JEE Advanced 2023

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $M = (a_{ij}), i, j \in {1, 2, 3}$ , be the $3 \times 3$ matrix such that $a_{ij} = 1$ if $j + 1$ is divisible by $i$ , otherwise $a_{ij} = 0$ . Then which of the following statements is (are) true ?

Q: Why does $MX = -X$ imply $|M+I|=0$?

This is the fundamental definition of an eigenvalue. If $MX = -X$, then $MX + IX = 0$, which factors to $(M+I)X = 0$. For a non-zero vector $X$ to satisfy this, the matrix $(M+I)$ must be singular, meaning its determinant must be zero.

Q: Is there a shortcut to finding the determinant?

For a $3 \times 3$ matrix, the Sarrus rule or cofactor expansion is the most reliable path. Don't look for shortcuts that might lead to errors. Precision is the hallmark of a JEE topper.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

The matrix is not given to you; it is hidden behind a simple divisibility rule. Before you jump into calculations, take a moment to carefully construct the

3 \times 3

matrix

M

. Remember,

i

is the row and

j

is the column. For each cell

(i, j)

, check if

i

divides

(j + 1)

Hint 2

Once you have the matrix, the key to unlocking all four options lies in the determinant. Recall that a matrix

M

is invertible if and only if

∣ M ∣ \neq = 0

. For the eigenvalue-like equation

M X = - X

, rewrite it as

(M + I) X = 0

. This implies a non-zero solution exists if and only if

∣ M + I ∣ = 0

Hint 3

Start by calculating

∣ M ∣

. If it is zero, the matrix is singular, and the null space is non-trivial. Then, construct

M + I

by adding

1

to the diagonal elements of

M

and calculate its determinant. If

∣ M + I ∣ = 0

, then Option B is correct.

Visualized Solution

Defining Matrix $M$

We need to construct a $3 \times 3$ matrix $M = (a_{ij})$ .
The condition is: $a_{ij} = 1$ if $i$ divides $(j + 1)$ .
Otherwise, $a_{ij} = 0$ .
Here, $i \in {1, 2, 3}$ is the row index and $j \in {1, 2, 3}$ is the column index.

Row 1 Elements ( $i = 1$ )

For the first row, $i = 1$ .
We check if $1$ divides $(j + 1)$ for $j = 1, 2, 3$ .
$j = 1 \Rightarrow j + 1 = 2$ . $1$ divides $2$ , so $a_{11} = 1$ .
$j = 2 \Rightarrow j + 1 = 3$ . $1$ divides $3$ , so $a_{12} = 1$ .
$j = 3 \Rightarrow j + 1 = 4$ . $1$ divides $4$ , so $a_{13} = 1$ .

Row 2 Elements ( $i = 2$ )

For the second row, $i = 2$ .
We check if $2$ divides $(j + 1)$ for $j = 1, 2, 3$ .
$j = 1 \Rightarrow j + 1 = 2$ . $2$ divides $2$ , so $a_{21} = 1$ .
$j = 2 \Rightarrow j + 1 = 3$ . $2$ does not divide $3$ , so $a_{22} = 0$ .
$j = 3 \Rightarrow j + 1 = 4$ . $2$ divides $4$ , so $a_{23} = 1$ .

Row 3 Elements and Matrix $M$

For the third row, $i = 3$ .
$j = 1 \Rightarrow j + 1 = 2$ . $3$ does not divide $2$ , so $a_{31} = 0$ .
$j = 2 \Rightarrow j + 1 = 3$ . $3$ divides $3$ , so $a_{32} = 1$ .
$j = 3 \Rightarrow j + 1 = 4$ . $3$ does not divide $4$ , so $a_{33} = 0$ .
$M = 110101110$

Determinant of $M$

Option A claims $M$ is invertible.
A matrix is invertible if and only if its determinant is non-zero ( $∣ M ∣ \neq = 0$ ).
$∣ M ∣ = 1 (0 - 1) - 1 (0 - 0) + 1 (1 - 0)$
$∣ M ∣ = - 1 - 0 + 1 = 0$
Since $∣ M ∣ = 0$ , $M$ is not invertible. Option A is false.

Homogeneous System $M X = 0$

Option C discusses the set ${X \in R^{3} : M X = 0}$ .
This represents a system of homogeneous linear equations.
It always has the trivial solution $X = 0$ .
Non-trivial solutions exist if and only if $∣ M ∣ = 0$ .

Evaluating Option C

We already established that $∣ M ∣ = 0$ .
This means the system $M X = 0$ has infinitely many non-trivial solutions.
Thus, the solution set is not just ${0}$ .
Therefore, ${X \in R^{3} : M X = 0} \neq = {0}$ is true. Option C is correct.

Analyzing $M X = - X$

Option B states there exists a non-zero $X$ such that $M X = - X$ .
We can rewrite this as $M X + X = 0$ .
Factoring out $X$ , we get $(M + I) X = 0$ , where $I$ is the identity matrix.
For a non-zero solution $X$ to exist, we must have $∣ M + I ∣ = 0$ .

Matrix $M + I$

Let's find the matrix $M + I$ .
$M + I = 110101110 + 100010001$
$M + I = 210111111$

Determinant of $M + I$

Now, calculate the determinant of $M + I$ .
$∣ M + I ∣ = 2 (1 - 1) - 1 (1 - 0) + 1 (1 - 0)$
$∣ M + I ∣ = 2 (0) - 1 (1) + 1 (1) = 0$
Since $∣ M + I ∣ = 0$ , a non-zero $X$ exists. Option B is true.

Matrix $M - 2 I$

Option D claims the matrix $(M - 2 I)$ is invertible.
$M - 2 I = 110101110 - 200020002$
$M - 2 I = - 1 10 1 - 2 1 11 - 2$

Determinant of $M - 2 I$

To check invertibility, we calculate $∣ M - 2 I ∣$ .
$∣ M - 2 I ∣ = - 1 (4 - 1) - 1 (- 2 - 0) + 1 (1 - 0)$
$∣ M - 2 I ∣ = - 1 (3) - 1 (- 2) + 1 (1)$
$∣ M - 2 I ∣ = - 3 + 2 + 1 = 0$
Since the determinant is $0$ , $(M - 2 I)$ is not invertible. Option D is false.

Final Answer

Option A is False ( $M$ is not invertible).
Option B is True (Non-zero $X$ exists for $M X = - X$ ).
Option C is True (Null space of $M$ is non-trivial).
Option D is False ( $(M - 2 I)$ is not invertible).
Correct Options: B and C

00:00 / 00:00

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Phase 1

The Code-Breaking Construction

Imagine you are standing before a

3 \times 3

grid. You have the power to fill it, but only if you obey the law:

a_{ij} = 1

i

divides

(j + 1)

, and

0

otherwise.

For the first row,

i = 1

. Since

1

divides every integer, every element in the first row is

1

. Our first row is

(1, 1, 1)

For the second row,

i = 2

. We check

j = 1, 2, 3

: For $j = 1$ , $j + 1 = 2$ , which is divisible by $2$ . For

j = 2

j + 1 = 3

, which is not. * For

j = 3

j + 1 = 4

, which is divisible by

2

. Thus, the second row is

(1, 0, 1)

For the third row,

i = 3

: For $j = 1$ , $j + 1 = 2$ (no). For

j = 2

j + 1 = 3

(yes). * For

j = 3

j + 1 = 4

(no). The third row is

(0, 1, 0)

Our matrix

M

stands revealed:

M = 110101110

Phase 2

The Gatekeeper

The determinant is the gatekeeper of matrix properties. It tells us if a matrix is invertible or singular. Let us calculate

∣ M ∣

Expanding along the first row:

∣ M ∣ = 1 (0 - 1) - 1 (0 - 0) + 1 (1 - 0) = - 1 - 0 + 1 = 0

The determinant is zero! This is a massive revelation. It means

M

is not invertible, instantly falsifying Option A.

Furthermore, because

∣ M ∣ = 0

, the homogeneous system

M X = 0

must have non-trivial solutions. This confirms Option C is true.

Phase 3

The Hidden Symmetry

Now, look at Option B:

M X = - X

. This is not just an equation; it is a question about eigenvalues. If we rearrange this, we get

M X + I X = 0

, or

(M + I) X = 0

For a non-zero vector

X

to exist, the matrix

(M + I)

must be singular, meaning

∣ M + I ∣ = 0

. Let us construct

M + I

M + I = 1 + 1 10 1 0 + 1 1 11 0 + 1 = 210111111

Now, calculate its determinant:

∣ M + I ∣ = 2 (1 - 1) - 1 (1 - 0) + 1 (1 - 0) = 2 (0) - 1 + 1 = 0

The determinant is zero! This confirms that a non-zero vector

X

exists. Option B is true.

Phase 4

The Final Verdict

Finally, we evaluate

M - 2 I

. We construct it:

M - 2 I = - 1 10 1 - 2 1 11 - 2

Calculating the determinant:

∣ M - 2 I ∣ = - 1 (4 - 1) - 1 (- 2 - 0) + 1 (1 - 0) = - 3 + 2 + 1 = 0

Since the determinant is zero,

M - 2 I

is not invertible. Therefore, Option D is false.

We have systematically dismantled the problem. The truth lies in Options B and C.

Common Traps & FAQs

Why does

M X = - X

imply

∣ M + I ∣ = 0

This is the fundamental definition of an eigenvalue. If

M X = - X

, then

M X + I X = 0

, which factors to

(M + I) X = 0

. For a non-zero vector

X

to satisfy this, the matrix

(M + I)

must be singular, meaning its determinant must be zero.

Why can't we just use energy conservation?

This is a linear algebra problem, not a physics problem! But I love the enthusiasm. In linear algebra, the 'conservation' principle is the determinant. If the determinant is zero, information is lost (the matrix collapses dimensions), which is exactly why the null space becomes non-trivial.

Is there a shortcut to finding the determinant?

For a

3 \times 3

matrix, the Sarrus rule or cofactor expansion is the most reliable path. Don't look for shortcuts that might lead to errors. Precision is the hallmark of a JEE topper.

Animated Solution for Mathematics - Matrices and Determinants: Let $M = (a_{ij}), i, j \in {1, 2, 3}$ , be the $3 \times 3$ matrix such that $a_{ij} = 1$ if $j + 1$ is divisible by $i$ , otherwise $a_{ij} = 0$ . Then which of the following statements is (are) true ?

Select Answer:

Visualized Solution

Defining Matrix $M$

Row 1 Elements ( $i = 1$ )

Row 2 Elements ( $i = 2$ )

Row 3 Elements and Matrix $M$

Determinant of $M$

Homogeneous System $M X = 0$

Evaluating Option C

Analyzing $M X = - X$

Matrix $M + I$

Determinant of $M + I$

Matrix $M - 2 I$

Determinant of $M - 2 I$

Final Answer

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

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

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

Row 2 Elements ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=2)

Row 3 Elements and Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

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

Homogeneous System .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MX=0

Evaluating Option C

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MX=−X

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

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

Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M−2I

Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M−2I

Final Answer

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Phase 1

Phase 2

Phase 3

Phase 4

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Defining Matrix $M$

Row 1 Elements ( $i = 1$ )

Row 2 Elements ( $i = 2$ )

Row 3 Elements and Matrix $M$

Determinant of $M$

Homogeneous System $M X = 0$

Analyzing $M X = - X$

Matrix $M + I$

Determinant of $M + I$

Matrix $M - 2 I$

Determinant of $M - 2 I$

Let $A$ and $B$ be two $3 \times 3$ non-zero real matrices such that $A B$ is a zero matrix. Then

The number of matrices in $A$ is

The number of matrices $A$ in $A$ for which the system of linear equations $A x y z = 100$ has a unique solution, is

The number of matrices $A$ in $A$ for which the system of linear equations $A x y z = 100$ is inconsistent, is

The number of $3 \times 3$ matrices $A$ whose entries are either 0 or 1 and for which the system $A x y z = 100$ has exactly two distinct solutions, is

Let $R^{2}$ denote $R \times R$ . Let $S = {(a, b, c) : a, b, c \in R and a x^{2} + 2 b x y + c y^{2} > 0 for all (x, y) \in R^{2} - {(0, 0)}}$ . Then which of the following statements is (are) TRUE?

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is