Let A = [[0, 1, 2], [a, 0, 3], [1, c, 0]], where a, c ∈ R. If A³ = A and the positive value of a bel

Q: Why don't we calculate the whole $A^3$ matrix? Isn't that safer?

In the high-stakes environment of JEE Advanced, time is your most precious currency. Calculating the entire $A^3$ matrix involves 27 multiplications and 18 additions. By identifying that we only need two equations to solve for two variables, we can extract exactly the information we need, saving precious minutes and reducing the probability of calculation errors.

Q: How do I know which elements to pick for the system of equations?

Great question! You want to pick elements that are 'rich' in the variables you need to find. Look at the first row of $A$. It contains $0, 1, 2$. When you multiply this by the columns of $A^2$, you get expressions involving $a$ and $c$. By choosing $(A^3)_{11}$ and $(A^3)_{12}$, we directly involve $a$ and $c$ in a way that creates a solvable system. Always look for the path of least resistance!

Q: Why do we reject the negative root for $a$?

The problem statement explicitly asks for the 'positive value of $a$'. In competitive exams, always re-read the question to ensure you are answering exactly what is asked. Rejecting the negative root isn't just about math; it's about following the constraints of the problem.

Question

Let $A = \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix}$, where $a, c \in \mathbb{R}$. If $A^3 = A$ and the positive value of $a$ belongs to the interval $(n-1, n]$, where $n \in \mathbb{N}$, then $n$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dive into a problem that might look like a tedious exercise in matrix multiplication, but is actually a beautiful dance of algebra and logic.

We are given a matrix 
$$A = \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix}$$ 
and the intriguing condition $A^3 = A$. Our mission is to find a natural number $n$ such that the positive value of $a$ lies in the interval $(n-1, n]$.

### The Art of Strategic Calculation

Many students, upon seeing $A^3 = A$, immediately start calculating the entire $A^3$ matrix. Stop! That is a trap. In JEE Advanced, efficiency is key.

We only need to find two equations to solve for our two unknowns, $a$ and $c$. To get there, we first need $A^2$. Let's compute it carefully:
$$A^2 = A \cdot A = \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix}$$

Performing the row-by-column multiplication, we find the first row of $A^2$:
$$(A^2)_{11} = 0(0) + 1(a) + 2(1) = a+2$$
$$(A^2)_{12} = 0(1) + 1(0) + 2(c) = 2c$$
$$(A^2)_{13} = 0(2) + 1(3) + 2(0) = 3$$

Following this, we complete the matrix:
$$A^2 = \begin{bmatrix} a+2 & 2c & 3 \ 3 & a+3c & 2a \ ca & 1 & 2+3c \end{bmatrix}$$
Take a moment to breathe and verify these entries. Accuracy here is the foundation of our success.

### The Algebraic Dance

Now, we use the condition $A^3 = A$. We don't need the whole matrix. Let's look at the first row of $A^3$.

The element $(A^3)_{11}$ is the first row of $A^2$ multiplied by the first column of $A$:
$$(A^3)_{11} = (a+2)(0) + (2c)(a) + (3)(1) = 2ac + 3$$
Since $A^3 = A$, this must equal $A_{11}$, which is $0$. Thus, $2ac + 3 = 0$.

Next, let's find $(A^3)_{12}$ by multiplying the first row of $A^2$ with the second column of $A$:
$$(A^3)_{12} = (a+2)(1) + (2c)(0) + (3)(c) = a + 2 + 3c$$
This must equal $A_{12}$, which is $1$. So, $a + 3c + 2 = 1$, or $a + 3c + 1 = 0$. We now have a beautiful system of two equations:
$$2ac = -3 \quad 	ext{and} \quad a + 3c = -1$$

### Solving for $a$

From the first equation, $c = -\frac{3}{2a}$. Substituting this into the second equation:
$$a + 3\left(-\frac{3}{2a}ight) + 1 = 0$$

Multiplying by $2a$ to clear the denominator, we get $2a^2 - 9 + 2a = 0$, or $2a^2 + 2a - 9 = 0$. Using the quadratic formula:
$$a = \frac{-2 \pm \sqrt{4 - 4(2)(-9)}}{4} = \frac{-2 \pm \sqrt{4 + 72}}{4} = \frac{-2 \pm \sqrt{76}}{4} = \frac{-1 \pm \sqrt{19}}{2}$$

### Final Calculation

We are looking for the positive value of $a$, so we take $a = \frac{\sqrt{19}-1}{2}$. Since $4^2 = 16$ and $5^2 = 25$, we know $4 < \sqrt{19} < 5$.

Therefore, $3 < \sqrt{19}-1 < 4$. Dividing by $2$, we get $1.5 < a < 2$.

This means $a$ lies in the interval $(1, 2]$. Comparing this with $(n-1, n]$, we see that **$n=2$**. You did it! The logic holds, the math is clean, and the answer is clear.

Animated Solution for Mathematics - Matrices and Determinants: 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 ____.

Enter Numerical Value:

Visualized Solution

Matrix $A$ and Condition $A^{3} = A$

Strategy: Computing $A^{2}$

Calculating $A^{2}$ (First Row)

Completing Matrix $A^{2}$

Finding $(A^{3})_{1, 1}$

Finding $(A^{3})_{1, 2}$

Setting up Equations for $a$ and $c$

Substituting $c$ in terms of $a$

Solving the Quadratic Equation

Estimating the Value of $a$

Final Conclusion for $n$

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Art of Strategic Calculation

The Algebraic Dance

Solving for $a$

Final Calculation

Similar Questions

Let $A = [21 - 1 1]$ . If the sum of the diagonal elements of $A^{13}$ is $3^{n}$ , then $n$ is equal to_________

Let $A = [a_{ij}]$ be $3 \times 3$ matrix such that $A 010 = 001$ , $A 413 = 010$ and $A 212 = 100$ , then $a_{23}$ equals:

Let $I$ be an identity matrix of order $2 \times 2$ and $P = [25 - 1 - 3]$ . Then the value of $n \in N$ for which $P^{n} = 5 I - 8 P$ is equal to ___

Let $A = cos θ 0 sin θ 010 - sin θ 0 cos θ$ . If for some $θ \in (0, π)$ , $A^{2} = A^{T}$ , then the sum of the diagonal elements of the matrix $(A + I)^{3} + (A - I)^{3} - 6 A$ is equal to _____ .

Let $A = 100 a 10 a b 1, a, b \in R$ . If for some $n \in N$ , $A^{n} = 10048102160961$ , then $n + a + b$ is equal to

The total number of matrices $A = 0 2 x 2 x 2 y y - y 1 - 1 1, (x, y \in R, x \neq = y)$ for which $A^{T} A = 3 I_{3}$ is :-

Let $A = [1 + i - i 10]$ where $i = - 1$ . Then, the number of elements in the set ${n \in {1, 2, \dots, 100} : A^{n} = A}$ is

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

Let $A = 211 - 1 0 - 1 - 1 - 1 0$ and $B = A - I$ . If $ω = \frac{3 i - 1}{2}$ , then the number of elements in the set ${n \in {1, 2, \dots, 100} : A^{n} + (ω B)^{n} = A + B}$ is equal to

Let A be a matrix such that $A \cdot [1023]$ is a scalar matrix and $∣3 A ∣ = 108$ . Then $A^{2}$ equals :

Enter Numerical Value:

Visualized Solution

Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3=A

Strategy: Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2 (First Row)

Completing Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A3)1,1​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A3)1,2​

Setting up Equations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solving the Quadratic Equation

Estimating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Final Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n