Ncert Solutions Maths class 12th

(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

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

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

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

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

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

= $\frac{47}{2}$ =23.5 sq. units

(iii) Area of triangle is given by,

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

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

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

$=\frac{1}{2}|-30|=\frac{30}{2}$  = 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.

LHS = $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill x^2\hfill \\ \hfill x^2\hfill & \hfill 1\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill 1+x^2+x\hfill & \hfill x+1+x^2\hfill & \hfill x^2+x+1\hfill \\ \hfill x^2\hfill & \hfill 1\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \end{array}\right|{}_{\mathrm{}}$R_1→ R₁ + R₂ + R₃

= (1 + x² + x) (1 -x)² [ (1 + x)× 1 - (-x) x].

= (1 + x² + x) (1 -x)² (1 + x + x²).

= { (1 + x² + x) (1 -x)}²

= {1 -x + x²-x³ + x-x²}²

= (1 -x³)² = R.H.S.

(i) LHS = $\left|\begin{array}{ccc}\hfill a-b-c\hfill & \hfill 2a\hfill & \hfill 2a\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill a-b-c+2b+2c\hfill & \hfill 2a+b-c-a+2c\hfill & \hfill 2a+2b+c-ab\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$R_1→ R₁ + R₂ + R₃

$=\left|\begin{array}{ccc}\hfill a+b+c\hfill & \hfill a+b+c\hfill & \hfill a+b+c\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$

= (a + b + c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$ Taking (a + b + c) common from R₁

= (a + b+ c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1-1\hfill & \hfill 1-1\hfill \\ \hfill 2b\hfill & \hfill b-c-a-2b\hfill & \hfill 2b-2b\hfill \\ \hfill 2c\hfill & \hfill 2c-2c\hfill & \hfill c-a-b-2c\hfill \end{array}\right|\begin{array}{cc}\hfill {}_{}\to {}_2-{}_1\hfill & \hfill {}_{}\to {}_3-{}_1\hfill \end{array}$

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

= (a + b + c) × 1. $\left|\begin{array}{cc}\hfill -(a+b+c)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -(a+b+c)\hfill \end{array}\right|$ Expand along R₁

= (a + b + c){(a + b + c)²- 0}

= (a + b +c)³ = R.H.S

(ii) LHS = ${\left|\begin{array}{ccc}\hfill x+y+2z\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|}_{\cdot }$

$=|\begin{array}{ccc}\hfill x+y+2z+x+y\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill z+y+z+2x+y\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill z+x+z+x+2y\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}$
C_1→ C₁ + C₂ + C₃.

$={\left|\begin{array}{ccc}\hfill 2(x+y+z)\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill 2(x+y+z)\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill 2(x+y+z)\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|}_{.}$

= 2 (k + y + z) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|$ Taking 2

(i) LHS = $\left|\begin{array}{ccc}\hfill x+4\hfill & \hfill 2x\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill x+4+2x+2x\hfill & \hfill 2x+x+4+2x\hfill & \hfill 2x+2x+x+4\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|$ R_1→R₁ + R₂ + R₃

$=\left|\begin{array}{ccc}\hfill 5x+4\hfill & \hfill 5x+4\hfill & \hfill 5x+4\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|$

Taking (5x + 4) common from R₁.

$=(5x+4)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|.$

= (5x + 4)[0 - (4 -x)(x- 4)]

= (5x + 4)(4 -x)(4 -x)

= (5x + 4)(4 -x)² = R.H.S.

= $\left|\begin{array}{ccc}\hfill y+k+y+y\hfill & \hfill y+y+k+y\hfill & \hfill y+y+y+k\hfill \\ \hfill y\hfill & \hfill y+k\hfill & \hfill y\hfill \\ \hfill y\hfill & \hfill y\hfill & \hfill y+k\hfill \end{array}\right|$ R_1→R₁ + R₂+ R₃

$=\left|\begin{array}{ccc}\hfill 3y+k\hfill & \hfill 3y+k\hfill & \hfill 3y+k\hfill \\ \hfill y\hfill & \hfill y+k\hfill & \hfill y\hfill \\ \hfill y\hfill & \hfill y\hfill & \hfill y+k\hfill \end{array}\right|$

= (3y + k) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y+k\hfill & \hfill y\hfill \\ \hfill y\hfill & \hfill y\hfill & \hfill y+k\hfill \end{array}\right|$