Matrices and Determinants — JEE Main practice

1. Find the characteristic equation of $A$ using its trace and determinant. For $A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$: trace $=3-1=2$, and $\det(A)=3(-1)-(-4)(1)=1$.

$A^2-2A+I=0$

2. This means $(A-I)^2=0$, i.e. $A-I$ is nilpotent. This special structure gives a simple closed formula for every power of $A$, provable by induction from $A^2=2A-I$.

$A^n=nA-(n-1)I$

3. Apply this formula for $n=15$.

$A^{15}=15A-14I$

4. Add matrix $B$ to this result to get the combined matrix.

$A^{15}+B=\begin{pmatrix}2&-11\\1&-11\end{pmatrix}$

5. Set up the equation $(A^{15}+B)(x,y)^T=(0,0)^T$ using this result. Using the first row:

$2x-11y=0\Rightarrow2x=11y$

6. Check which option satisfies this relation: for $x=11, y=2$: $2(11)=22$ and $11(2)=22$. ✓ These match.

Hence, the answer is C ($x=11, y=2$).

1. For the system to have no solution, the coefficient determinant must be zero (a necessary condition consistent with inconsistency in this system).

$\Delta=\begin{vmatrix}3&1&4\\2&\alpha&-1\\1&2&1\end{vmatrix}=0$

2. Solving this determinant equation for $\alpha$ gives:

$\alpha=19$

Hence, the answer is C (19).

1. Compute $\det(A)$ for the given $2\times2$ matrix.

$|A|=2(5)-3(3)=1$

2. Factor the expression by pulling out the common power $A^{2023}$.

$A^{2025}-3A^{2024}+A^{2023}=A^{2023}(A^2-3A+I)$

3. Use the determinant product property $\det(XY)=\det(X)\det(Y)$, along with $\det(A^{2023})=\det(A)^{2023}$.

$|A^{2023}(A^2-3A+I)|=|A|^{2023}\cdot|A^2-3A+I|$

4. Since $|A|=1$, this simplifies to just the determinant of the second factor.

$=|A^2-3A+I|$

5. Directly computing $A^2$, then $A^2-3A+I$, and taking its determinant gives:

$|A^2-3A+I|=16$

Hence, the answer is D (16).