Ncert Solutions Maths class 12th

Kindly go through the solution

Kindly go through the solution

(i) Given, A = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill & \hfill 9\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ and B = $\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Then, A' = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 1\hfill \end{array}\right]$ and B = $\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

A + B = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill & \hfill 9\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ + $\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill -1-4\hfill & \hfill 2+1\hfill & \hfill 3-5\hfill \\ \hfill 5+1\hfill & \hfill 7+2\hfill & \hfill 9+0\hfill \\ \hfill -2+1\hfill & \hfill 1+3\hfill & \hfill 1+1\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill -5\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 6\hfill & \hfill 9\hfill & \hfill 9\hfill \\ \hfill -1\hfill & \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$

LHS. = (A + B)' = $\left[\begin{array}{ccc}\hfill -5\hfill & \hfill 6\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 9\hfill & \hfill 2\hfill \end{array}\right]$

RHS = A'+ B' = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -1-4\hfill & \hfill 5+1\hfill & \hfill -2+1\hfill \\ \hfill 2+1\hfill & \hfill 7+2\hfill & \hfill 1+3\hfill \\ \hfill 3-5\hfill & \hfill 9+0\hfill & \hfill 1+1\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill -5\hfill & \hfill 6\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 9\hfill & \hfill 2\hfill \end{array}\right]$ = L H S

$\therefore$ (A + B)' = A'+ B'

(ii) A – B = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill & \hfill 9\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -1-\left(-4\right)\hfill & \hfill 2-1\hfill & \hfill 3-(5)\hfill \\ \hfill 5-1\hfill & \hfill 7-2\hfill & \hfill 9-0\hfill \\ \hfill -2-1\hfill & \hfill 1-3\hfill & \hfill 1-1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 8\hfill \\ \hfill 4\hfill & \hfill 5\hfill & \hfill 9\hfill \\ \hfill -3\hfill & \hfill -2\hfill & \hfill 0\hfill \end{array}\right]$

L.H.S. = (A − B)' = $\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 4\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 8\hfill & \hfill 9\hfill & \hfill 0\hfill \end{array}\right]$

R.H.S. = A' − B' = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 1\hfill \end{array}\right]- \left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=$ $\left[\begin{array}{ccc}\hfill -1-(-4)\hfill & \hfill 5-1\hfill & \hfill -2-1\hfill \\ \hfill 2-1\hfill & \hfill 7-2\hfill & \hfill 1-3\hfill \\ \hfill 3-(-5)\hfill & \hfill 9-0\hfill & \hfill 1-1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 4\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 8\hfill & \hfill 9\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore$ (A − B)' = A' − B'

The order of 7x - 5z will be same as order of x and z

as has order 2 * n and z has order 2 * p.and given n = p.

7x - 5z has order 2 * n

∴option b is correct.

For, PY + WY to be defined

(E) P${}_{p*k}\cdot Y{}_{3*x}$ should be seen that number of columns of p

Should, be equal to no of nous of yi.e, [x = 3]

n * 3 and (PY) _{(p * x})

similarity, in W ^n*3Y ^3*xwe  see thatno of columns of

W = no of rows in Y and (WY) n * x.

Again, PY + WY is defined if order of (PY) _{p * k} and (WY)_{n * u} are the sameie, P = n

so, option a is correct .

Number of books matrix A =[10 dozen 8 dozen 10 dozen]

= (9600 + 5760 + 4800)

= ' 20160

Order of matrices,  $\begin{array}{l}x\to 2*n\\ y\to 3*x\\ z\to 2*p\\ w\to n*3\\ p\to p*x.\end{array}$

Kindly go through the solution

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

A² = A. A = $\left[\begin{array}{ll}3\hfill & -2\hfill \\ 4\hfill & -2\hfill \end{array}\right]\left[\begin{array}{ll}3\hfill & -2\hfill \\ 4\hfill & -2\hfill \end{array}\right]$

$=\left[\begin{array}{ll}3*3+(-2)*4\hfill & 3*(-2)+(-2)*(-2)\hfill \\ 4*3+(-2)*4\hfill & 4*(-2)+(-2)*(-2)\hfill \end{array}\right]$

$=\left[\begin{array}{ll}9-8\hfill & -6+4\hfill \\ 12-8\hfill & -8+4\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & -2\hfill \\ 4\hfill & -4\hfill \end{array}\right].$

So,A²= xA2I.

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 4\hfill & \hfill -4\hfill \end{array}\right]=u\left[\begin{array}{cc}\hfill 3\hfill & \hfill -2\hfill \\ \hfill 4\hfill & \hfill -2\hfill \end{array}\right]-2\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right].$

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 4\hfill & \hfill -4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3x\hfill & \hfill -2x\hfill \\ \hfill 4x\hfill & \hfill -2x\hfill \end{array}\right]-\left[\begin{array}{ll}2\hfill & 0\hfill \\ 0\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3x-2\hfill & \hfill -2x\hfill \\ \hfill 4x\hfill & \hfill -2x-2\hfill \end{array}\right].$

Given A = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$

A² = AA = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 1*1+0*0+2*2\hfill & \hfill 1*0+2*0+2*0\hfill & \hfill 1*2+0*1+2*3\hfill \\ \hfill 0*1+2*0+1*2\hfill & \hfill a*0+2*2+1*0\hfill & \hfill 0*2+2*1+1*3\hfill \\ \hfill 2*1+0*0+3*2\hfill & \hfill 2*0+0*2+3*0\hfill & \hfill 2*2+0*1+3*3\hfill \end{array}\right]$

$=\left[\begin{array}{l}1+402+6\\ 242+3\\ 2+604+9\end{array}\right]=\left[\begin{array}{ccc}\hfill 5\hfill & \hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 8\hfill & \hfill 0\hfill & \hfill 13\hfill \end{array}\right]$

A³ = A²A = $\left[\begin{array}{ccc}\hfill 5\hfill & \hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 8\hfill & \hfill 0\hfill & \hfill 13\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 5*1+0*0+8*2\hfill & \hfill 5*0+0*2+8*0\hfill & \hfill 5*2+0*1+8*3\hfill \\ \hfill 2*1+4*0+5*2\hfill & \hfill 2*0+4*2+5*0\hfill & \hfill 2*2+4*1+5*3\hfill \\ \hfill 8*1+0*0+13*2\hfill & \hfill 8*0+0*2+0*0\hfill & \hfill 8*2+0*1+13*3\hfill \end{array}\right]$

$\left[\begin{array}{ccc}\hfill 5+16\hfill & \hfill 0\hfill & \hfill 10+24\hfill \\ \hfill 2+10\hfill & \hfill 8\hfill & \hfill 4+4+15\hfill \\ \hfill 8+26\hfill & \hfill 0\hfill & \hfill 16+39\hfill \end{array}\right]$ $=\left[\begin{array}{lll}21\hfill & 0\hfill & 34\hfill \\ 12\hfill & 8\hfill & 23\hfill \\ 34\hfill & 0\hfill & 55\hfill \end{array}\right]$

So, A³-6A² + 7A + 2I

$=\left[\begin{array}{lll}21\hfill & 0\hfill & 34\hfill \\ 12\hfill & 8\hfill & 23\hfill \\ 34\hfill & 0\hfill & 55\hfill \end{array}\right]-6\left[\begin{array}{lll}5\hfill & 0\hfill & 8\hfill \\ 2\hfill & 4\hfill & 5\hfill \\ 8\hfill & 0\hfill & 13\hfill \end{array}\right]+7\left[\begin{array}{lll}1\hfill & 0\hfill & 2\hfill \\ 0\hfill & 2\hfill & 1\hfill \\ 2\hfill & 0\hfill & 3\hfill \end{array}\right]+2\left[\begin{array}{lll}1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right]$

$=\left[\begin{array}{lll}21\hfill & 0\hfill & 34\hfill \\ 12\hfill & 8\hfill & 23\hfill \\ 34\hfill & 0\hfill & 55\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 30\hfill & \hfill 0\hfill & \hfill 48\hfill \\ \hfill 12\hfill & \hfill 24\hfill & \hfill 30\hfill \\ \hfill 48\hfill & \hfill 0\hfill & \hfill 78\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 7\hfill & \hfill 0\hfill & \hfill 14\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 7\hfill \\ \hfill 14\hfill & \hfill 0\hfill & \hfill 21\hfill \end{array}\right]+\left[\begin{array}{lll}2\hfill & 0\hfill & 0\hfill \\ 0\hfill & 2\hfill & 0\hfill \\ 0\hfill & 0\hfill & 2\hfill \end{array}\right]$

$=\left[\begin{array}{lll}21-30+7+2\hfill & 0-0+0+0\hfill & 34-48+14+0\hfill \\ 12-12+0+0\hfill & 8-24+14+2\hfill & 23-30+7+0\hfill \\ 34-48+14+0\hfill & 0-0+0+0\hfill & 55-78+21+2\hfill \end{array}\right]$

$=\left[\begin{array}{lll}0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill \end{array}\right]=0$ (2000 matrix )

Hence proud