Matrices and Determinants

$A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

A² = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$ \Rightarrow $ A² = $\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$

Similarly, Aⁿ = $\left[ {\matrix{ {\cos n\theta } & {\sin n\theta } \cr { - \sin n\theta } & {\cos n\theta } \cr } } \right]$

$ \therefore $ B = A + A⁴

= $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$ + $\left[ {\matrix{ {\cos 4\theta } & {\sin 4\theta } \cr { - \sin 4\theta } & {\cos 4\theta } \cr } } \right]$

= $\left[ {\matrix{ {\cos 4\theta + \cos \theta } & {\sin 4\theta + \sin \theta } \cr { - \sin 4\theta - \sin \theta } & {\cos 4\theta + \cos \theta } \cr } } \right]$

detB = (cos4$\theta $ + cos$\theta $)² + (sin4$\theta $ + sin$\theta $)²

= cos²4$\theta $ + cos²$\theta $ + 2cos4$\theta $ cos$\theta $
+ sin²4$\theta $ + sin2$\theta $ + 2sin4$\theta $ –sin$\theta $

= 2 + 2 ( cos4$\theta $ cos$\theta $ + sin4$\theta $ sin$\theta $)

$ \Rightarrow $ detB = 2 + 2 cos3$\theta $

at $\theta $ = ${\pi \over 5}$

detB = 2 + 2cos ${{3\pi } \over 5}$

= 2(1 - sin18)

= 2(1 - ${{\sqrt 5 - 1} \over 4}$)

= 2$\left( {{{5 - \sqrt 5 } \over 4}} \right)$

= ${{{5 - \sqrt 5 } \over 2}}$ $ \simeq $ 1.385

$ \therefore $ detB $ \in $ (1, 2)

For infinite many solutions

D = D₁ = D₂ = D₃ = 0

Now D = $\left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \lambda \cr } } \right|$ = 0

$ \Rightarrow $ 1. (2$\lambda $ – 9) –1.($\lambda $ – 3) + 1.(3 – 2) = 0

$ \Rightarrow $ $\lambda $ = 5

Now D₁ = $\left| {\matrix{ 2 & 1 & 1 \cr 5 & 2 & 3 \cr \mu & 3 & 5 \cr } } \right|$ = 0

$ \Rightarrow $ 2(10 – 9) –1(25 – 3$\mu $) + 1(15 – 2$\mu $) = 0

$ \Rightarrow $ $\mu $ = 8

$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$

R₁ $ \to $ R₁ – R₂, R₂ $ \to $ R₂ – R₃

$\left| {\matrix{ { - 1} & 1 & 0 \cr 1 & 0 & { - 1} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$

= –1(sin² x) – 1(1 + sin2x + cos² x)

= - sin2x - 2

$ \therefore $ minimum value when sin2x = 1

m = - 2 - 1 = -3

$ \therefore $ Maximum value when sin2x = –1

M = -2 + 1 = -1

$ \therefore $ (m, M) = (–3, –1)

x + y + 3z = 0 .....(i)
x + 3y + k²z = 0 .........(ii)
3x + y + 3z = 0 ......(iii)

$\left| {\matrix{ 1 & 1 & 3 \cr 1 & 3 & {{k^2}} \cr 3 & 1 & 3 \cr } } \right|$ = 0

$ \Rightarrow $ 9 + 3 + 3k² – 27 – k² – 3 = 0

$ \Rightarrow $ k² = 9

Perform (i) – (iii),

–2x = 0 $ \Rightarrow $ x = 0

Now from (i), y + 3z = 0

$ \Rightarrow $ ${y \over z} = - 3$

$ \therefore $ x + $\left( {{y \over z}} \right)$ = -3

$\left| {\matrix{ x & {a + y} & {x + a} \cr y & {b + y} & {y + b} \cr z & {c + y} & {z + c} \cr } } \right|$

C₃ $ \to $ C₃ – C₁

= $\left| {\matrix{ x & {a + y} & a \cr y & {b + y} & b \cr z & {c + y} & c \cr } } \right|$

C₂ $ \to $ C₂ – C₃

= $\left| {\matrix{ x & y & a \cr y & y & b \cr z & y & c \cr } } \right|$

R₃ $ \to $ R₃ – R₁, R₂ $ \to $ R₂ – R₁

= $\left| {\matrix{ x & y & a \cr {y - x} & 0 & {b - a} \cr {z - x} & 0 & {c - a} \cr } } \right|$

= (–y)[(y – x) (c – a) – (b – a) (z – x)]

Given, a + x = b + y = c + z + 1

= (–y)[(a – b) (c – a) + (a – b) (a – c – 1)]

= (–y)[(a – b) (c – a) + (a – b) (a – c) + b – a)

= –y(b – a) = y(a – b)

D = $\left| {\matrix{ 2 & { - 4} & \lambda \cr 1 & { - 6} & 1 \cr \lambda & { - 10} & 4 \cr } } \right|$ = 0

$ \Rightarrow $ $\lambda $ = 3, $ - {2 \over 3}$

D₁ = $\left| {\matrix{ 1 & { - 4} & \lambda \cr 2 & { - 6} & 1 \cr 3 & { - 10} & 4 \cr } } \right|$

= 14 + 4(5) + $\lambda $(–2)

= –2$\lambda $ + 6

D₂ = $\left| {\matrix{ 2 & 1 & \lambda \cr 1 & 2 & 1 \cr \lambda & 3 & 4 \cr } } \right|$

= –2($\lambda $ – 3)($\lambda $ + 1)

D₃ = $\left| {\matrix{ 2 & { - 4} & 1 \cr 1 & { - 6} & 2 \cr \lambda & { - 10} & 3 \cr } } \right|$

= – 2$\lambda $ + 6

When , $\lambda $ = 3 then

D = D₁ = D₂ = D₃ = 0

$ \Rightarrow $ Infinite many solution

When $\lambda $ = $ - {2 \over 3}$ then D₁, D₂, D₃ none of them is zero so equations are inconsistant.

$ \therefore $ $\lambda $ = $ - {2 \over 3}$

Given
$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$

C₁ $ \to $ C₁ – C₂ , C₃ $ \to $ C₃ + C₂

= $\left| {\matrix{ 1 & { - 1 - {{\sin }^2}\theta } & { - {{\sin }^2}\theta } \cr 1 & { - 1 - {{\cos }^2}\theta } & { - {{\cos }^2}\theta } \cr 2 & {10} & 8 \cr } } \right|$

C₂ $ \to $ C₂ – C₃

= $\left| {\matrix{ 1 & { - 1} & { - {{\sin }^2}\theta } \cr 1 & { - 1} & { - {{\cos }^2}\theta } \cr 2 & 2 & 8 \cr } } \right|$

= 1(2cos²$\theta $ – 8) + (8 + 2cos²$\theta $) – 4sin²$\theta $

= 4cos²$\theta $ - 4cos²$\theta $

= 4 cos 2$\theta $

$\theta $ $ \in $ $\left[ {{\pi \over 4},{\pi \over 2}} \right]$

$ \Rightarrow $ 2$\theta $ $ \in $ $\left[ {{\pi \over 2},{\pi }} \right]$

$ \Rightarrow $ cos 2$\theta $ $ \in $ [-1, 0]

$ \Rightarrow $ 4cos 2$\theta $ $ \in $ [-4, 0]

$ \Rightarrow $ $f\left( \theta \right)$ $ \in $ [-4, 0]

$ \therefore $ (m, M) = (–4, 0)

Let A = $\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]$

For Ax₁ = b₁ :

$ \Rightarrow $ $\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$

$ \therefore $ ${a_1} + {a_2} + {a_3} = 1$ ....(1)

${a_4} + {a_5} + {a_6} = 0$ ......(2)

${a_7} + {a_8} + {a_9} = 0$ .....(3)

For Ax₂ = b₂ :

$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$

$ \therefore $ $2{a_2} + {a_3} = 0$ .....(4)

$2{a_5} + {a_6} = 2$ ....(5)

$2{a_8} + {a_9} = 0$ ....(6)

For Ax₃ = b₃ :

$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$

$ \therefore $ ${{a_3} = 0}$

${{a_6} = 0}$

${{a_9} = 2}$

Putting value of ${{a_3}}$ in equation (4), we get

${{a_2}}$ = 0

Putting value of ${{a_6}}$ in equation (5), we get

${{a_5}}$ = 1

Putting value of ${{a_9}}$ in equation (6), we get

${{a_8}}$ = -1

Putting value of ${{a_2}}$ and ${{a_3}}$ in equation (1), we get

${{a_1}}$ = 1

Putting value of ${{a_5}}$ and ${{a_6}}$ in equation (6), we get

${{a_4}}$ = -1

Putting value of ${{a_5}}$ and ${{a_6}}$ in equation (6), we get

${{a_4}}$ = -1

Putting value of ${{a_8}}$ and ${{a_9}}$ in equation (6), we get

${{a_7}}$ = -1

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

So, |A| = 2(1) = 2

$D = 0\,\left| {\matrix{ 1 & 1 & 1 \cr 2 & 4 & { - 1} \cr 3 & 2 & \lambda \cr } } \right| = 0$

$ \Rightarrow $ $(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$

$ \Rightarrow $ $4\lambda + 2$ $ - 2\lambda - 3$$ - $8$ = 0$

$ \Rightarrow $ $2\lambda = 9 \Rightarrow \lambda = {9 \over 2}$

${D_x} = \left| {\matrix{ 2 & 1 & 1 \cr 6 & 4 & { - 1} \cr \mu & 2 & { - 9/2} \cr } } \right| = 0$

$ \Rightarrow \mu = 5$

By checking all the options we find (C) is correct

$2\lambda + \mu = 14$

$ \because $ $A = \left[ {\matrix{ {\cos \theta } & {i\sin \theta } \cr {i\sin \theta } & {\cos \theta } \cr } } \right]$

$ \therefore $ ${A^n} = \left[ {\matrix{ {\cos \,n\theta } & {i\sin \,n\theta } \cr {i\sin \,n\theta } & {\cos \,n\theta } \cr } } \right],n \in N$

$ \therefore $${A^5} = \left[ {\matrix{ {\cos \,5\theta } & {i\sin \,5\theta } \cr {i\sin \,5\theta } & {\cos \,5\theta } \cr } } \right] = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$

$ \therefore $ $a = \cos 5\theta ,\,b = i\sin 5\theta = c,\,d = \cos 5\theta $

$ \therefore $ ${a^2} - {b^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$

${a^2} - {c^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$

${a^2} - {d^2} = {\cos ^2}5\theta - {\cos ^2}5\theta = 1$

${a^2} + {b^2} = {\cos ^2}5\theta - {\sin ^2}5\theta = \cos 10\theta = \cos {{10\pi } \over {24}}$

and $0 < \cos {{5\pi } \over {12}} < 1 $

$\Rightarrow 0 \le {a^2} + {b^2} \le 1$

$adj\,A = \left[ {\matrix{ 2 & { - 1} & 1 \cr { - 1} & 0 & 2 \cr 1 & { - 2} & { - 1} \cr } } \right]$

$B = adj\,(adj\,A)$

$ = |A{|^{n - 2}}A$

$ = |A{|^{3 - 2}}.A$ [As here n = 3]

$ = |A|.A$ .....(1)

Now, $|adj\,A| = \left[ {\matrix{ 2 & { - 1} & 1 \cr { - 1} & 0 & 2 \cr 1 & { - 2} & { - 1} \cr } } \right]$

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

$ = 8 - 1 + 2$

$ = 9$

Also we know, |adj A| = |A|^n$-$1

$ \therefore $ Here |adj A| = |A|^{3 $-$ 1} = |A|²

$ \therefore $ |A|² = 9

$ \Rightarrow $ |A| = $ \pm $3

Given, |A| = $\lambda $

$ \therefore $ $\lambda $ = $ \pm $3

$ \Rightarrow $ |$\lambda $| = 3

From (1)

B = ($ \pm $3)A

Given, |(B^$-$1)^T| = $\mu $

$ \Rightarrow $ |(B^T)^$-$1| = $\mu $

$ \Rightarrow $ ${1 \over {|{B^T}|}} = \mu $

$ \Rightarrow $ ${1 \over {|B|}} = \mu $ [As |B^T| = |B|]

$ \Rightarrow $ ${1 \over {| \pm 3A|}} = \mu $

$ \Rightarrow $ ${1 \over {{{\left( { \pm 3} \right)}^3}\left| A \right|}} = \mu $ [As |KA| = Kⁿ |A|]

$ \Rightarrow {1 \over {( \pm 27)|A|}} = \mu $

$ \Rightarrow \mu = {1 \over {( \pm 27)( \pm 3)}} = \pm {1 \over {81}}$

$ \therefore $ $(|\lambda |,\,\mu ) = \left( {3,\, \pm {1 \over {81}}} \right)$

Here correct option will be $\left( {3,\,{1 \over {81}}} \right)$

$\Delta $ = $\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {2x - 3} & {3x - 4} & {4x - 5} \cr {3x - 5} & {5x - 8} & {10x - 17} \cr } } \right|$

R₂ $ \to $ R₂ – R₁
R₃ $ \to $ R₃ – R₂

= $\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {x - 1} & {x - 1} & {x - 1} \cr {x - 2} & {2\left( {x - 2} \right)} & {6\left( {x - 2} \right)} \cr } } \right|$

= $\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr 1 & 1 & 1 \cr 1 & 2 & 6 \cr } } \right|$

C₁ $ \to $ C₁ - C₂
C₂ $ \to $ C₂ - C₃

= $\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ { - x + 1} & { - x + 1} & {3x - 4} \cr 0 & 0 & 1 \cr { - 1} & { - 4} & 6 \cr } } \right|$

= -(x - 1)(x - 2)[-4(1 - x) + 1(1 - x)]

= -(x² - 3x + 2)[3x - 3]

= -3x³ + 9x² - 6x + 3x² - 9x + 6

= -3x³ + 12x² - 15x + 6 = Ax³ + Bx² + Cx + D

$ \therefore $ A = -3, B = 12, C = -15

$ \therefore $ B + C = 12 – 15 = – 3

Given,
${a^3} + {b^3} + {c^3} = 2$

${A^T}A = I$

$ \Rightarrow \left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right]\left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

$ \Rightarrow {a^2} + {b^2} + {c^2} = 1$

and ab + bc + ca = 0

Now (a + b + c)² = 1

$ \Rightarrow (a + b + c) = \pm 1$

So, ${a^3} + {b^3} + {c^3} - 3abc$

$ \Rightarrow (a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$

$ \Rightarrow \pm 1(1 - 0) = \pm 1$

$ \therefore 2 - 3abc = \pm 1$

$ \Rightarrow 3abc = 2 \pm 1 = 3,1$

$ \Rightarrow abc = 1,{1 \over 3}$

Let $X = \left[ {\matrix{ x \cr y \cr z \cr } } \right]$

PX = O

$\left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

x + 2y + z = 0........(1)

-2x + 3y – 4z = 0....(2)

x + 9y - z = 0..........(3)

from (1) & (3)
$ \Rightarrow $ 2x+11y =0

from (1) & (2)
$ \Rightarrow $ 2x + 11y = 0

from (2) & (3)
–6x –33y = 0
$ \Rightarrow $ 2x +11y = 0

putting value of x in (1), we get
–7y + 2z = 0

Now ${\left( {{{11y} \over 2}} \right)^2} + {y^2} + {\left( {{{7y} \over 2}} \right)^2} = 1$

y²(121 + 1 + 49) = 4

y²(171) = 4

$y = \pm {2 \over {\sqrt {171} }}$

$ \Rightarrow x = \pm {7 \over {\sqrt {171} }}$

$ \Rightarrow z = \pm {{11} \over {\sqrt {171} }}$

$ \therefore $ So, there are 2 solution set of (x, y, z)

For no solution :

$\Delta $ = 0 and $\Delta $₁/$\Delta $₂/$\Delta $₃ $ \ne $ 0

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

$ \Rightarrow $ 2(–2 – $\lambda $²) + 1 (1 – $\lambda $) + 2($\lambda $ + 2) = 0

$ \Rightarrow $ –2$\lambda $² + $\lambda $ + 1 = 0

$ \Rightarrow $ $\lambda $ = 1, $ - {1 \over 2}$

When $\lambda $ = 1

2x – y + 2z = 2 ...(1)

x – 2y + z = –4 ...(2)

x + y + z = 4 ...(3)

Adding (2) and (3), we get

2x – y + 2z = 0 (contradiction) hence no solution.

$ \therefore $ $\lambda $ = 1 belongs to set S.

When $\lambda $ = $ - {1 \over 2}$

2x – y + 2z = 2 ...(1)

x – 2y $ - {1 \over 2}$z = –4 ...(2)

x $ - {1 \over 2}$y + z = 4 ...(3)

(1) and (3) contradict each other, hence no solution.

$ \therefore $ $\lambda $ = $ - {1 \over 2}$ belongs to set S.

Let A = $\left[ {\matrix{ a & b \cr c & d \cr } } \right]$, where a, b, c, d $ \in $ {0, 1}

$ \Rightarrow $ |A| = ad – bc

$ \therefore $ ad = 0 or 1 and bc = 0 or 1

So possible values of |A| are 1, 0 or –1

(P) If A $ \ne $ I₂ then |A| is either 0 or –1

(Q) If |A| = 1 then ad = 1 and bc = 0

$ \Rightarrow $ a = d = 1 $ \Rightarrow $ Tr(A) = 2

Given
7x + 6y – 2z = 0 .......(1)
3x + 4y + 2z = 0 ......(2)
x – 2y – 6z = 0 .......(3)

$\Delta $ = $\left| {\matrix{ 7 & 6 & { - 2} \cr 3 & 4 & 2 \cr 1 & { - 2} & { - 6} \cr } } \right|$

= 7(–24 + 4) – 6(–18 – 2) – 2(–6 – 4) = 0

$ \therefore $ $\Delta $ = 0

The system of equation has infinite non-trival solution.

Also adding equation (1) and 3$ \times $(3), we get

10x = 20z

$ \Rightarrow $ x = 2z

A = $\left[ {\matrix{ 1 & 1 & 2 \cr 1 & 3 & 4 \cr 1 & { - 1} & 3 \cr } } \right]$

$ \Rightarrow $ |A| = 6

${{\left| {adjB} \right|} \over {\left| C \right|}}$

= ${{\left| {adj\left( {adjA} \right)} \right|} \over {\left| {3A} \right|}}$

= ${{{{\left| A \right|}^4}} \over {{3^3}\left| A \right|}}$

= ${{{{\left| A \right|}^3}} \over {{3^3}}}$

= ${{{6^3}} \over {{3^3}}}$ = 8

For planes to intersect on a line there should be infinite solution of the given system of equations.

For infinite solutions

$\Delta $ = $\left| {\matrix{ 1 & 4 & { - 2} \cr 1 & 7 & { - 5} \cr 1 & 5 & \alpha \cr } } \right|$ = 0

$ \Rightarrow $ 1(7$\alpha $ + 25) – 4($\alpha $ + 5) – 2(5 – 7) = 0

$ \Rightarrow $ 7$\alpha $ + 25 – 4$\alpha $ – 20 + 4 = 0

$ \Rightarrow $ 3$\alpha $ + 9 = 0

$ \Rightarrow $ $\alpha $ = -3

Also $\Delta $_z = 0

$ \Rightarrow $ $\left| {\matrix{ 1 & 4 & 1 \cr 1 & 7 & \beta \cr 1 & 5 & 5 \cr } } \right|$ = 0

$ \Rightarrow $ 1(35 – 5$\beta $) – 4(5 – $\beta $) + 1(5 – 7) = 0

$ \Rightarrow $ 35 - 5$\beta $ - 20 + 4$\beta $ - 2 = 0

$ \Rightarrow $ $\beta $ = 13

$ \therefore $ $\alpha $ + $\beta $ = -3 + 13 = 10

According to Cayley Hamilton equation
|A – $\lambda $I| = 0

$ \Rightarrow $ $\left| {\matrix{ {2 - \lambda } & 2 \cr 9 & {4 - \lambda } \cr } } \right|$ = 0

$ \Rightarrow $ (2 – $\lambda $)(4 – $\lambda $) – 18 = 0

$ \Rightarrow $ 8 – 2$\lambda $ – 4$\lambda $ + $\lambda $² – 18 = 0

$ \Rightarrow $ $\lambda $² – 6$\lambda $ – 10 = 0

$ \therefore $ A² – 6A– 10 = 0

$ \Rightarrow $ A^–1(A²) – 6A^–1A – 10A^–1 = 0

$ \Rightarrow $ A – 6I – 10A^–1 = 0

$ \Rightarrow $ 10A^–1 = A – 6I

$\Delta $ = $\left| {\matrix{ \lambda & 2 & 2 \cr {2\lambda } & 3 & 5 \cr 4 & \lambda & 6 \cr } } \right|$

= $\lambda $ ( 18 – 5$\lambda $) – 2(12$\lambda $ – 20) + 2(2$\lambda $² – 12)

= 18$\lambda $ – 5$\lambda $² – 24$\lambda $ + 40 + 4$\lambda $² – 24

= – $\lambda $² – 6$\lambda $ + 16

= – ($\lambda $ + 8)($\lambda $ – 2)

For no solutions $\lambda $ = 0 $ \Rightarrow $ $\lambda $ = – 8, $\lambda $ = 2

when $\lambda $ = 2

$\Delta $_x = $\left| {\matrix{ 5 & 2 & 2 \cr 8 & 3 & 5 \cr {10} & 2 & 6 \cr } } \right|$

= 5 (18 – 10) – 2 (48 – 50) + 2 (16 – 30)

= 40 + 4 – 28 $ \ne $ 0

So no solution for $\lambda $ = 2

For inconsistent system we need

$\Delta $ = 0 and atleast one of $\Delta $x, $\Delta $y, $\Delta $z $ \ne $ 0

$ \therefore $ $\Delta $ = $\left| {\matrix{ 1 & 2 & 3 \cr 3 & 4 & 5 \cr 4 & 4 & 4 \cr } } \right|$ = 0

$\Delta $_x = $\left| {\matrix{ 1 & 2 & 3 \cr \mu & 4 & 5 \cr \delta & 4 & 4 \cr } } \right|$

= (-4) - 2($\mu $ - 5$\delta $) + 3(4$\mu $ - 4$\delta $)

$ \Rightarrow $ 2$\mu $ $ \ne $ $\delta $ + 2 ....(1)

Only ($\mu $, $\delta $) = (4, 3) does satisfy the equation (1).

|B| = $\left| {\matrix{ {{b_{11}}} & {{b_{12}}} & {{b_{13}}} \cr {{b_{21}}} & {{b_{22}}} & {{b_{23}}} \cr {{b_{31}}} & {{b_{32}}} & {{b_{33}}} \cr } } \right|$

= $\left| {\matrix{ {{3^0}{a_{11}}} & {{3^1}{a_{12}}} & {{3^2}{a_{13}}} \cr {{3^1}{a_{21}}} & {{3^2}{a_{22}}} & {{3^3}{a_{23}}} \cr {{3^2}{a_{31}}} & {{3^3}{a_{32}}} & {{3^4}{a_{33}}} \cr } } \right|$

= ${3.3^2}\left| {\matrix{ {{a_{11}}} & {{3^1}{a_{12}}} & {{3^2}{a_{13}}} \cr {{a_{21}}} & {{3^1}{a_{22}}} & {{3^2}{a_{23}}} \cr {{a_{31}}} & {{3^1}{a_{32}}} & {{3^2}{a_{33}}} \cr } } \right|$

= ${3.3^2}{.3.3^2}\left| {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right|$

= 3⁶.|A|

$ \therefore $ 3⁶.|A| = 81

$ \Rightarrow $ |A| = ${1 \over 9}$

x² + x + 1 = 0

$ \Rightarrow $ x = ${{ - 1 + i\sqrt 3 } \over 2}$ = $\omega $ or ${{ - 1 - i\sqrt 3 } \over 2}$ = ${\omega ^2}$

Let $\alpha $ = $\omega $

$ \therefore $ A = ${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^4}} \cr } } \right]$

A² = ${1 \over 3}\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 0 & 3 \cr 0 & 3 & 0 \cr } } \right]$ = $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$

Now A⁴ = $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$ = I

$ \therefore $ A³¹ = A²⁸A³ = A³

For non-zero solution

$\left| {\matrix{ 2 & {2a} & a \cr 2 & {3b} & b \cr 2 & {4c} & c \cr } } \right| = 0$

$ \Rightarrow $ $\left| {\matrix{ 1 & {2a} & a \cr 0 & {3b - 2a} & {b - a} \cr 0 & {4c - 2a} & {c - a} \cr } } \right| = 0$

$ \Rightarrow $ (3b – 2a) (c –a) – (b – a) (4c – 2a) = 0

$ \Rightarrow $ 2ac = bc + ab

$ \Rightarrow $ ${2 \over b} = {1 \over a} + {1 \over c}$

$ \therefore $ ${1 \over a},{1 \over b},{1 \over c}$ are in A.P.

It is given that matrix M be a 3 $ \times $ 3 invertible matrix, such that

M^$-$1 = adj(adj M) $ \Rightarrow $ M^$-$1 = |M| M

($ \because $ for a matrix A of order 'n' adj(adjA) = |A|^n$-$2 A}

$ \Rightarrow $ M^$-$1 M = |M|M² $ \Rightarrow $ M²|M| = I .....(i)

$ \because $ det(M² |M|) = det(I) = 1

$ \Rightarrow $ |M|³|M|² = 1 $ \Rightarrow $ |M| = 1 .....(ii)

from Eqs. (i) and (ii), we get

M² = I

As, adj M = |M|M^$-$1 = M

$ \Rightarrow $ (adj M)² = M² (adj M)² = I

Given that,

$ a_{i j}= \begin{cases}1, & \text { if } i+j=7 \\ 0, & \text { if } i+j \neq 7\end{cases} $

Now, $\left(A(\operatorname{adj} A) A^{-1}\right) A^2$

$ \begin{array}{lrr} =\left(A(\operatorname{Adj} A) A^{-1}\right) A^2 & \\ =\left((|A| I) A^{-1}\right) A^2 & {[\because A \operatorname{adj} A=|A| I]} \\ =-1\left(A^{-1} A\right) A & {[\because|A|=-1]} \\ =-I A & {\left[\because A^{-1} A=I\right]} \\ =-A & \end{array} $

Given system of equations has a non-trivial solution

$ \begin{aligned} & \Pi_1 \equiv x+a y+a z=0 \\ & \Pi_2 \equiv b x+y+b z=0 \\ & \Pi_3 \equiv c x+c y+z=0 \end{aligned} $

It can be written as $A X=0$

Where, $A=\left[\begin{array}{lll}1 & a & a \\ b & 1 & b \\ c & c & 1\end{array}\right]$

So, the system has non-trivial solutions hence it has infinitely many solutions or unique solutions.

$ \therefore \quad|A|=0 $

Now, let the system has infinitely many solutions

So, let $y=t ; t \in \mathbf{R}$

Then from Eqs. (i) and (ii), we get

$ x+a z=-a t \Rightarrow b x+b z=-t $

Solving these, we get

$ z=\frac{t(1-a b)}{b(a-1)}, x=\frac{a}{b} t \frac{(1-b)}{(1-a)} $

Hence, if $a \neq b \neq c$, then

$ \begin{aligned} & x=\frac{a}{b} t \frac{(1-b)}{(1-a)}, y=t \\ & z=\frac{t}{b} \frac{(1-a b)}{(a-1)}, t \in \mathbf{R} \end{aligned} $

If $a=b=c$

Let $y=t_1, z=t_2, t_1, t_2 \in \mathbf{R}$

Then by Eq. (i)

$ x=-a\left(t_1+t_2\right) $

So, the system of equations has unique solution.

Given that, Rank of matrix $A=$ rank of matrix $B$

$ f(A)=f(B) \text { and } f(B)=3 $

∴ Order of matrix $B \geq$ Rank of matrix $B$.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} &\left|\begin{array}{ccc} 2 a-2 b-4 & 4 a & 4 a \\ 4 & 2-b-a & 4 \\ 2 b & 2 b & b-a-2 \end{array}\right| \\ &= 2\left|\begin{array}{ccc} a-b-2 & 2 a & 2 a \\ 4 & 2-b-a & 4 \\ 2 b & 2 b & b-a-2 \end{array}\right| \\ &= 2\left|\begin{array}{ccc} -(a+b+2) & 0 & 2 a \\ a+b+2 & -(a+b+2) & 4 \\ 0 & a+b+2 & b-a-2 \end{array}\right| \\ & C_1 \rightarrow C_1-C_2 \text { and } C_2 \rightarrow C_2-C_3 \end{aligned} \end{aligned} $

$ \begin{aligned} & =2(a+b+2)^2\left|\begin{array}{ccc} -1 & 0 & 2 a \\ 1 & -1 & 4 \\ 0 & 1 & b-a-2 \end{array}\right| \\ & =2(a+b+2)^2(-1(-b+a+2-4)+2 a(1) \\ & =2(a+b+2)^2(a+b+2)=2(a+b+2)^3 \\ & =2\left[(a+b)^3+6(a+b)^2+12(a+b)+8\right] \end{aligned} $

We have,

$ \begin{aligned} A & =\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right]\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 2 & -2 & -16-4 x \\ -1 & 3 & 16+4 x \\ 4+x & -8-2 x & -12+x^2 \end{array}\right] \end{aligned} $

Given, $A^2=A$

$ \begin{array}{ll} \because & 4+x=1 \Rightarrow x=-3 \\ \therefore & A=\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right] \\ & |A|=2(-9+8)+2(3-4)-4(2-3) \\ & =-2-2+4=0 \end{array} $

Rank of $A=2 \Rightarrow r+x=2-3=-1$

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \Delta_1 & =\left|\begin{array}{ll} m & b \\ n & d \end{array}\right|=m d-b n \\ \Delta_2 & =\left|\begin{array}{ll} a & m \\ c & n \end{array}\right|=a n-m c \\ \Delta_3 & =\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=a d-b c \\ x^a y^b & =e^m, x^c y^d=e^n \\ y & =\left(\frac{e^m}{x^a}\right)^{1 / b} \text { and } y=\left(\frac{e^n}{x^c}\right)^{1 / d} \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore & \frac{e^{m / b}}{x^{a / b}} & =\frac{e^{n / d}}{x^{c / d}} \\ \Rightarrow & & x^{c / d-a / b} & =e^{n / d-m / b} \\ \Rightarrow & & x & =\left(e^{n / d-m / b}\right)^{\frac{1}{c}-\frac{a}{b}} \\ \Rightarrow & & x & =e^{\left(\frac{b n-m d}{b d} \times \frac{b d}{b c-a d}\right)} \\ \Rightarrow & & x & =e^{\frac{b n-m d}{b c-a d}} \\ \Rightarrow & & x & =e^{\frac{m d-b n}{a d-b c}}=e^{\frac{\Delta_1}{\Delta_3}} \end{aligned} $

Putting the value of $x$ in $y=\frac{e^{m / b}}{x^{a / b}}$ we get $\quad y=e^{\frac{a n-m c}{a d-b c}}=e^{\frac{\Delta_2}{\Delta_3}}$

We have, $A=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]$ and

$ \begin{aligned} & B=\left[\begin{array}{lll} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right]=\left[\begin{array}{ccc} 2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right] \\ & \therefore A B=\left[\begin{array}{ccc} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right] \end{aligned} $

$ =\left[\begin{array}{cc} 2+4 b_{21}+2 b_{31} & 0+4 b_{22}+2 b_{32} \\ 4-b_{21}+4 b_{31} & 0+b_{22}+4 b_{32} \\ -6+7 b_{21}-6 b_{31} & 0+7 b_{22}-6 b_{32} \\ & -2+4 b_{23}+2 b_{33} \\ & -4-b_{23}+4 b_{33} \\ & 6+7 b_{23}-6 b_{33} \end{array}\right] $

$ =\left[\begin{array}{ccc} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{array}\right] $

On solving above equal matrices with corresponding elements, we get

$ \begin{aligned} & b_{21}=b_{31}=0, b_{22}=3, b_{32}=1, b_{23}=0 \\ & \text { and } b_{33}=-1 \\ & \therefore B=\left[\begin{array}{ccc} 2 & 0 & -2 \\ 0 & 3 & 0 \\ 0 & 1 & -1 \end{array}\right] \\ & \therefore|B|=2(-3-0)+(-2)(0-0)=-6 \\ & \text { and Trace }(B)=2+3-1=4 \\ & \therefore|B|+\operatorname{trace}(B)=-6+4=-2 \end{aligned} $

Given, $A$ is $m \times n$ matrix of rank 4

$A$ contains $m$ th order of non-singular sub matrix.

$A$ is non-singular matrix

$\therefore A$ is a square matrix of order $M$

∴ Rank of $A$ is $M$

∴ order of $A=4$       [rank of $A=4$ ]

Hence, rows of $A=4$.

Given, $C$ and $D$ are non-singular matrix of order $n$

$ \quad \begin{aligned} \because\,\,\,\,\,\,\, |C| & \neq 0,|D| \neq 0 \\ C D & =-D C \\ |C D| & =|-D C| \\ |C||D| & =(-1)^n|D||C| \\ 1 & =(-1)^n \end{aligned} $

$\therefore n$ is even integer.

It is given that the matrices $A$ and $B$ are non-singular of order 3 and $|B|=k, a$ positive integer, so

$ \begin{aligned} \left|k^{-1} A^{-1}\right|=\left(k^{-1}\right)^3\left|A^{-1}\right| & =\frac{1}{k^3|A|} \\ & {\left[\because\left|A^{-1}\right|=\frac{1}{|A|}\right] } \end{aligned} $

$ \begin{gathered} =\frac{1}{|B|^3|A|} \\ \because\left|\operatorname{adj}\left(A^{-1}\right)\right|=\left|A^{-1}\right|^{3-1}=\left|A^{-1}\right|^2=\frac{1}{|A|^2} \end{gathered} $

$\left[\because|\operatorname{adj}(A)|=|A|^{n-1}\right.$, where $n$ is the order of matrix $A$ ]

Now, since it is given that $B A B^{-1}=I$

$ \begin{aligned} & \Rightarrow \quad A B^{-1}=B^{-1} \\ & \therefore B A^k B^{-1}=B A^{k-1}\left(A B^{-1}\right)=B A^{k-1} B^{-1} \\ & =B A^{k-2}\left(A B^{-1}\right)=B A^{k-2} B^{-1} \\ & \therefore B A^k B^{-1}=B A B^{-1}=I \\ & \text { and }, B \frac{\operatorname{adj}(B)}{|B|}=B B^{-1}=I \end{aligned} $

Therefore, $B A^k B^{-1}=B \frac{\operatorname{adj}(B)}{|B|}$, if

$ B A B^{-1}=I $

Now, the adj $\left(\operatorname{adj}\left(A^{-1}\right)\right)$

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

$\left[\because \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A\right.$ where $n$ is the order of matrix $A$ ]

$ =\frac{1}{|A|}\left(A^{-1}\right) $

We have,

$ \begin{array}{r} 2 x+p y+6 z=8 \\ x+2 y+q z=5 \\ x+y+3 z=4 \end{array} $

Now, if we express above equation in matrix form, We have,

$ A X=B $

Where,

$ A=\left[\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}\right], B=\left[\begin{array}{l} 8 \\ 5 \\ 4 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] $

For no solution $|A|=0$

$ \begin{aligned} & \Rightarrow \quad\left|\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}\right|=0 \\ & \Rightarrow \quad 2(6-q)-p(3-q)+6(1-2)=0 \\ & \Rightarrow \quad 12-2 q-3 p+p q-6=0 \\ & \Rightarrow \quad 3 p+2 q-p q-6=0 \\ & \Rightarrow \quad 3 p-6+2 q-p q=0 \\ & \Rightarrow \quad 3(p-2)-q(p-2)=0 \\ & \Rightarrow \quad(p-2)(3-q)=0 \Rightarrow \\ & \vec{p}=2,3 \end{aligned} $

But for $p=2$ equations $2 x+p y+6 z=8$ and $x+y+3 z=4$ are same.

So, $p \neq 2, q=3$

We have,

$ \begin{array}{r} (\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0 \\ (\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0 \\ 2 x+(3 \lambda+1) y+3(\lambda-1) z=0 \end{array} $

Now, it can be express as $\quad A X=0$

where, $A=\left[\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3(\lambda-1)\end{array}\right]$ and $X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$

For non-zero solution $|A|=0$

$ \left[\begin{array}{ccc} \lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3(\lambda-1) \end{array}\right]=0 $

On applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$,

$ \left[\begin{array}{ccc} \lambda-1 & 3 \lambda+1 & 2 \lambda \\ 0 & \lambda-3 & -\lambda+3 \\ -\lambda+3 & 0 & \lambda-3 \end{array}\right]=0 $

$ \begin{aligned} & \Rightarrow(\lambda-1)\left[(\lambda-3)^2\right]+(-\lambda+3)((3 \lambda+1) \\ & \quad(-\lambda+3)-2 \lambda(\lambda-3))=0 \\ & \Rightarrow(\lambda-1)(\lambda-3)^2-(\lambda-3) \\ & \quad[(\lambda-3)(-3 \lambda-1-2 \lambda))=0 \\ & \Rightarrow(\lambda-3)^2[\lambda-1+5 \lambda+1]=0 \\ & \Rightarrow(\lambda-3)^2(6 \lambda)=0 \Rightarrow \lambda=3,0 \\ & \therefore p=3, q=0 \\ & \therefore p^2+q^2-p q=9+0-0=9 \end{aligned} $

$\left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$

R₁ $ \to $ R₁ - R₂, R₂ $ \to $ R₂ - _R3

$ \Rightarrow \left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$

C₂ $ \to $ C₂ + C₁

$ \Rightarrow \left| {\matrix{ 1 & 0 & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & 1 & {1 + 4\cos 6\theta } \cr } } \right| = 0$

$ \Rightarrow 1 + 4\cos 6\theta + 1 = 0$

$ \Rightarrow 2\cos 6\theta = - 1 \Rightarrow \cos 6\theta = - {1 \over 2}$ = $\cos {{2\pi } \over 3}$

$ \Rightarrow 6\theta = 2n\pi \pm {{2\pi } \over 3}$

$ \Rightarrow \theta = {{n\pi } \over 3} \pm {\pi \over 9}\,\,\,n \in 1$

$ \Rightarrow \theta = {\pi \over 9},{{2\pi } \over 9},{{4\pi } \over 9}$

Given |A| + 1 = 0

$ \Rightarrow $ |A| = -1

$\left| B \right| = \left| {{A^{ - 1}}} \right| = {1 \over {\left| A \right|}} = - 1$

$\left| {\matrix{ 5 & {2\alpha } & 1 \cr 0 & 2 & 1 \cr \alpha & 3 & { - 1} \cr } } \right| $ = -1

$ \Rightarrow $ $ 5( - 2 - 3) + 2\alpha (\alpha ) + 1( - 2\alpha ) = - 1$

$ \Rightarrow $ $2{\alpha ^2} - 2\alpha - 24 = 0$

$ \therefore $ sum of value of $\alpha $ = ${{ - ( - 2)} \over 2} = 1$

$A + B = \left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right] = P(say)$

Now $A = {{P + {P^T}} \over 2}\& B = {{P - {P^T}} \over 2}$

So $A = {1 \over 2}\left( {\left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right] + \left[ {\matrix{ 2 & 5 \cr 3 & { - 1} \cr } } \right]} \right) = \left[ {\matrix{ 2 & 4 \cr 4 & { - 1} \cr } } \right]$

$B = {1 \over 2}\left( {\left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right] - \left[ {\matrix{ 2 & 5 \cr 3 & { - 1} \cr } } \right]} \right) = \left[ {\matrix{ 0 & { - 1} \cr 1 & 0 \cr } } \right]$

So $AB = \left( {\left[ {\matrix{ 2 & 4 \cr 4 & { - 1} \cr } } \right]\left[ {\matrix{ 0 & { - 1} \cr 1 & 0 \cr } } \right]} \right) = \left[ {\matrix{ 4 & { - 2} \cr { - 1} & { - 4} \cr } } \right]$

$\Delta = 0$

$\left| {\matrix{ 1 & 1 & 1 \cr 4 & \lambda & { - \lambda } \cr 3 & 2 & { - 4} \cr } } \right| = 0$

On solving we get $\lambda $ = 3

x(-3x $ \times $ (x + 2) - 2x(x - 3)) + (– 6) (2(x + 2) + 3 (x – 3)) + (–1) (4x + 3 (–3x))

$ \Rightarrow $ – 5x³ + 30x –30 + 5x = 0

$ \Rightarrow $ x³ – 7x + 6 = 0

$ \therefore $ sum of roots = 0

x + y + z = 5

x + 2y + 2z = 6

x + 3y + $\lambda $z = $\mu $ have infinite solution

$\Delta $ = 0, $\Delta $x = $\Delta $y = $\Delta $z = 0

$\Delta = \left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 2 \cr 1 & 3 & \lambda \cr } } \right| = 0$

$ \Rightarrow 1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2) = 0$

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

$ \Rightarrow \lambda = 3$

Now, $\Delta x = \left| {\matrix{ 5 & 1 & 1 \cr 6 & 2 & 2 \cr \mu & 3 & 3 \cr } } \right| = 0$,

$\Delta $y = $\left| {\matrix{ 1 & 5 & 1 \cr 1 & 6 & 2 \cr 1 & \mu & 3 \cr } } \right| = 0$

$ \Rightarrow \left| {\matrix{ 1 & 5 & 1 \cr 0 & 1 & 1 \cr 0 & {\mu - 5} & 2 \cr } } \right| = 0$

$ \Rightarrow \mu = 7$

$\Delta z = \left| {\matrix{ 1 & 1 & 5 \cr 1 & 2 & 6 \cr 1 & 3 & \mu \cr } } \right| = \left| {\matrix{ 1 & 1 & 5 \cr 0 & { - 1} & { - 1} \cr 0 & 2 & {\mu - 5} \cr } } \right|$

$ \Rightarrow 1(5 - \mu + 2) = 0$

$ \Rightarrow \mu = 7$

So, $\lambda + \mu = 10$

${\Delta _1} = \left| {\matrix{ x & {\sin \theta } & {\cos \theta } \cr { - \sin \theta } & { - x} & 1 \cr {\cos \theta } & 1 & x \cr } } \right|$

= x(–x² –1) – sin$\theta $(–xsin$\theta $ – cos$\theta $) + cos$\theta $(–sin$\theta $+ xcos$\theta $)

= –x³ – x + xsin²$\theta $ + sin$\theta $cos$\theta $ – cos$\theta $sin$\theta $ + xcos²$\theta $

= –x³ – x + x = –x³

Similarly ${\Delta _2} = - {x^3}$

${\Delta _1} + {\Delta _2} = - 2{x^3}$

Given 2x + 3y – z = 0,

x + ky – 2z = 0

2x – y + z = 0

For non trivial solution

$\Delta = 0 \Rightarrow \left| {\matrix{ 2 & 3 & { - 1} \cr 1 & k & { - 2} \cr 2 & { - 1} & 1 \cr } } \right| = 0$

$ \Rightarrow k = {9 \over 2}$

$ \therefore $ Equations are 2x + 3y – z = 0 ...(i)

2x – y + z = 0 ...(ii)

2x + 9y – 4z = 0 ...(iii)

By (i) – (ii) we get,

4y - 2z = 0

$ \Rightarrow $ 2y = z .......(iv)

$ \Rightarrow $ ${y \over z} = {1 \over 2}$

From equation (i) and (iv)

2x + 3y - 2y = 0

$ \Rightarrow $ 2x + y = 0

$ \Rightarrow $ ${x \over y} = - {1 \over 2}$

$ \Rightarrow $ ${x \over y} \times {y \over z} = - {1 \over 2} \times {1 \over 2} = - {1 \over 4}$

$ \Rightarrow $ ${z \over x} = - 4$

$ \therefore $ ${x \over y} + {y \over z} + {z \over x} + k = {{ - 1} \over 2} + {1 \over 2} - 4 + {9 \over 2}$ = ${1 \over 2}$