Class 12th

The given system of equation can be written in the form of AK = B

Now,  = 3 * 1 - 2 * 2 = 3 - 4 = -1 ≠ 0.

∴ The system has unique solⁿ  and hence equation are consistent

Given, A is an invertible matrix of order 2.

∴ Option (B) is correct.

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

Explanation- power = 1W

Zener breakdown voltage = 5V

Minimum voltage = 3V

Maximum voltage = 7V

Current = power/voltage= 1/5=0.2A

The value of R for safe operation would be R = max voltage – Zener voltage/current

= 7- 5/0.2 = 2/0.2 = 10ohm

Kindly go through the solution

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

Explanation- this is AND GATE so truth table for this will be

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

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

Explanation- In elemental semiconductor, the band gap is such that the emission are in infrared region and not in visible region.

Kindly go through the solution

⇒ b = 1

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