Matrices and Determinants

$ \begin{aligned} & \text { } \Delta=\left|\begin{array}{ccc} 2 \sin A \cos A & \sin C & \sin B \\ \sin C & 2 \sin B \cos B & \sin A \\ \sin B & \sin A & 2 \sin C \cos C \end{array}\right| \\ & =\left|\begin{array}{ccc} 2 k a \cos A & k c & k b \\ k c & 2 k b \cos B & k a \\ k b & k a & 2 k c \cos C \end{array}\right| \\ & {\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\right]} \end{aligned} $

$ =k^3\left|\begin{array}{ccc} 2 a \cos A & c & b \\ c & 2 b \cos B & a \\ b & a & 2 c \cos c \end{array}\right| $

$ =k^3 \left\lvert\, \begin{array}{ll} a \cos A+a \cos A & a \cos B+b \cos A \\ a \cos B+b \cos A & b \cos B+b \cos B \\ c \cos A+a \cos C & b \cos C+c \cos B \end{array}\right. $

$ \left.\begin{aligned} & c \cos A+a \cos C \\ & b \cos C+c \cos B \\ & c \cos C+c \cos C \end{aligned} \right\rvert\, $

$ \begin{aligned} & \left.=k^3\left|\begin{array}{ccc} \cos A & a & 0 \\ \cos B & b & 0 \\ \cos C & c & 0 \end{array}\right|| | \begin{array}{ccc} a & \cos A & 0 \\ b & \cos B & 0 \\ c & \cos C & 0 \end{array} \right\rvert\, \\ & =k^3 \times 0=0 \end{aligned} $

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 $|A| = 15$.

To determine the value of $\lambda$ for which the given system of linear equations has no solution, we need to analyze the consistency of these equations. The given system is:

$3x - 2y + z = 2$ (Equation 1)

$4x - 3y + 3z = -5$ (Equation 2)

$7x - 5y + \lambda z = 9$ (Equation 3)

For a system of linear equations to have no solution, the equations must be inconsistent, meaning they do not intersect at any point. This happens when the determinant of the coefficient matrix is zero, and the augmented matrix creates a contradiction.

First, let's form the coefficient matrix and calculate its determinant. The coefficient matrix $A$ of the system is:

$A = \begin{pmatrix} 3 & -2 & 1 \\ 4 & -3 & 3 \\ 7 & -5 & \lambda \end{pmatrix}$

The determinant of matrix $A$ is given by:

$\text{det}(A) = \begin{vmatrix} 3 & -2 & 1 \\ 4 & -3 & 3 \\ 7 & -5 & \lambda \end{vmatrix}$

Expanding this determinant along the first row, we get:

$\text{det}(A) = 3\begin{vmatrix} -3 & 3 \\ -5 & \lambda \end{vmatrix} - (-2)\begin{vmatrix} 4 & 3 \\ 7 & \lambda \end{vmatrix} + 1\begin{vmatrix} 4 & -3 \\ 7 & -5 \end{vmatrix}$

Now, calculate these 2x2 determinants:

$\begin{vmatrix} -3 & 3 \\ -5 & \lambda \end{vmatrix} = (-3)(\lambda) - (3)(-5) = -3\lambda + 15$

$\begin{vmatrix} 4 & 3 \\ 7 & \lambda \end{vmatrix} = (4)(\lambda) - (3)(7) = 4\lambda - 21$

$\begin{vmatrix} 4 & -3 \\ 7 & -5 \end{vmatrix} = (4)(-5) - (-3)(7) = -20 + 21 = 1$

Substitute these into the expansion to get the determinant:

$\text{det}(A) = 3(-3\lambda + 15) + 2(4\lambda - 21) + 1(1)$

$= 3(-3\lambda + 15) + 2(4\lambda - 21) + 1$

$= 3(-3\lambda) + 3(15) + 2(4\lambda) - 2(21) + 1$

$= -9\lambda + 45 + 8\lambda - 42 + 1$

$= -\lambda + 4$

For the system to have no solution, $\text{det}(A)$ must be zero, because that's when the coefficient matrix does not have full rank and the system is inconsistent:

$-\lambda + 4 = 0$

$\lambda = 4$

Therefore, the value of $\lambda$ that makes the system of linear equations have no solution is $4$.

Answer: (A) 4

Since, $B$ is the inverse of $A$.

$\begin{aligned} & \therefore \quad A \times B=I \\ & \Rightarrow\left[\begin{array}{lll} 3 & 2 & 3 \\ 4 & 1 & 0 \\ 2 & 5 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 / 49 & 13 / 49 & -3 / 49 \\ -4 / 49 & -3 / 49 & 12 / 49 \\ \alpha / 49 & -11 / 49 & -5 / 49 \end{array}\right] \\ & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} (3-8+3 \alpha / 49 & (39-6-33) / 49 & (-9+24-15) / 49 \\ (4-4+0) / 49 & (52-3+0) / 49 & (-12+12+0) / 49 \\ (2-20+\alpha) / 49 & (26-15-11) / 49 & (-6+60-5) / 49 \end{array}\right] \\ & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} (-5+3 \alpha) / 49 & 0 & 0 \\ 0 & 1 & 0 \\ (-18+\alpha) / 49 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{-5+3 \alpha}{49}=1 \text { and } \frac{-18+\alpha}{49}=0 \\ & \Rightarrow \quad 3 \alpha=54 \text { and } \alpha=18 \\ & \Rightarrow \quad \alpha=18 \text { and } \alpha=18 \end{aligned}$

Thus, $\alpha=18$

$\text { Given, } A=\left[\begin{array}{cc} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]$

$ \begin{aligned} & \Rightarrow \quad \operatorname{adj} A=\left[\begin{array}{cc} \sin \theta & -\cos \theta \\ -(-\cos \theta) & \sin \theta \end{array}\right] \\ & {\left[\because \text { If } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \text {, then adj } A=\left[\begin{array}{cc} d & -c \\ -b & a \end{array}\right]\right] } \end{aligned} $

$\begin{aligned} & \Rightarrow \operatorname{adj} A=A \\ & \begin{aligned} \operatorname{Now,} A\left(\operatorname{adj}(A)^{-1}\right) & =A(A)^{-1} \\ = & I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad\left[\because A A^{-1}=A^{-1} A=I\right] \end{aligned} \end{aligned}$

Given, system of equations has a non-trivial solution

$\therefore \quad\left|\begin{array}{ccc} a-1 & -1 & -1 \\ -1 & b-1 & -1 \\ -1 & -1 & c-1 \end{array}\right|=0$

On applying $C_2 \rightarrow C_2-C_3$, we get

$\left|\begin{array}{ccc} a-1 & 0 & -1 \\ -1 & b & -1 \\ -1 & -c & c-1 \end{array}\right|=0$

On applying $R_2 \rightarrow R_2-R_3$, we get

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

On applying $R_3 \rightarrow R_3-R_1$, we get

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

On expanding along $R_1$, we get

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

Since, the system has no solution.

$\left| {\matrix{ 2 & { - 1} & 2 \cr 1 & { - 2} & { - 1} \cr 1 & 1 & \lambda \cr } } \right| = 0$

$ \Rightarrow 2( - 2\lambda + 1) + 1(\lambda + 1) + 2(3) = 0$

$ \Rightarrow - 4\lambda + 2 + \lambda + 1 + 6 = 0$

$ \Rightarrow 3\lambda = 9 \Rightarrow \lambda = 3$

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

$\therefore$ $\left| A \right| = \left| {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right|$

$ = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 0$

$ \Rightarrow Adj\,A = {\left[ {\matrix{ 4 & { - 5} & 1 \cr 2 & 0 & { - 2} \cr 2 & 5 & 3 \cr } } \right]^T} = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & 5 \cr 1 & { - 2} & 3 \cr } } \right]$

$ \Rightarrow B = {A^{ - 1}} = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & 5 \cr 1 & { - 2} & 3 \cr } } \right]$ [$\because$ ${A^{ - 1}} = {1 \over {|A|}}Adj\,A$]

$ \Rightarrow 10B = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & 5 \cr 1 & { - 2} & 3 \cr } } \right]$

Hence, $\alpha = 5$

Given,

$\left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right]{\left[ {\matrix{ 1 & {\tan \theta } \cr { - \tan \theta } & 1 \cr } } \right]^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right]\,.\,{1 \over {1 + {{\tan }^2}\theta }}\left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow {1 \over {1 + {{\tan }^2}\theta }}\left[ {\matrix{ {1 - {{\tan }^2}\theta } & { - 2\tan \theta } \cr {2\tan \theta } & {1 - {{\tan }^2}\theta } \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {{{1 - {{\tan }^2}\theta } \over {1 + {{\tan }^2}\theta }}} & {{{2\tan \theta } \over {1 + {{\tan }^2}\theta }}} \cr {{{2\tan \theta } \over {1 + {{\tan }^2}\theta }}} & {{{1 - {{\tan }^2}\theta } \over {1 + {{\tan }^2}\theta }}} \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {\cos 2\theta } & { - \sin 2\theta } \cr {\sin 2\theta } & {\cos 2\theta } \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow a = \cos 2\theta ,b = \sin 2\theta $

As the given system of equations has a non-trivial solution.

$\Delta = \left| {\matrix{ p & a & a \cr b & q & b \cr c & c & r \cr } } \right| = 0$

Applying ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}$

$\Delta = \left| {\matrix{ p & {a - p} & {a - p} \cr b & {q - b} & 0 \cr c & 0 & {r - c} \cr } } \right| = 0$

Expanding along C₃, we get

$(a - p)\left| {\matrix{ b & {q - b} \cr c & 0 \cr } } \right| + (r - c)\left| {\matrix{ p & {a - p} \cr b & {q - b} \cr } } \right| = 0$

$ \Rightarrow (a - p)( - c)(q - b) + (r - c)\{ p(q - b) - b(a - p)\} = 0$

$ \Rightarrow (p - a)(q - b)c + p(r - c)(q - b) + b(r - c)(p - a) = 0$

Dividing $(p - a)(q - b)(r - c)$, we get

${c \over {r - c}} + {p \over {p - a}} + {b \over {q - b}} = 0$

$ \Rightarrow {p \over {p - a}} + {q \over {q - b}} + {r \over {r - c}} = {{q - b} \over {q - b}} + {{r - c} \over {r - c}} = 2$

Given, $\left| {\matrix{ 0 & {x - p} & {x - q} \cr {x + p} & 0 & {x - r} \cr {x + q} & {x - r} & 0 \cr } } \right| = 0$

$ - (x - p)\left| {\matrix{ {x + p} & {x - r} \cr {x + q} & 0 \cr } } \right| + (x - q)\left| {\matrix{ {x + p} & 0 \cr {x + q} & {x - r} \cr } } \right| = 0$

$ \Rightarrow (x - p)(x + q)(x - r) + (x - q)(x + p)(x - r) = 0$

$ \Rightarrow (x - r)[(x - p)(x + q) + (x - q)(x + p)] = 0$

$ \Rightarrow (x - r)[{x^2} - px + qx - pq + {x^2} - qx + px - pq] = 0$

$ \Rightarrow (x - r)[2{x^2} - 2pq] = 0 \Rightarrow x - r = 0$ or ${x^2} - pq = 0$

A . adj A = | A | I

| A | = xyz $-$ 8x $-$ 3(z $-$ 8) + 2(2 $-$ 2y)

| A | = xyz $-$ (8x + 3z + 4y) + 28

$\Rightarrow$ 60 $-$ 20 + 28 = 68

$\Rightarrow$ (adj A)^$-$1 always exists whenever (A)^$-$1 exists.

$\therefore$ A . adj A = | A | I

$ = 68\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ {68} & 0 & 0 \cr 0 & {68} & 0 \cr 0 & 0 & {68} \cr } } \right]$

We have,

$(1 + \alpha )x + \beta y + z = 2$

$\alpha x + (1 + \beta )y + z = 3$

$\alpha x + \beta y + 2z = 2$

For unique solution $\left| {\matrix{ {1 + \alpha } & \beta & 1 \cr \alpha & {1 + \beta } & 1 \cr \alpha & \beta & 2 \cr } } \right| \ne 0$

Apply R₁ $\to$ R₁ - R₂ and R₂ $\to$ R₂ - R₃, we get

$\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr \alpha & \beta & 2 \cr } } \right| \ne 0$

Expanding along R₁, we get

$1(2 + \beta ) + 1(0 + \alpha ) \ne 0$

$ = \alpha + \beta + 2 \ne 0$

Only option (c) satisfied above equation.

We have,

$A = \left[ {\matrix{ 2 & 1 \cr 1 & 2 \cr } } \right]$

We know that, $\left| {A \to \lambda I} \right| = 0$

$\left| {\left[ {\matrix{ 2 & 1 \cr 1 & 2 \cr } } \right] - \left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]} \right| = 0$

$ = \left| {\matrix{ {2 - \lambda } & 1 \cr 1 & {2 - \lambda } \cr } } \right| = 0$

$ \Rightarrow {(2 - \lambda )^2} - 1 = 0$

$ \Rightarrow 4 - 4\lambda + {\lambda ^2} - 1 = 0$

$ \Rightarrow {\lambda ^2} - 4\lambda + 3 = 0$

$\therefore$ ${A^2} - 4A + 3I = 0$

Multiply by A^$-$1, we get

$A - 4I + 3{A^{ - 1}} = 0$

$ \Rightarrow {A^{ - 1}} = {{ - 1} \over 3}A + {4 \over 3}I$

$\therefore$ $\alpha = - {1 \over 3}$ and $\beta = {4 \over 3}$

$\alpha + \beta = - {1 \over 3} + {4 \over 3} = {3 \over 3} = 1$