Let $\beta$ be a real number. Consider the matrix $A = \begin{pmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{pmatrix}$. If $A^7 - (\beta - 1)A^6 - \beta A^5$ is a singular matrix, then the value of $9\beta$ is ______.

### Analyzing the Setup Welcome, future engineer! Today we are going to dismantle a problem that looks like a monster but is actually a masterpiece of algebraic elegance. We are given a matrix $A$ and a polynomial expression $M = A^7 - (\beta - 1)A^6 - \beta A^5$. The sight of $A^7$ might make your heart race, but take a deep breath. In the world of JEE Advanced, high powers are rarely a call to calculate; they are a call to factor. ### Factoring the Expression Notice how $A^5$ is common to every term? By pulling it out, we transform the expression into: $$M = A^5(A^2 - (\beta - 1)A - \beta I)$$ Now, look at that quadratic inside the parenthesis. It is begging to be factored! By expanding the middle term, we get $A^2 - \beta A + A - \beta I$. Grouping these terms gives us $(A + I)(A - \beta I)$. Suddenly, our terrifying $M$ is just a product of three matrices: $A^5$, $(A + I)$, and $(A - \beta I)$. ### Applying Determinant Properties The problem tells us $M$ is singular, which means its determinant is zero. Using the property $|XYZ| = |X||Y||Z|$, we know that: $$|A|^5 \cdot |A + I| \cdot |A - \beta I| = 0$$ We calculate $|A|$ and find it is $-1$. We calculate $|A + I|$ and find it is $-4$. Since neither is zero, the only way for the product to be zero is if $|A - \beta I| = 0$. ### Final Calculation Constructing $A - \beta I$ and calculating its determinant leads us to the equation: $$3\beta - 1 = 0$$ Solving this gives $\beta = \frac{1}{3}$. Finally, the question asks for $9\beta$, which is: $$9 \times \frac{1}{3} = 3$$ **The final answer is 3.** You did it! You didn't just solve a problem; you mastered a technique.

JEE Advanced 2022

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $β$ be a real number. Consider the matrix $A = β 23 011 1 - 2 - 2$ . If $A^{7} - (β - 1) A^{6} - β A^{5}$ is a singular matrix, then the value of $9 β$ is ______.

Enter Numerical Value:

Progressive Hints

Hint 1

Don't let the power of 7 scare you. In JEE Advanced, whenever you see high powers of a matrix, it is almost never about direct calculation. It is a signal to look for a pattern or a common factor. Can you see a common matrix term hidden in all three parts of the expression

A^{7} - (β - 1) A^{6} - β A^{5}

Hint 2

Once you have factored the expression, remember the property of determinants: the determinant of a product is the product of the determinants. If

M = X \cdot Y \cdot Z

, then

∣ M ∣ = ∣ X ∣ \cdot ∣ Y ∣ \cdot ∣ Z ∣

. Since

M

is singular,

∣ M ∣ = 0

. This means at least one of the factors must have a determinant of zero.

Hint 3

You should arrive at the factorization

M = A^{5} (A + I) (A - β I)

. Now, calculate the determinants of

A^{5}

and

(A + I)

individually. You will find they are non-zero constants. This forces the determinant of the final term,

∣ A - β I ∣

, to be zero. Set that determinant to zero and solve for

β

Visualized Solution

Analyzing the Given Matrix Expression

Given matrix: $A = β 23 011 1 - 2 - 2$
We are given that the matrix $M = A^{7} - (β - 1) A^{6} - β A^{5}$ is singular.
Our goal is to find the value of $9 β$ .

Factoring out $A^{5}$

Let $M = A^{7} - (β - 1) A^{6} - β A^{5}$
Notice that $A^{5}$ is common in all terms.
Factoring out $A^{5}$ :
$M = A^{5} (A^{2} - (β - 1) A - β I)$

Expanding the Inner Term

Focus on the quadratic part: $A^{2} - (β - 1) A - β I$
Expand the middle term:
$A^{2} - β A + A - β I$

Factorizing by Grouping

Group the terms: $(A^{2} - β A) + (A - β I)$
Factor out $A$ from the first group and $I$ from the second:
$A (A - β I) + I (A - β I)$
Factor out the common term $(A - β I)$ :
$(A + I) (A - β I)$
Therefore, $M = A^{5} (A + I) (A - β I)$

Condition for Singular Matrix

We are given that $M$ is a singular matrix.
A matrix is singular if its determinant is zero: $∣ M ∣ = 0$
Substitute the factorized form of $M$ :
$∣ A^{5} (A + I) (A - β I) ∣ = 0$
Using the property $∣ X Y Z ∣ = ∣ X ∣∣ Y ∣∣ Z ∣$ :
$∣ A ∣^{5} \cdot ∣ A + I ∣ \cdot ∣ A - β I ∣ = 0$

Evaluating Determinant of $A$

Let's check if $∣ A ∣ = 0$ .
$A = β 23 011 1 - 2 - 2$
Expanding along the first row:
$∣ A ∣ = β (1 (- 2) - (- 2) (1)) - 0 + 1 (2 (1) - 1 (3))$
$∣ A ∣ = β (- 2 + 2) + 1 (2 - 3)$
$∣ A ∣ = β (0) - 1 = - 1$
Since $∣ A ∣ = - 1 \neq = 0$ , the term $∣ A ∣^{5} \neq = 0$ .

Forming Matrix $A + I$

Next, let's check the term $∣ A + I ∣$ .
First, construct the matrix $A + I$ :
$A + I = β 23 011 1 - 2 - 2 + 100010001$
$A + I = β + 1 23 021 1 - 2 - 1$

Evaluating Determinant of $A + I$

Now, calculate $∣ A + I ∣$ :
$∣ A + I ∣ = β + 1 23 021 1 - 2 - 1$
Expanding along the first row:
$∣ A + I ∣ = (β + 1) (2 (- 1) - (- 2) (1)) - 0 + 1 (2 (1) - 2 (3))$
$∣ A + I ∣ = (β + 1) (- 2 + 2) + 1 (2 - 6)$
$∣ A + I ∣ = (β + 1) (0) - 4 = - 4$
Since $∣ A + I ∣ = - 4 \neq = 0$ , this term is also not zero.

The Crucial Deduction

We established that:
$∣ A ∣^{5} \cdot ∣ A + I ∣ \cdot ∣ A - β I ∣ = 0$
$∣ A ∣^{5} = (- 1)^{5} = - 1 \neq = 0$
$∣ A + I ∣ = - 4 \neq = 0$
For the product to be zero, the third factor must be zero.
Therefore, $∣ A - β I ∣ = 0$

Forming Matrix $A - β I$

Let's construct the matrix $A - β I$ :
$A - β I = β 23 011 1 - 2 - 2 - β 00 0 β 0 00 β$
$A - β I = 023 0 1 - β 1 1 - 2 - 2 - β$

Evaluating Determinant of $A - β I$

Calculate $∣ A - β I ∣$ by expanding along the first row:
$∣ A - β I ∣ = 023 0 1 - β 1 1 - 2 - 2 - β$
$∣ A - β I ∣ = 0 - 0 + 1 \cdot (2 (1) - 3 (1 - β))$
$∣ A - β I ∣ = 2 - (3 - 3 β)$
$∣ A - β I ∣ = 2 - 3 + 3 β = 3 β - 1$

Solving for $β$

We know that $∣ A - β I ∣ = 0$
Substitute the evaluated determinant:
$3 β - 1 = 0$
Transpose the constant term:
$3 β = 1$
Solve for $β$ :
$β = \frac{1}{3}$

Final Calculation

The question asks for the value of $9 β$ .
Substitute $β = \frac{1}{3}$ :
$9 β = 9 (\frac{1}{3})$
$9 β = 3$
Final Answer: $3$

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Welcome, future engineer! Today we are going to dismantle a problem that looks like a monster but is actually a masterpiece of algebraic elegance.

We are given a matrix

A

and a polynomial expression

M = A^{7} - (β - 1) A^{6} - β A^{5}

. The sight of

A^{7}

might make your heart race, but take a deep breath.

In the world of JEE Advanced, high powers are rarely a call to calculate; they are a call to factor.

Factoring the Expression

Notice how

A^{5}

is common to every term? By pulling it out, we transform the expression into:

M = A^{5} (A^{2} - (β - 1) A - β I)

Now, look at that quadratic inside the parenthesis. It is begging to be factored! By expanding the middle term, we get

A^{2} - β A + A - β I

Grouping these terms gives us

(A + I) (A - β I)

. Suddenly, our terrifying

M

is just a product of three matrices:

A^{5}

(A + I)

, and

(A - β I)

Applying Determinant Properties

The problem tells us

M

is singular, which means its determinant is zero. Using the property

∣ X Y Z ∣ = ∣ X ∣∣ Y ∣∣ Z ∣

, we know that:

∣ A ∣^{5} \cdot ∣ A + I ∣ \cdot ∣ A - β I ∣ = 0

We calculate

∣ A ∣

and find it is

- 1

. We calculate

∣ A + I ∣

and find it is

- 4

Since neither is zero, the only way for the product to be zero is if

∣ A - β I ∣ = 0

Final Calculation

Constructing

A - β I

and calculating its determinant leads us to the equation:

3 β - 1 = 0

Solving this gives

β = \frac{1}{3}

. Finally, the question asks for

9 β

, which is:

9 \times \frac{1}{3} = 3

The final answer is 3. You did it! You didn't just solve a problem; you mastered a technique.

Common Traps & FAQs

Wait, why can't I just calculate

A^{7}

directly? It's just matrix multiplication, right?

Ah, the classic trap! While technically possible, calculating

A^{7}

is a recipe for disaster in an exam setting. It is computationally heavy and prone to arithmetic errors. JEE Advanced problems are designed to test your insight, not your stamina. Always look for algebraic structures—like factoring—before you start grinding out calculations.

Why does an identity matrix

I

suddenly appear when we factor out

A

This is a fundamental rule of matrix algebra! When you factor a matrix

A

out of a term like

A - β A

, you are left with

A (I - β I)

. Wait, let's be precise:

A - β A

A (I - β I)

? No, the term was

A - β I

. When you factor

A

out of

A

, you are left with the identity matrix

I

, because

A \cdot I = A

. Think of

I

as the '1' of the matrix world.

How do I know which determinant is zero? Do I have to check all of them?

Great question! Yes, you check them systematically. If you find that

∣ A ∣ \neq = 0

and

∣ A + I ∣ \neq = 0

, then the only remaining possibility for the product to be zero is that

∣ A - β I ∣ = 0

. It is a process of elimination. Don't guess; calculate and verify.

Animated Solution for Mathematics - Matrices and Determinants: Let $β$ be a real number. Consider the matrix $A = β 23 011 1 - 2 - 2$ . If $A^{7} - (β - 1) A^{6} - β A^{5}$ is a singular matrix, then the value of $9 β$ is ______.

Enter Numerical Value:

Visualized Solution

Analyzing the Given Matrix Expression

Factoring out $A^{5}$

Expanding the Inner Term

Factorizing by Grouping

Condition for Singular Matrix

Evaluating Determinant of $A$

Forming Matrix $A + I$

Evaluating Determinant of $A + I$

The Crucial Deduction

Forming Matrix $A - β I$

Evaluating Determinant of $A - β I$

Solving for $β$

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Factoring the Expression

Applying Determinant Properties

Final Calculation

Similar Questions

Let $P = - 30 90120 20140605611214$ and $A = 2 - 1 0 7 - ω - ω ω^{2} 1 - ω + 1$ where $ω = \frac{- 1 + i 3}{2}$ , and $I_{3}$ be the identity matrix of order 3. If the determinant of the matrix $(P^{- 1} A P - I_{3})^{2}$ is $α ω^{2}$ , then the value of $α$ is equal to ________

Let $A = 100 0 α β 0 β α$ and $∣2 A ∣^{3} = 2^{21}$ where $α, β \in Z$ , Then a value of $α$ is

If matrix $A = a b c b c a c a b$ where $a, b, c$ are real positive numbers, $ab c = 1$ and $A^{T} A = I$ , then find the value of $a^{3} + b^{3} + c^{3}$ .

If $A = [α 2 2 α]$ and $∣ A^{3} ∣ = 125$ then the value of $α$ is

The value of the determinant $111 a b c a^{2} - b c b^{2} - c a c^{2} - ab$ is .........

Which of the following values of $α$ satisfy the equation $(1 + α)^{2} (2 + α)^{2} (3 + α)^{2} (1 + 2 α)^{2} (2 + 2 α)^{2} (3 + 2 α)^{2} (1 + 3 α)^{2} (2 + 3 α)^{2} (3 + 3 α)^{2} = - 648 α$ ?

Let $A = 500 5 α α 0 α 5 α 5$ . If $∣ A^{2} ∣ = 25$ , then $∣ α ∣$ equals

For $α, β \in R$ and a natural number $n$ , let $A_{r} = r 2 r 3 r - 2 123 \frac{n ^{2}}{2} + α n^{2} - β \frac{n ( 3 n - 1 )}{2}$ . Then $2 A_{10} - A_{8}$ is equal to :

Let $ω \neq = 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $1 ω ω^{2} a 1 ω b c 1$ where each of $a, b$ and $c$ is either $ω$ or $ω^{2}$ . Then the number of distinct matrices in the set $S$ is

Enter Numerical Value:

Visualized Solution

Analyzing the Given Matrix Expression

Factoring out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A5

Expanding the Inner Term

Factorizing by Grouping

Condition for Singular Matrix

Evaluating Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Forming Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+I

Evaluating Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+I

The Crucial Deduction

Forming Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A−βI

Evaluating Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A−βI

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

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Factoring the Expression

Applying Determinant Properties

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​100​0αβ​0βα​​ and ∣2A∣3=221 where α,β∈Z, Then a value of α is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If matrix A=​abc​bca​cab​​ where a,b,c are real positive numbers, abc=1 and ATA=I, then find the value of a3+b3+c3.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=[α2​2α​] and ∣A3∣=125 then the value of α is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the determinant ​111​abc​a2−bcb2−cac2−ab​​ is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following values of α satisfy the equation ​(1+α)2(2+α)2(3+α)2​(1+2α)2(2+2α)2(3+2α)2​(1+3α)2(2+3α)2(3+3α)2​​=−648α?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​500​5αα0​α5α5​​. If ∣A2∣=25, then ∣α∣ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For α,β∈R and a natural number n, let Ar​=​r2r3r−2​123​2n2​+αn2−β2n(3n−1)​​​. Then 2A10​−A8​ is equal to :

Factoring out $A^{5}$

Evaluating Determinant of $A$

Forming Matrix $A + I$

Evaluating Determinant of $A + I$

Forming Matrix $A - β I$

Evaluating Determinant of $A - β I$

Solving for $β$

Let $A = 100 0 α β 0 β α$ and $∣2 A ∣^{3} = 2^{21}$ where $α, β \in Z$ , Then a value of $α$ is

If matrix $A = a b c b c a c a b$ where $a, b, c$ are real positive numbers, $ab c = 1$ and $A^{T} A = I$ , then find the value of $a^{3} + b^{3} + c^{3}$ .

If $A = [α 2 2 α]$ and $∣ A^{3} ∣ = 125$ then the value of $α$ is

The value of the determinant $111 a b c a^{2} - b c b^{2} - c a c^{2} - ab$ is .........

Which of the following values of $α$ satisfy the equation $(1 + α)^{2} (2 + α)^{2} (3 + α)^{2} (1 + 2 α)^{2} (2 + 2 α)^{2} (3 + 2 α)^{2} (1 + 3 α)^{2} (2 + 3 α)^{2} (3 + 3 α)^{2} = - 648 α$ ?

Let $A = 500 5 α α 0 α 5 α 5$ . If $∣ A^{2} ∣ = 25$ , then $∣ α ∣$ equals

For $α, β \in R$ and a natural number $n$ , let $A_{r} = r 2 r 3 r - 2 123 \frac{n ^{2}}{2} + α n^{2} - β \frac{n ( 3 n - 1 )}{2}$ . Then $2 A_{10} - A_{8}$ is equal to :