If the matrix A = [[0, 2], [K, -1]] satisfies A(A³ + 3I) = 2I then the value of K is :

Q: Wait, why can't I just calculate $A^2, A^3, A^4$ by multiplying the matrix directly? It seems safer.

That is a classic trap! While you *could* do it, it is computationally heavy and prone to arithmetic errors. JEE Advanced is not testing your ability to do tedious multiplication; it is testing your ability to recognize the underlying structure. Using the Cayley-Hamilton theorem is the elegant, professional way to handle matrix polynomials.

Q: How do I know I can just equate the coefficients of $A$ and $I$ to zero at the end?

Excellent question. We can equate coefficients because $I$ and $A$ are linearly independent (unless $A$ is a scalar multiple of $I$). If you have an equation of the form $c_1 A + c_2 I = 0$, then for this to hold for a general matrix $A$, the coefficients $c_1$ and $c_2$ must both be zero. It is the same logic as saying $ax + b = 0$ implies $a=0$ and $b=0$ if $x$ is a variable.

Q: What if the characteristic equation was cubic instead of quadratic? Would the method change?

The core philosophy remains identical! If the characteristic equation were cubic, say $A^3 + aA^2 + bA + cI = 0$, you would simply express $A^3$ as a linear combination of $A^2, A,$ and $I$. You would then use this to reduce any higher power of $A$ down to a quadratic expression. The degree of the reduction tool always matches the dimension of the matrix.

Question

If the matrix $A = \begin{pmatrix} 0 & 2 \ K & -1 \end{pmatrix}$ satisfies $A(A^3 + 3I) = 2I$ then the value of $K$ is :

Options:

(A) $1/2$

(B) $-1/2$

(C) $-1$

(D) $1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dismantle a problem that often intimidates students at first glance. We are given a matrix $A = \begin{pmatrix} 0 & 2 \ K & -1 \end{pmatrix}$ and a seemingly daunting equation: $A(A^3 + 3I) = 2I$.

Many students see $A^3$ and $A^4$ and immediately reach for their pens to start multiplying matrices. I want you to pause and take a deep breath. In the world of JEE Advanced, brute force is rarely the intended path.

There is a hidden elegance here, a mathematical shortcut that transforms a tedious calculation into a beautiful, logical flow. This shortcut is the **Cayley-Hamilton Theorem**.

### The DNA of a Matrix

Every square matrix has a 'DNA'—a characteristic equation that defines its behavior. This equation is found by calculating the determinant of $A - \lambda I$ and setting it to zero.

For our matrix $A$, we set up the determinant:
$$\begin{vmatrix} -\lambda & 2 \ K & -1 - \lambda \end{vmatrix} = 0$$

Expanding this, we get $(-\lambda)(-1 - \lambda) - (2)(K) = 0$, which simplifies to:
$$\lambda^2 + \lambda - 2K = 0$$

Now, here is the magic: the **Cayley-Hamilton theorem** states that every matrix satisfies its own characteristic equation. This means we can replace $\lambda$ with $A$:
$$A^2 + A - 2KI = 0$$

This is our 'reduction tool'. It allows us to express $A^2$ as:
$$A^2 = 2KI - A$$

### The Reduction Strategy

Now, let us look at the equation given in the problem: $A(A^3 + 3I) = 2I$. If we distribute the $A$, we get:
$$A^4 + 3A = 2I$$

We have an $A^4$ term. We know $A^4 = (A^2)^2$. Substituting our reduction tool, we get:
$$A^4 = (2KI - A)^2$$

Expanding this carefully using the identity $(x - y)^2 = x^2 - 2xy + y^2$:
$$A^4 = (2KI)^2 - 2(2KI)(A) + A^2$$
$$A^4 = 4K^2 I - 4KA + A^2$$

We still have an $A^2$ term, but we know exactly what to do with it. We substitute $A^2 = 2KI - A$ once more:
$$A^4 = 4K^2 I - 4KA + (2KI - A)$$
$$A^4 = (4K^2 + 2K)I - (4K + 1)A$$

### The Final Comparison

We take this simplified expression for $A^4$ and plug it back into our main equation $A^4 + 3A = 2I$:
$$[(4K^2 + 2K)I - (4K + 1)A] + 3A = 2I$$

Let us combine the terms involving $A$:
$$-(4K + 1)A + 3A = -(4K + 1 - 3)A = -(4K - 2)A$$

So, our equation becomes:
$$(4K^2 + 2K)I - (4K - 2)A = 2I$$

Now, look at the right-hand side. It is just $2I$. This means the coefficient of $A$ on the left must be zero, because there is no $A$ on the right.

Therefore, $4K - 2 = 0$. Solving this, we get $4K = 2$, which means:
**$$K = \frac{1}{2}$$**

### The Takeaway

We have found our answer, $K = \frac{1}{2}$. But more importantly, look at what we achieved. We didn't just solve a problem; we navigated a complex algebraic landscape using the power of the Cayley-Hamilton theorem.

We turned a high-degree matrix polynomial into a simple linear comparison. This is the essence of JEE Advanced mathematics. It is not about how fast you can multiply; it is about how clearly you can see the structure of the problem.

Animated Solution for Mathematics - Matrices and Determinants: If the matrix $A = (0 K 2 - 1)$ satisfies $A (A^{3} + 3 I) = 2 I$ then the value of $K$ is :

Select Answer:

Visualized Solution

The Matrix and the Equation

The Characteristic Equation

Setting up the Determinant

Expanding the Determinant

Cayley-Hamilton Theorem

Expanding the Given Equation

Calculating $A^{4}$

Expanding the Square

Substituting $A^{2}$ Again

Grouping Terms for $A^{4}$

Back to the Main Equation

Final Simplification

Comparing Coefficients

Solving for $K$

Verification and Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The DNA of a Matrix

The Reduction Strategy

The Final Comparison

The Takeaway

Similar Questions

If for the matrix, $A = [1 α - α β]$ , $A A^{T} = I_{2}$ , then the value of $α^{4} + β^{4}$ is:

Let $A = [a_{ij}]$ be a real matrix of order $3 \times 3$ , such that $a_{i 1} + a_{i 2} + a_{i 3} = 1$ , for $i = 1, 2, 3$ . Then, the sum of all the entries of the matrix $A^{3}$ is equal to:

Let $α$ and $β$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^{2} = 3 A + α I$ . If $A^{4} = 21 A + β I$ , then

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If $det (A^{2} - A) = 4$ , the sum of all possible values of $α$ is equal to

Let $A = 0 a 1 10 c 230$ , where $a, c \in R$ . If $A^{3} = A$ and the positive value of $a$ belongs to the interval $(n - 1, n]$ , where $n \in N$ , then $n$ is equal to ____.

Suppose A is any $3 \times 3$ non-singular matrix and $(A - 3 I) (A - 5 I) = 0$ , where $I = I_{3}$ and $O = O_{3}$ . If $α A + β A^{- 1} = 4 I$ , then $α + β$ is equal to :-

Let $A = 210 a 35 01 b$ . If $A^{3} = 4 A^{2} - A - 21 I$ , where $I$ is the identity matrix of order $3 \times 3$ , then $2 a + 3 b$ is equal to:

Let $A$ be a $3 \times 3$ real matrix such that $A^{2} (A - 2 I) - 4 (A - I) = O$ , where $I$ and $O$ are the identity and null matrices, respectively. If $A^{5} = α A^{2} + β A + γ I$ , where $α, β$ and $γ$ are real constants, then $α + β + γ$ is equal to:

Let $A$ be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $∣ A - x I ∣ = 0$ be -1 and 3, then the sum of the diagonal elements of the matrix $A^{2}$ is.

Let $B = [1135]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1}$ . If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O$ , then $2 β - α$ is equal to

Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the matrix A=(0K​2−1​) satisfies A(A3+3I)=2I then the value of K is :