Determinants

We have,

$A_{11}={(-1)}^{1+1}*\left|\begin{array}{cc}\hfill 3\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=(3*1-5*0)=3$

$A_{12}={(-1)}^{1+2}*\left|\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right|=(-1)(2*1-5*(-2)=-1(2+10)=-12$

$A_{13}={(-1)}^{1+3}*\left|\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right|=[2*0-(-2)*3]=6$

We have,

$A_{11}={(-1)}^{1+1}*4=4$

$A_{12}={(-1)}^{1+2}*3=-3$

$A_{21}={(-1)}^{2+1}*2=-2$

$A_{22}={(-1)}^{2+2}*1=1.$

Option D is correct.

Given,  $\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill yz\hfill \\ \hfill 1\hfill & \hfill y\hfill & \hfill zx\hfill \\ \hfill 1\hfill & \hfill z\hfill & \hfill xy\hfill \end{array}\right|$

Co-factor of elements of third column

∴ Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

 = yz (z-y) + zx [- (z-x)] + xy (y-x)

 = y z²-y²z-z²x+ zx² + xy²-x²y.

 = yz²-y²z+ (xy²-xz²) + (zx²-x²y)

 = yz (z-y) + x (y² – z²) -x² (y-z)

 = -yz (y-z) + x (y + z) (y-z) -x² (y-z)

= (y-z) [-yz + x (y + z) -x²]

 = (y-z) [-yz + xy + xz-x²]

 = (y-z) [-y (z-x) + x (z-x)]

 = (y-z) (z-x) (x -y)

Given,  $?=\left|\begin{array}{lll}5\hfill & 3\hfill & 8\hfill \\ 2\hfill & 0\hfill & 1\hfill \\ 1\hfill & 2\hfill & 3\hfill \end{array}\right|$

Co-factors of elements of second row,

? ? = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

= 2 * 7 + 0 * 7 + 1 * (-7) = 14 + 0 - 7 = 7.

(i) Given, A = $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|$

So,

(ii)Given,  $\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|$

(i) We know that,

Minor of element a_ij is m_ij and its co-factor is A_ij = (–1)^i+j M_ij

So,

M₁₁ = 3 and A₁₁ = (- 1)^{1 + 1} M₁₁ = 1 * 3 = 3

M₁₂ = 0 and A₁₂ = (-1)¹⁺² M₁₂ = -1 * 0 = 0

M₂₁ = -4 and A₂₁ = (-1)²⁺¹ M₂₁ = (-1) * (-4) = 4

M₂₂ = 2 and A₂₂ = (-1)²⁺² M₂₂ = 1 * 2 = 2

(ii) Given A = $\left|\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right|$

So,

M₁₁ = d and A₁₁ = (-1)¹⁺¹ M₁₁ = 1 * d = d

M₁₂ = b and A₁₂ = (-1)¹⁺² M₁₂ = (-1) * b = -b

M₂₁ = c and A₂₁ = (-1)^{2+ 1} M₂₁ = (–1) * c = –c

M₂₂ = a and A₂₂ = (-1)²⁺² M₂₂ = 1 * a = a

Given,

Area of triangle = 35 sq. Units

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill k\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|=35.$

∴ Option D is correct.

(i) Let P (x, y) be any point on line joining A (1, 2) & B (3, 6)

Then, area of triangle (ABP) = 0 {the point are collinear

$\frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill 1\hfill \end{array}\right|=0$

(i) Area of the triangle = 4 sq. units (given)

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill k\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=4$

(ii) Area of the triangle = 4 sq units