infty$ solution

$\lambda=-2 \Rightarrow$ no solution

$\sum\limits_{\lambda \in S}|\lambda|^{2}+|\lambda|=6$