Ncert Solutions Maths class 12th

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${}_{p\times k}\cdot Y{}_{3\times x}$ 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

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

= (9600 + 5760 + 4800)

= ' 20160

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

let x be the investment in the first bond out of total `30,000.

(M) then, investment use for the second bond = ` 30,000 - x.

Hence, investment matrix A = ${[x30,000-x]}_{1\times 2}$

As there is 5% and 7% interest paid the first and second , (per year)

$\begin{array}{l}={\left[x\times \frac{5}{100}+(30000-x)\times \frac{7}{100}\right]}_{1\times 1}\\ =\frac{5x+2,10,000-7x}{100}=\frac{2,10,000-2x}{100.}\end{array}$

Manual total interest = ` $\frac{210,000-2x}{100.}$

(a) Given, manual total interest = ` 1800.

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

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

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

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

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

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

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

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

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

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

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

A² = AA² = $\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]$

= $\left[\begin{array}{l}2\times 2+0\times 2+1\times 12\times 0+0\times 1+1\times (1)2\times 1+0\times 3+1\times 0\\ 2\times 2+1\times 2+3\times 12\times 0+1\times 1+3\times (-1)2\times 1+1\times 3+3\times 0\\ 1\times 2+(-1)\times 2+0\times 11\times 0+(-1)\times 1+0\times (-1)1\times 1+(-1)\times 3+0\times 0\end{array}\right]$

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

Kindly go through the solution

Given,  $f(x)=\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

So, F (y) $=\left[\begin{array}{ccc}\hfill cosy\hfill & \hfill -siny\hfill & \hfill 0\hfill \\ \hfill siny\hfill & \hfill cosy\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\therefore f(x)f(y)=\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$