Maths

We have, AB = BA. (given)

(E) P (n):AB' = B'A.

P (i):AB¹ = B¹A. Þ AB = BA

so, the result is true for n = 1.

Let the result be true for n = k.

P (k):AB^k = B^kA

Then,

P (k + 1) : AB^{k + 1} = A. B^k. B = B^kA.B = B^k.BA $\{?\}\; AB=BA$

= Bk + 1.A .

So, AB^{k + 1} = B^{k + 1}A.

The result also holds for n = k + 1.

Hence, AB^n = B^n A^n holds for all natural number 'n'.

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

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

∴ 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,

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

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

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

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

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.