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JEE Main · Mathematics · Matrices & Determinants
MCQ · Mathematics · Matrices
Q. For the matrices $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}$, if $(A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, then among the following which one is true?
A$x = 16, y = 3$
B$x = 5, y = 7$
C$x = 11, y = 2$ ✓
D$x = 18, y = 11$
✅ Correct Answer: (C) $x = 11, y = 2$
Step-by-Step Solution
1
Find the pattern for $A^n$
Let’s split $A$ into Identity matrix $I$ and another matrix $C$:
$$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} = I + C$$
Now, check $C^2$:
$$C^2 = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
Since $C^2 = O$, using Binomial expansion:
$$A^n = (I + C)^n = I + nC + 0 = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$$
2
Calculate $A^{15}$
Put $n=15$ in the formula:
$$A^{15} = \begin{bmatrix} 1+2(15) & -4(15) \\ 15 & 1-2(15) \end{bmatrix} = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix}$$
3
Find $(A^{15} + B)$
$$A^{15} + B = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix} + \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}$$
$$= \begin{bmatrix} 31-29 & -60+49 \\ 15-13 & -29+18 \end{bmatrix} = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$$
4
Solve for $x$ and $y$
The equation is $\begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.
This gives:
$$2x – 11y = 0 \implies 2x = 11y$$
Checking option (C): $x=11, y=2 \Rightarrow 2(11) = 22$ and $11(2) = 22$. ✅
Final Answer: $x=11, y=2$ → Option (C) ✓
Related Theory: Powers of Matrices & System of Equations
📌 Nilpotent Matrices and Binomial Theorem
A matrix $C$ is called nilpotent of index $k$ if $C^k = O$ but $C^{k-1} \neq O$.
In JEE problems, often a matrix $A$ can be written as $A = \lambda I + C$, where $C$ is nilpotent.
This allows us to calculate $A^n$ easily:
$$(\lambda I + C)^n = (\lambda I)^n + n(\lambda I)^{n-1}C + \frac{n(n-1)}{2}(\lambda I)^{n-2}C^2 + \dots$$
If $C^2 = O$, the series terminates after just two terms.
📌 Cayley-Hamilton Theorem
Every square matrix satisfies its own characteristic equation $|A – \lambda I| = 0$.
For $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$:
$$\begin{vmatrix} 3-\lambda & -4 \\ 1 & -1-\lambda \end{vmatrix} = 0 \implies \lambda^2 – 2\lambda + 1 = 0 \implies (\lambda-1)^2 = 0$$
By theorem, $(A-I)^2 = O$. This confirms that $C = A-I$ is nilpotent of index 2.
📌 Homogeneous System of Linear Equations
An equation of the form $MX = O$ represents a homogeneous system.
• If $|M| \neq 0$, it has only the trivial solution $(0,0)$.
• If $|M| = 0$, it has infinitely many solutions. In our case, $M = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$, and $|M| = (2)(-11) – (-11)(2) = 0$. Thus, $x$ and $y$ are related by the ratio $x/y = 11/2$.
• If $|M| \neq 0$, it has only the trivial solution $(0,0)$.
• If $|M| = 0$, it has infinitely many solutions. In our case, $M = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$, and $|M| = (2)(-11) – (-11)(2) = 0$. Thus, $x$ and $y$ are related by the ratio $x/y = 11/2$.
📌 Matrix Addition and Multiplication Properties
$A + B = B + A$ (Commutative)
$(A+B)C = AC + BC$ (Distributive)
$A^n \cdot A^m = A^{n+m}$
$|A^n| = |A|^n$
For our matrix $A$, $|A| = (3)(-1) – (-4)(1) = 1$. This implies $|A^{15}| = 1^{15} = 1$.
📌 Shortcut for $2 \times 2$ Matrix Powers
If a matrix $A$ has repeated eigenvalues $\lambda, \lambda$, then $A^n$ can be expressed as:
$$A^n = n\lambda^{n-1}A – (n-1)\lambda^n I$$
Applying this to our matrix where $\lambda = 1, n = 15$:
$$A^{15} = 15(1)^{14}A – (14)(1)^{15}I = 15A – 14I$$
$$A^{15} = 15\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} – \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = \begin{bmatrix} 45-14 & -60 \\ 15 & -15-14 \end{bmatrix} = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix}$$
This confirms our previous result.
Frequently Asked Questions
1. Why did we check $C^2$ first?
Checking if a matrix is nilpotent is the fastest way to compute high powers like $A^{15}$.
2. Can we solve this using eigenvalues?
Yes, the eigenvalue is 1 (repeated). Since it’s not diagonalizable, we use the $A^n = nA – (n-1)I$ shortcut.
3. What if $2x – 11y = 0$ had multiple options?
The problem would usually provide another constraint or ask for a ratio. Here, only one option satisfies the ratio.
4. Is $|A^{15}+B|$ equal to 0?
Yes, because the system has non-trivial solutions, the determinant of the coefficient matrix must be zero.
Related JEE Main Questions
Q. Let $A = \{2, 3, 5, 7, 9\}$. Relation $R$ on $A$ defined by $xRy \iff 2x \le 3y$. Find $l+m$ where $l$ is elements in $R$ and $m$ is elements added for symmetry.
✅ Correct Answer: 25
Q. Roots $\alpha, \beta$ of $x^2 + 2ax + (3a+10) = 0$ satisfy $\alpha < 1 < \beta$. Find set of values of $a$.
✅ Correct Answer: $(-\infty, -11/5)$
Q. Locus of centroid of $\triangle OPA$ where $P$ is on $y^2=12x$ and $\angle OPA=90°$.
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