JAMIA 2026 — Mathematics PYQ

Step 1: Calculate $A^2$

Let's multiply matrix $A$ by itself:

$$A^2=A\cdot A=\left[\begin{array}{cc}0& amp;1\\ 1& amp;0\end{array}\right]\left[\begin{array}{cc}0& amp;1\\ 1& amp;0\end{array}\right]$$

Using standard matrix multiplication rules:

$$A^2=\left[\begin{array}{cc}(0\times 0+1\times 1)& amp;(0\times 1+1\times 0)\\ (1\times 0+0\times 1)& amp;(1\times 1+0\times 0)\end{array}\right]$$

$$A^2=\left[\begin{array}{cc}1& amp;0\\ 0& amp;1\end{array}\right]=I$$

Where $I$ is the Identity Matrix.

Step 2: Identify the Cycle Pattern

Since $A^2=I$, let's observe what happens with higher powers:

• $A^3=A^2\cdot A=I\cdot A=A$

• $A^4=A^2\cdot A^2=I\cdot I=I$

• $A^5=A^4\cdot A=I\cdot A=A$

From this, we can establish a clear general rule for any integer power $k$:

1. If the power is even, the result is the Identity matrix:

   $$A^{2k}=I$$

2. If the power is odd, the result is the original matrix $A$:

   $$A^{2k+1}=A$$

Step 3: Evaluate $A^{37}$

The given exponent is $37$, which is an odd number. We can break down the exponent mathematically as follows:

$$A^{37}=A^{36+1}$$

$$A^{37}=A^{36}\cdot A^1$$

$$A^{37}=(A^2{)}^{18}\cdot A$$

Since $A^2=I$:

$$A^{37}=(I{)}^{18}\cdot A$$

$$A^{37}=I\cdot A$$

$$A^{37}=A$$

Substituting the value of matrix $A$ back into the result:

$$A^{37}=\left[\begin{array}{cc}0& amp;1\\ 1& amp;0\end{array}\right]$$

Conclusion

The matrix value for $A^{37}$ is equal to $\left[\begin{array}{cc}0& amp;1\\ 1& amp;0\end{array}\right]$.

Therefore, the correct option is (c).