Ncert Solutions Maths class 12th

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

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

So; $A^2=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{ll}3\times 3+1\times (-1)\hfill & 3\times 1+1\times 2\hfill \\ -1\times 3+2\times (-1)\hfill & -1\times 1+2\times 2\hfill \end{array}\right]$

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

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

$=\left[\begin{array}{cc}\hfill 8\hfill & \hfill 5\hfill \\ \hfill -5\hfill & \hfill 3\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 15\hfill & \hfill 5\hfill \\ \hfill -5\hfill & \hfill 10\hfill \end{array}\right]+\left[\begin{array}{ll}7\hfill & 0\hfill \\ 0\hfill & 7\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill 8-15+7\hfill & \hfill 5-5+0\hfill \\ \hfill -5+5+0\hfill & \hfill 3-10+7\hfill \end{array}\right]=\left[\begin{array}{ll}0\hfill & 0\hfill \\ 6\hfill & 6\hfill \end{array}\right]=0.$

Hence Showed.

Kindly go through the solution

Given, A¹${}^{}$$\left[\bcancel{\begin{array}{ccc}\hfill 2\hfill & \hfill 2y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill -y\hfill & \hfill z\hfill \end{array}}\right]$Then, $\left[\bcancel{\begin{array}{ccc}\hfill 0\hfill & \hfill x\hfill & \hfill z\hfill \\ \hfill 2y\hfill & \hfill y\hfill & \hfill -y\hfill \\ \hfill z\hfill & \hfill -z\hfill & \hfill z\hfill \end{array}}\right]$

Since, ${\prime }^{}=$ we can write,

$\Rightarrow \left[\bcancel{\begin{array}{ccc}\hfill 2\hfill & \hfill 2y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill -y\hfill & \hfill z\hfill \end{array}}\right]\left[\bcancel{\begin{array}{ccc}\hfill 0\hfill & \hfill x\hfill & \hfill z\hfill \\ \hfill 2y\hfill & \hfill y\hfill & \hfill -y\hfill \\ \hfill z\hfill & \hfill -z\hfill & \hfill z\hfill \end{array}}\right]$

$\Rightarrow \left[\begin{array}{ccc}\hfill 0\hfill & \hfill x\hfill & \hfill x\hfill \\ \hfill 2y\hfill & \hfill y\hfill & \hfill -y\hfill \\ \hfill z\hfill & \hfill -z\hfill & \hfill z\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill -z\hfill \\ \hfill x\hfill & \hfill -y\hfill & \hfill z\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}\hfill 0+x^2+x^2\hfill & \hfill 0+xy-xy\hfill & \hfill 0-xz+xz\hfill \\ \hfill 0+xy-xy\hfill & \hfill 4y^2+y^2+y^2\hfill & \hfill 2yz-yz-yz\hfill \\ \hfill 0-2x+2x\hfill & \hfill 2yz-yz-yz\hfill & \hfill z^2+z^2+z^2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}\hfill 2x^2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 6y^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3z^2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

We have,

(E) (B'AB)' = [B' (AB]'

= (AB)' (B')'

= B'A'B.

When A is symmetric, A' = A

(B'AB)' = B'AB

ie, B'AB is symmetric.

And when A is skew-symmetric, A¹ = -A

(B'AB)' = -B'AB.

ie, B'AB is skew-symmetric.

Given, A and B are symmetric matrices.

(E) Then A' = A and B' = B.

Now, (AB - BA)' = (AB)' - (BA)'

= B'A' - A'B'

= BA - AB

= - (AB - BA).

Hence, AB BA is skew-symmetric matrix

We have,

$=\left[\begin{array}{cc}\hfill 3(1+2k)+(-4k)\times 1\hfill & \hfill -4(1+2k)+(-4k)(-1)\hfill \\ \hfill 3k+1\times (1-2k)\hfill & \hfill -4\times k+(1-2k)(-1)\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill 3+6k-4k\hfill & \hfill -4-8k+4k\hfill \\ \hfill 3k+1-2k\hfill & \hfill -4k+2k-1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3+2k\hfill & \hfill -4-4k\hfill \\ \hfill 1+k\hfill & \hfill -1-2k\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill 1+2+2k\hfill & \hfill -4(1+k)\hfill \\ \hfill 1+k\hfill & \hfill 1-2-2x\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill 1+2\left(1+k\right)\hfill & \hfill -4(1+k)\hfill \\ \hfill 1+k\hfill & \hfill 1-2(1+k)\hfill \end{array}\right]$

The results also holds for n = k + 1. Hence, An $=\left[\begin{array}{cc}\hfill 1+2n\hfill & \hfill -4n\hfill \\ \hfill n\hfill & \hfill 1-2n\hfill \end{array}\right]$

Holds for all natural number n.

We have,

(E) P (n) : If A = $[$
$\begin{array}{l}\mathrm{}\\ \\ \end{array}<math><mfenced\; open="["\; close="]"\; separators="|"><mrow><mrow><mtable><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr></mtable></mrow></mrow></mfenced></math>$ ]. then Aⁿ = $\left[\begin{array}{l}{3}^{n-1}{3}^{n-1}{3}^{n-1}\\ 3^{n-1}{3}^{n-1}{3}^{n-1}\\ 3^{n-1}{3}^{n-1}{3}^{n-1}\end{array}\right]$ n ? N.

P (1) : A¹= $\left[\begin{array}{l}{3}^{1-1}{3}^{1-1}{3}^{1-1}\\ 3^{1-1}{3}^{1-1}{3}^{1-1}\\ 3^{1-1}{3}^{1-1}{3}^{1-1}\end{array}\right]$ = $\left[\begin{array}{l}{3}^0{3}^0{3}^0\\ 3^0{3}^0{3}^0\\ 3^0{3}^0{3}^0\end{array}\right]$ =$[$
$\begin{array}{l}\mathrm{}\\ \\ \end{array}<math><mfenced\; open="["\; close="]"\; separators="|"><mrow><mrow><mtable><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr></mtable></mrow></mrow></mfenced></math>$ ]

So, the result holds true for n = 1.

Let the result be for n = k. So,

P (k): A^u= $\left[\begin{array}{l}{3}^{k-1}{3}^{k-1}{3}^{k-1}\\ 3^{k-1}{3}^{k-1}{3}^{k-1}\\ 3^{k-1}{3}^{k-1}{3}^{k-1}\end{array}\right]$

Then P (k+1): A^{k + 1} = A^k. A = $\left[\begin{array}{l}{3}^{k-1}{3}^{k-1}{3}^{k-1}\\ 3^{k-1}{3}^{k-1}{3}^{k-1}\\ 3^{k-1}{3}^{k-1}{3}^{k-1}\end{array}\right]$ $[$
$\begin{array}{l}\mathrm{}\\ \\ \end{array}<math><mfenced\; open="["\; close="]"\; separators="|"><mrow><mrow><mtable><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd><mtd><mrow><mn>1</mn></mrow></mtd></mtr></mtable></mrow></mrow></mfenced></math>$ ]

= $\left[\begin{array}{ccc}\hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill & \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill & \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill \\ \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill & \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill & \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill \\ \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill & \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill & \hfill 3^{k-1}+3^{k-1}+3^{k-1}\hfill \end{array}\right]$

= A^{k + 1} = $\left[\begin{array}{l}{3}^{\left(k+1\right)-1}{3}^{\left(k+1\right)-1}{3}^{\left(k+1\right)-1}\\ 3^{\left(k+1\right)-1}{3}^{\left(k+1\right)-1}{3}^{\left(k+1\right)-1}\\ 3^{\left(k+1\right)-1}{3}^{\left(k+1\right)-1}{3}^{\left(k+1\right)-1}\end{array}\right]$

The result holds for n = k + 1. Hence,

Aⁿ = $\left[\begin{array}{l}{3}^{n-1}{3}^{n-1}{3}^{n-1}\\ 3^{n-1}{3}^{n-1}{3}^{n-1}\\ 3^{n-1}{3}^{n-1}{3}^{n-1}\end{array}\right]$ holds fo