Ncert Solutions Maths class 12th

(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

The area of triangle from by the given points is area ( ΔABC) = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill a\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill b\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill c\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill a+b+c+1\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill b+c+a+1\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill c+a+b+1\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$C_1→ C₁ + C₂ + C₃

$=\frac{1}{2}(a+b+c+1)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$ Taking (a + b + c + 1) common from C₁

$=\frac{(a+b+c+1)}{2}*0\left\{?c=c_3\right\}$

= 0

Hence the points A, B C are collinear.

(i) Area of triangle is given by,

Δ = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill x_1\hfill & \hfill y_1\hfill & \hfill 1\hfill \\ \hfill x_2\hfill & \hfill y_2\hfill & \hfill 1\hfill \\ \hfill x_3\hfill & \hfill y_3\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 6\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}|[-3+18]|=\frac{15}{2}$ =7.5sq. units.

(ii) Area of the triangle is given by,

Δ = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 10\hfill & \hfill 8\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}|[2(1-8)-7(1-10)+1(8-10)]|$

$=\frac{1}{2}|(2*(-7)-7*(-9)+21*(-2)]|$

$=\frac{1}{2}|[-14+63-2]|=\frac{1}{2}\left|47\right|$

= $\frac{47}{2}$ =23.5 sq. units

(iii) Area of triangle is given by,

Δ = $\frac{1}{2}$ ${\left|\begin{array}{ccc}\hfill -2\hfill & \hfill -3\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -8\hfill & \hfill 1\hfill \end{array}\right|}_{.}$

$=\frac{1}{2}|[-2*10+3(4)+(-24+2)]|$

$=\frac{1}{2}|[-20+12-22]|$

$=\frac{1}{2}|-30|=\frac{30}{2}$  = 15 sq. units.

Option 'C' is correct as determinant is a number associated to a square matrix.

Then, KA = $k\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill k{a}_{11}\hfill & \hfill k{a}_{12}\hfill & \hfill k{a}_{13}\hfill \\ \hfill k{a}_4\hfill & \hfill k{a}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill k{a}_{31}\hfill & \hfill k{a}_{32}\hfill & \hfill k{a}_{33}\hfill \end{array}\right]$

$\therefore |KA|=\left|\begin{array}{ccc}\hfill {}_{a11}\hfill & \hfill k{a}_{12}\hfill & \hfill {}_{13}\hfill \\ \hfill {}_{a_{21}}\hfill & \hfill {}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill {}_{31}\hfill & \hfill {}_{a_{32}}\hfill & \hfill k_{a_{33}}\hfill \end{array}\right|$

$=k{}^3\left|\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{22}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right|$

$=k^3|A|$

So, option c is correct.

LHS = ${\left|\begin{array}{ccc}\hfill a^2+1\hfill & \hfill ab\hfill & \hfill ac\hfill \\ \hfill ab\hfill & \hfill b^2+1\hfill & \hfill bc\hfill \\ \hfill ca\hfill & \hfill cb\hfill & \hfill c^2+1\hfill \end{array}\right|}_{.}$

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

$=\frac{abc}{abc}\left[\begin{array}{ccc}\hfill a^2+1\hfill & \hfill b^2\hfill & \hfill c^2\hfill \\ \hfill a^2\hfill & \hfill b^2+1\hfill & \hfill c^2.\hfill \\ \hfill a^2\hfill & \hfill b^2\hfill & \hfill c^2+1\hfill \end{array}\right]$Taking a, b&c common from R₁, R₂&R₃

$\left[\begin{array}{ccc}\hfill 1+a^2+b^2+c^2\hfill & \hfill b^2\hfill & \hfill c^2\hfill \\ \hfill a^2+b^2+1+c^2\hfill & \hfill b^2+1\hfill & \hfill c^2\hfill \\ \hfill a^2+b^2+c^2+1\hfill & \hfill b^2\hfill & \hfill c^2+1\hfill \end{array}\right]$ C_1→ C₂ + C₃.

LHS = ${\left|\begin{array}{ccc}\hfill 1+a^2-b^2\hfill & \hfill 2ab\hfill & \hfill -2b\hfill \\ \hfill 2ab\hfill & \hfill 1-a^2+b^2\hfill & \hfill 2a\hfill \\ \hfill 2b\hfill & \hfill -2a\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|}_{.}$

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

$=\left|\begin{array}{ccc}\hfill 1+a^2-b^2+2b^2\hfill & \hfill 2ab-2ab\hfill & \hfill -2b\hfill \\ \hfill 2ab-2ab\hfill & \hfill 1-a^2+b^2+2a^2\hfill & \hfill 2a\hfill \\ \hfill 2b-b+a^{2}b+b^3\hfill & \hfill -2a+a-a^3-a{b}^2\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|$

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

= (1 + a² + b²)² [(1 -a² + b²) - 2a (-a)]

= (1 + a² + b²)² (1 -a² + b² + 2a²)

= (1 + a² + b²)² (1 + a² + b²)

= (1 + a² + b²)³ = R.H.S.