Ncert Solutions Maths class 12th

= 1 [1× (−9) – 4× (−3)]–1 [2× (–9)–5× (– 3)]– 2 [2×4 −5×1]

= 1 [− 9 + 12] −1 [− 18+15] − 2 [8 − 5]

= 3 + 3 −6

= 0

$=3\left[0(-5)(-1)\right]+1(0-3(-1)-2[0\times (-5)-0\times 3]$

=3 [1×1− (−2)×3]+4 [1×1− (−2)×2]+5 [3×1−2×1]

= 3 [1+6]+4 [1+4]+5 (3−2)

= 3×7+4×5+5×1

= 21+20+5

= 46

= 0 –1 [ − 1×0 – (−2)× (−3)] + 2 [−1×3 – (−2)×0]

= (− 1)× ( − 6)+2 (− 3)

= 6 – 6

= 0

= 2 [2×0− ( − 5)× (−1)]+3 [ (−1)× (− 1) – (− 2)×2]

= 2 (−5)+3 (1+4)

= 2 (−5) + 3× (5)

= −10+15

= 5

=3 [36 – 6×0]

= 108

RHS = 27. |A| = 27 4 = 108

∴ LHS = RHS

Given, A= $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$

So, 2A= 2 $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 4\hfill \end{array}\right]$ .

L.H.S. = $\left|2\right|$ = $\left|\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 4\hfill \end{array}\right|$ = 2×4 – 4×8 = 8−32 = −24

R.H.S. = 4 $\left|\right|$ = 4 $\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right|$ = 4 (1×2 – 2×4)

= 4 [2 − 8]

= 4 [−6] = 24.

LHS = RHS

(i) $\left|\begin{array}{cc}\hfill cos\theta \hfill & \hfill -sin\theta \hfill \\ \hfill sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right|$

$=cos\theta \times cos\theta -(-sin\theta )\times sin\theta$

= cos²θ + sin²θ

=1

(ii) $\left|\begin{array}{cc}\hfill x^2-x+1\hfill & \hfill x-1\hfill \\ \hfill x+1\hfill & \hfill x+1\hfill \end{array}\right|$

$=(x^2-x+1)(x+1)-(x-1)(x+1)$

= $x^3$ + $x^2$ − $x^2$ − $x$ + $x$ +1 – $x^2$  + 1

= $x^3$ - $x^2$+ 2

$\left|\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill -5\hfill & \hfill -1\hfill \end{array}\right|=$ 2× (−1) – 4× (−5) = −2+20 = 18

Given, A²= A.

(E) we need to calculate,

(I + A)³- 7A = I³ + A³ + 3IA (I + A) - 7A { (x + y)³ = x³ + y³ + 3xy (a + y)}

Given, A is both symmetric and skew-symmetric.

(E) Then, A' = A ____ (1) and A' = -A ____ (2)

So using (2), A' = -A.

A = -A {eqⁿ (I)}

A + A = 0

2A = 0

A = 0.

A is a zero matrix

So, option B is correct.

Kindly go through the solution

We have, AB = BA. (given)

(E) P (n):AB’ = B’A.

P (i):AB¹ = B¹A. Þ AB = BA

so, the result is true for n = 1.

Let the result be true for n = k.

P (k):AB^k = B^kA

Then,

P (k + 1) : AB^{k + 1} = A. B^k. B = B^kA.B = B^k.BA $\{?\}\; AB=BA$

= Bk + 1.A .

So, AB^{k + 1} = B^{k + 1}A.

The result also holds for n = k + 1.

Hence, AB^n = B^n A^n ho