Class 12th

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

we have, A² = A?A = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill 3*3+1*(-1)\hfill & \hfill 3*1+1*2\hfill \\ \hfill (-1)*3+(-1)*2\hfill & \hfill (-1)*1+2*2\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill 9-1\hfill & \hfill 3+2\hfill \\ \hfill -3-2\hfill & \hfill -1+4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 8\hfill & \hfill 5\hfill \\ \hfill -5\hfill & \hfill 3\hfill \end{array}\right]$

Hence, A² – 5A+71= $\left[\begin{array}{cc}\hfill 8\hfill & \hfill 5\hfill \\ \hfill -5\hfill & \hfill 3\hfill \end{array}\right]-5\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]+7\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill 8-5*3+7*1\hfill & \hfill 5-5*1+7*0\hfill \\ \hfill -5-5*(1)+7*0\hfill & \hfill 3-5*2+7*1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=0$

Now, A² – 5A+7I=0.

A?A – 5A= –7I.

A A(A^–1)–5(AA^–1)= –7I(A^–1)          (Multiplying by A^-1on both sides)

AI – 5I= –7A^–1

7A^–1= –(AI – 5I)

A^–1= $\frac{1}{7}[-A+5I]$

= $\frac{1}{7}\left\{(-1)\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]+5\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\right\}$

= $\frac{1}{7}\left\{\left[\begin{array}{cc}\hfill (-1)*3+5*1\hfill & \hfill -1+5*0\hfill \\ \hfill (-1)*(-1)+5*0\hfill & \hfill (-1)*2+5*1\hfill \end{array}\right]\right\}$

A^–1= $\frac{1}{7}\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right]$

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- 

OR GATE gives the desired output.

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This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- as base current I_B = $\frac{V_{BB}-V_{BE}}{R_i}$

As R_i is increased I_B is decreased.

And collector current is I_c = I_B as I_B decreased I_c also decreased and the rading of voltmeter and ammeter also decreased.

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This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- E=hc/ $\lambda$ = $\frac{6.6*{10}^{-34}*3*{10}^8}{6000*{10}^{-10}*1.6*{10}^{-19}}$ = 2.06eV

The incident radiation which is detected by the photodiode having energy should be greater

Than the band-gap. So, it is only valid for diode D2.Then, diode D2 will detect this radiation.

Let A = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 4\hfill \end{array}\right]$ Then |A| = $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 4\hfill \end{array}\right|$

$\Rightarrow \left|\right|=1*\left|\begin{array}{cc}\hfill 2\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 4\hfill \end{array}\right|-(-1)\left|\begin{array}{cc}\hfill 0\hfill & \hfill -3\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right|+2\left|\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill -2\hfill \end{array}\right|$

= (8 - 6) + (0 - (- 9) + 2 (0 - 6)

= 2 + 9 - 12 = - 1 ¹ 0

So, A^-1 exist

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This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- In CE transistor amplifier, the power gain is very high.

In this circuit, the extra power required for amplified output is obtained from DC source.

Thus, the circuit used does not violet the law of conservation.