GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 24

Answer 1

Clearly $ w = 3u + 2v$ …...eq1

As we now if x is an eigenVector of A then

$Ax= \lambda x$ (Now Multipy  A on both side )

$A^{2}x =  \lambda Ax $ ( But $Ax= \lambda x $ )

So $A^{2}x =  \lambda^{2} x$…...and Similarly $A^{n}x =  \lambda^{n} x$

Now  $A^{5}w =  \lambda^{5} w$

$ A^{5}(u) =  2^{5} u$(as $\lambda$ for u = 2) …..eq2

$ A^{5}(v) =  (-1)^{5}v $(as $\lambda$ for v = -1)….eq3

Adding 3 * eq2 and 2 * eq3

$ A^{5}(3u +2v) =  3 ( 2^{5}) u +2 (-1)^{5}v $

As 3u +2v = w by eq1

$ A^{5}w =  96 u  –  2v $

And $ 96 u  –  2v $ comes out to be $\begin{bmatrix} 92 \\ -2 \\-96 \end{bmatrix}$
Option C is correct

Comments

Solution is correct but you can not say Aw⁵ = (lemda)⁵w bcz in question it is not metion that w is eigen vector of matrix A.

---

3u+2V=w os only required

Answer 2

From a quick glance, we can see that
$$\begin{array}\ 3u + 2v = w \end{array}$$

Now,
$$\begin{array}\ Ax = kx \\ A.Ax = k.Ax \\ A^2x = k.kx \\ A^2x = k^2x \end{array}$$

It means that if k is the eigen value of A, then k^m will be the eigen value of A^m 

$$\begin{array}\ Now,\ its\ given\ that.\ Au=2u\ and\ Av=3v \\ So,\ A^5u=2^5u = 32u\\ A^5v=(-1)^5v = -1v \end{array}$$

Now, according to the question, we have to find A⁵w

$$\begin{array}\ A^5w \\ A^5(3u + 2v) \\ 3.(A^5u) + 2.(A^5v) \\ Putting\ the\ values\ - \\ 3.(32u) + 2(-1v) \\ 96u - 2v \end{array}$$

Putting the values of u and v - 

$$96\begin{bmatrix}\ 1 \\ 0 \\ -1 \end{bmatrix} - 2 \begin{bmatrix}\ 2 \\ 1 \\ 0 \end{bmatrix}$$

$$\begin{bmatrix}\ 96 \\ 0 \\ -96 \end{bmatrix} - \begin{bmatrix}\ 4 \\ 2 \\ 0 \end{bmatrix}$$

$$\begin{bmatrix}\ 92 \\ -2 \\ -96 \end{bmatrix}$$

Comments

your answer is better than most voted, there he wrongly assumed w to be eigen vector which was not mentioned in the question at all.