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.

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

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

Given A^TA = 3I₃

$ \Rightarrow $ $\left[ {\matrix{ 0 & {2x} & {2x} \cr {2y} & y & { - y} \cr 1 & { - 1} & 1 \cr } } \right]\left[ {\matrix{ 0 & {2y} & 1 \cr {2x} & y & { - 1} \cr {2x} & { - y} & 1 \cr } } \right]$

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

$ \Rightarrow $ $\left[ {\matrix{ {8{x^2}} & 0 & 0 \cr 0 & {6{y^2}} & 0 \cr 0 & 0 & 3 \cr } } \right]$ = $\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 3 & 0 \cr 0 & 0 & 3 \cr } } \right]$

$ \therefore $ 8x² = 3, 6y² = 3

$ \Rightarrow $ x = $ \pm \sqrt {{3 \over 8}} $, y = $ \pm \sqrt {{1 \over 2}} $

Total possible combination of x and y = 2 $ \times $ 2 = 4

Given

$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$....$\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$

$ \Rightarrow $ $\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$ = $\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$

$ \Rightarrow $ $\left[ {\matrix{ 1 & 6 \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$ = $\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$

$ \Rightarrow $ $\left[ {\matrix{ 1 & {1 + 2 + 3} \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$ = $\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$
.
.
.
.
$ \Rightarrow $ $\left[ {\matrix{ 1 & {1 + 2 + 3 + .... + \left( {n - 1} \right)} \cr 0 & 1 \cr } } \right]$ = $\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$

By comparing both sides we get,

1 + 2 + 3 + ........+ (n - 1) = 78

$ \Rightarrow $ ${{n\left( {n - 1} \right)} \over 2}$ = 78

$ \Rightarrow $ n = 13, - 12(not possible)

$ \therefore $ The inverse of $\left[ {\matrix{ 1 & 13 \cr 0 & 1 \cr } } \right]$ = $\left[ {\matrix{ 1 & -13 \cr 0 & 1 \cr } } \right]$

$\alpha $ and $\beta $ are the roots of the equation x² + x + 1 = 0.

$ \therefore $ $\alpha $ = $\omega $ and $\beta $ = ${\omega ^2}$

$\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right|$

= $\left| {\matrix{ {y + 1} & \omega & {{\omega ^2}} \cr \omega & {y + {\omega ^2}} & 1 \cr {{\omega ^2}} & 1 & {y + \omega } \cr } } \right|$

C₁ $ \to $ C₁ + C₂ + C₃

= $\left| {\matrix{ {y + 1 + \omega + {\omega ^2}} & \omega & {{\omega ^2}} \cr {y + 1 + \omega + {\omega ^2}} & {y + {\omega ^2}} & 1 \cr {y + 1 + \omega + {\omega ^2}} & 1 & {y + \omega } \cr } } \right|$

= $\left| {\matrix{ y & \omega & {{\omega ^2}} \cr y & {y + {\omega ^2}} & 1 \cr y & 1 & {y + \omega } \cr } } \right|$

As $1 + \omega + {\omega ^2}$ = 0

= $y\left| {\matrix{ 1 & \omega & {{\omega ^2}} \cr 1 & {y + {\omega ^2}} & 1 \cr 1 & 1 & {y + \omega } \cr } } \right|$

R₂ $ \to $ R₂ - R₁
R₃ $ \to $ R₃ - R₁

= $y\left| {\matrix{ 1 & \omega & {{\omega ^2}} \cr 0 & {y + {\omega ^2} - \omega } & {1 - {\omega ^2}} \cr 0 & {1 - \omega } & {y + \omega - {\omega ^2}} \cr } } \right|$

= y$\left[ {\left( {y + {\omega ^2} - \omega } \right)\left( {y + \omega - {\omega ^2}} \right) - \left( {1 - {\omega ^2}} \right)\left( {1 - \omega } \right)} \right]$

= y(y²) = y³

2, b, c are in AP.

Let common difference = d

$ \therefore $ b = 2 + d and c = 2 + 2d

|A| = $\left[ {\matrix{ 1 & 1 & 1 \cr 2 & b & c \cr 4 & {{b^2}} & {{c^2}} \cr } } \right]$

C₂ = C₂ - C₁

C₃ = C₃ - C₁

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

= $\left( {b - 2} \right)\left( {c - 2} \right)\left| {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 1 \cr 4 & {b + 2} & {c + 2} \cr } } \right|$

= $\left( {b - 2} \right)\left( {c - 2} \right)\left[ {c + 2 - b - 2} \right]$

= $\left( {b - 2} \right)\left( {c - 2} \right)\left( {c - b} \right)$

[ As b = 2 + d and c = 2 + 2d, then b - 2 = 4, c - 2 = 2d and c - b = d]

= (d) (2d) (d)

= 2d³

Given |A| $ \in $ [2, 16]

$ \therefore $ 2d³ $ \in $ [2, 16]

$ \Rightarrow $ d³ $ \in $ [1, 8]

$ \Rightarrow $ d $ \in $ [1, 2]

As c = 2 + 2d

then c $ \in $ [4, 6]

If the system of equations has non-trivial solutions, then

D = 0

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

$ \Rightarrow $ (1 - c²) + c(-c - c²) - c(c² + c) = 0

$ \Rightarrow $ (1 + c)(1 - c - 2c²) = 0

$ \Rightarrow $ (1 + c)² (1 - 2c) = 0

$ \Rightarrow $ c = -1 or ${1 \over 2}$

$ \therefore $ Greatest value of c is ${1 \over 2}$.

From here sin 32$\alpha $ = 1 and cos 32$\alpha $ = 0

$ \therefore $ 32$\alpha $ = 2n$\pi $ + ${\pi \over 2}$

$ \Rightarrow $ $\alpha $ = ${\pi \over {64}} + {{n\pi } \over {16}}$

Putting n = 0, $\alpha $ = ${\pi \over {64}}$

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

$ \Rightarrow {\left( {\lambda - 1} \right)^3} = 0 \Rightarrow \lambda = 1$

$\left| A \right| = \left| {\matrix{ 1 & {\sin \theta } & 1 \cr { - \sin \theta } & 1 & {\sin \theta } \cr { - 1} & { - \sin \theta } & 1 \cr } } \right|$

= 2(1 + sin²$\theta $)

$\theta $ $ \in $ $\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right) \Rightarrow {1 \over {\sqrt 2 }} < \sin \theta < {1 \over {\sqrt 2 }}$

       $ \Rightarrow $  0 $ \le $ sin²$\theta $ < ${1 \over 2}$

$ \therefore $  $\left| A \right| \in \left[ {2,3} \right)$

$P = \left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 0 \cr 9 & 3 & 1 \cr } } \right]$

${P^2} = \left[ {\matrix{ 1 & 0 & 0 \cr {3 + 3} & 1 & 0 \cr {9 + 9 + 9} & {3 + 3} & 1 \cr } } \right]$

${P^3} = \left[ {\matrix{ 1 & 0 & 0 \cr {3 + 3 + 3} & 1 & 0 \cr {6.9} & {3 + 3 + 3} & 1 \cr } } \right]$

${P^n} = \left[ {\matrix{ 1 & 0 & 0 \cr {3n} & 1 & 0 \cr {{{n\left( {n + 1} \right)} \over 2}{3^2}} & {3n} & 1 \cr } } \right]$

${P^5} = \left[ {\matrix{ 1 & 0 & 0 \cr {5.3} & 1 & 0 \cr {15.9} & {5.3} & 1 \cr } } \right]$

$Q = {P^5} + {{\rm I}_3}$

$Q = \left[ {\matrix{ 2 & 0 & 0 \cr {15} & 2 & 0 \cr {135} & {15} & 2 \cr } } \right]$

${{{q_{21}} + {q_{31}}} \over {{q_{32}}}} = {{15 + 135} \over {15}} = 10$

Aliter

$P = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right) + \left( {\matrix{ 0 & 0 & 0 \cr 3 & 0 & 0 \cr 9 & 3 & 0 \cr } } \right)$

$P = {\rm I} + X$

$X = \left( {\matrix{ 0 & 0 & 0 \cr 3 & 0 & 0 \cr 9 & 3 & 0 \cr } } \right)$

${X^2} = \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 9 & 0 & 0 \cr } } \right)$

${{X_3} = 0}$
${{P^5} = {\rm I} + 5X + 10{X^2}}$
${Q = {P^5} + {\rm I} = 2{\rm I} + 5X + 10{X^2}}$

$Q = \left( {\matrix{ 2 & 0 & 0 \cr 0 & 2 & 0 \cr 0 & 0 & 2 \cr } } \right) + \left( {\matrix{ 0 & 0 & 0 \cr {15} & 0 & 0 \cr {15} & {15} & 0 \cr } } \right) + \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr {90} & 0 & 0 \cr } } \right)$

$ \Rightarrow \,\,Q = \left( {\matrix{ 2 & 0 & 0 \cr {15} & 2 & 0 \cr {135} & {15} & 2 \cr } } \right)$

For unique solution

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

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

$\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

R₁ $ \to $ R₁ + R₂ + R₃

$ = \left| {\matrix{ {a + b + c} & {a + b + c} & {a + b + c} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

$ = \left( {a + b + c} \right)\left| {\matrix{ 1 & 0 & 0 \cr {2b} & { - \left( {a + b + c} \right)} & 0 \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

$=$ (a + b + c) (a + b + c)²

$ \Rightarrow $  x $=$ $-$ 2(a + b + c)

${\left| A \right|^2}.\left| B \right| = 8$

and ${{\left| A \right|} \over {\left| B \right|}} = 8 \Rightarrow \left| A \right| = 4$

and $\left| B \right| = {1 \over 2}$

$ \therefore $  det(BA^$-$1. B^T) $ = {1 \over 4} \times {1 \over 4} = {1 \over {16}}$

P₁ : 2x + 2y + 3z = a

P₂ : 3x $-$ y + 5z = b

P₃ : x $-$ 3y + 2z = c

We find

P₁ + P₃ = P₂ $ \Rightarrow $ a + c = b

A is orthogonal matrix

$ \Rightarrow $  0² + p² + p² = 1

$ \Rightarrow $  $\left| p \right| = {1 \over {\sqrt 2 }}$

A = $\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$ (b > 0)

$\left| A \right|$ = 2(2b² + 2 $-$ b²) $-$ b(2b $-$ b) + 1(b₂ $-$ b₂ $-$ 1)

$\left| A \right|$ = 2(b² + 2) $-$ b² $-$ 1

$\left| A \right|$ = b² + 3

${{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3 $

$b + {3 \over b} \ge 2\sqrt 3 $

$\left| {\matrix{ 1 & 3 & 7 \cr { - 1} & 4 & 7 \cr {\sin 3\theta } & {\cos 2\theta } & 2 \cr } } \right| = 0$

(8 $-$ 7 cos 2$\theta $) $-$ 3($-$2 $-$ 7 sin 3$\theta $)

      +7 ($-$ cos 2$\theta $ $-$ 4 sin 3$\theta $) = 0

14 $-$ 7 cos 2$\theta $ + 21 sin 3$\theta $ $-$ 7 cos 2$\theta $

      $-$ 28 sin 3$\theta $ = 0

14 $-$ 7 sin 3$\theta $ $-$ 14 cos 2$\theta $ = 0

14 $-$ 7 (3 sin $\theta $ $-$ 4 sin³$\theta $ ) $-$ 14 (1 $-$ 2 sin² $\theta $) = 0

$-$ 21 sin $\theta $ + 28 sin³ $\theta $ + 28 sin² $\theta $ = 0

7 sin $\theta $ [$-$ 3 + 4 sin² $\theta $ + 4 sin $\theta $] = 0 sin$\theta $,

4 sin² $\theta $ + 6 sin $\theta $ $-$ 2 sin $\theta $ $-$ 3 = 0

2 sin $\theta $(2 sin $\theta $ + 3) $-$ 1 (2 sin $\theta $ + 3) = 0

sin $\theta $ = ${{ - 3} \over 2}$; sin$\theta $ = ${1 \over 2}$

Hence, 2 solutions in (0, $\pi $)

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

      $ = \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)$

for infinite solutions $D = 0 \Rightarrow \alpha = 5$

${D_x} = 0 \Rightarrow \left| {\matrix{ 5 & 1 & 1 \cr 9 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$

$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr { - 1} & { - 1} & 3 \cr {\beta - 15} & { - 2} & 5 \cr } } \right| = 0$

$ \Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0$

on $\beta = 13$ we get ${D_y} = {D_z} = 0$

$\alpha = 5,\beta = 13$