GATE CSE 2025 | Set 2 | Question: 1

Question

​​​​If $A=\left(\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right)$, then which ONE of the following is $A^{8}$ ?

• A. $\left(\begin{array}{cc}25 & 0 \\ 0 & 25\end{array}\right)$
• B. $\left(\begin{array}{cc}125 & 0 \\ 0 & 125\end{array}\right)$
• C. $\left(\begin{array}{cc}625 & 0 \\ 0 & 625\end{array}\right)$
• D. $\left(\begin{array}{cc}3125 & 0 \\ 0 & 3125\end{array}\right)$

Comments

Another approach

Find the Eigen values of the given matrix , they are $ +\sqrt{5}$ and $-\sqrt{5}$

Property - Eigen value of $A^k$ is $\lambda^k$

Thus, the Eigen values of $A^8$ is $\sqrt{5}^8 = 5^4 = 625$

Similarly , the other eigen value of  $A^8$ is $-\sqrt{5}^8 = 5^4 = 625$

From the given options , option C is the answer because it is a diagonal matrix, and

eigen values of the diagonal matrix are on the diagonal itself.

Thus, correct option is C

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Finding A^2 = [ 5   0          =  5*I..
                         0   5]    

So, A^4 = 25*I
A^8 = 625*I

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Use Diagonalization method : 

For a matrix $A$ that is diagonalizable, we can write it as $A = PDP^{-1}$, where:

• $D$ is the diagonal matrix of eigenvalues.

• $P$ is the matrix of eigenvectors.

Then, $A^n = PD^nP^{-1}$.

 Find the Eigenvalues

To diagonalize $A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}$, we find the characteristic equation $\det(A - \lambda I) = 0$:

$$\begin{vmatrix} 1-\lambda & 2 \\ 2 & -1-\lambda \end{vmatrix} = 0$$

$$(1-\lambda)(-1-\lambda) - 4 = 0$$

$$-(1-\lambda^2) - 4 = 0 \implies \lambda^2 - 5 = 0 \implies \lambda = \pm\sqrt{5}$$

Form $D$ and calculate $D^8$

Since the eigenvalues are $\lambda_1 = \sqrt{5}$ and $\lambda_2 = -\sqrt{5}$, the diagonal matrix is:

$$D = \begin{pmatrix} \sqrt{5} & 0 \\ 0 & -\sqrt{5} \end{pmatrix}$$

Now, raise $D$ to the power of 8:

$$D^8 = \begin{pmatrix} (\sqrt{5})^8 & 0 \\ 0 & (-\sqrt{5})^8 \end{pmatrix} = \begin{pmatrix} 5^4 & 0 \\ 0 & 5^4 \end{pmatrix} = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix} = 625I$$

 Find $A^8$

Using the formula $A^8 = PD^8P^{-1}$:

$$A^8 = P(625I)P^{-1} = 625(PIP^{-1}) = 625I = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix}$$

 

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Option C is correct

Answer 1

\[A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}\]\[\det(A - \lambda I) = \begin{vmatrix} 1 - \lambda & 2 \\ 2 & -1 - \lambda \end{vmatrix}\]\[(1 - \lambda)(-1 - \lambda) - (2 \times 2) = \lambda^2 - 1 - 4 = \lambda^2 - 5\]Characteristic Equation,\[A^2-5I =0 \Rightarrow A^2= 5I\]\[ A^4 = (A^2)^2 = (5I)^2 = 25I \]\[ A^8 = (A^4)^2 = (25I)^2 = 625I \]\[ A^8 = \begin{bmatrix} 625 & 0 \\ 0 & 625 \end{bmatrix} \]

Comments

Below you can find some clarity for the above answer given :

$\lambda = \sqrt5$

$|A - \lambda I| = 0$
$|A| = |\lambda I|$
$|A|= |\sqrt5 I|$​​​​​​​

$\texttt{square}$ on both sides

$|A|^2= |\sqrt5 I|^2$​​​​​​​​​​​​​​
$|A|^2= |(\sqrt5)^2 I^2|$​​​​​​​​​​​​​​​​​​​​​
$|A|^2= |25 I|$​​​​​​​​​​​​​​​​​​​​​
$|A|^2= 25$​​​​​​​​​​​​​​​​​​​​​

similarly $|A|^n= |(\sqrt5)^n I^n|$​​​​​​​​​​​​​​​​​​​​​

now subtitute n = 8
$|A|^8= 625$​​​​​​​​​​​​​​​​​​​​​​​​​​​​

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cant do det(A-B)=det(A)-det(B)

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Would like to add the following-

Part_I: Char. Equation- λ^2-5 = 0

Then, by Cayley–Hamilton theorem: Every square matrix satisfies its own characteristic equation.

So we can write  A^2 - 5I = 0.

The rest of the deducion follows.

Answer 2

Option - c

Answer 3

OPTION C)

Answer 4

For the given matrix A if we try to find the Determinant it would be

Det(A) = 1 x (-1) - 2 x 2 = -1 -4 = -5.

Det(A) = -5

So they asked to find the A^8 .

WKT     Det(A^n) = ( Det(A) ) ^ n

Therefore for the matrix A^8 the determinant would be (-5) ^ 8 which is 390625.

Now checking the options if we try to find the determinant of Option C it would be same as 390625.

Hence Option C is right.

Answer 5

POV: As we can see in the options, all are diagonal matrices.  
 

So the first thing that comes to mind is that the determinant is very easy to calculate, because for a diagonal matrix.

The best part is that all the options have different determinant values, so there is no confusion here.  
Then we start by calculating the determinant of the given matrix $A$:  

$$\det(A) = 1 \times (-1) - 2 \times 2 = -1 - 4 = -5$$  

Next, we use the property that for any square matrix $A$:  
$$\det(A^n) = (\det(A))^n$$  

So for $A^8$, we get:  
$$\det(A^8) = (-5)^8 = 390625$$  

Finally, we check the options. Option C has determinant $390625$, which matches our calculation.  

Therefore, the correct answer is Option C.
 

Answer 6

Good Luck