Let $A=[a_{ij}]$ be a matrix of order $3\times3$, with $a_{ij}=(\sqrt{2})^{i+j}$. If the sum of all the elements in the third row of $A^{2}$ is $\alpha+\beta\sqrt{2}, \alpha, \beta\in Z$, then $\alpha+\beta$ is equal to : Options: (A) 280 (B) 224 (C) 210 (D) 168

### Analyzing the Setup We are given a $3 \times 3$ matrix $A$ where the elements are defined by $a_{ij} = (\sqrt{2})^{i+j}$. Let us construct the matrix explicitly: For the first row ($i=1$): $a_{11} = (\sqrt{2})^{1+1} = 2$, $a_{12} = (\sqrt{2})^{1+2} = 2\sqrt{2}$, and $a_{13} = (\sqrt{2})^{1+3} = 4$. For the second row ($i=2$): $a_{21} = 2\sqrt{2}$, $a_{22} = 4$, and $a_{23} = 4\sqrt{2}$. For the third row ($i=3$): $a_{31} = 4$, $a_{32} = 4\sqrt{2}$, and $a_{33} = 8$. Observing these rows, we see that every row is a scalar multiple of the first row. This confirms that $A$ is a **rank-1 matrix**. ### The Power of Rank-1 Decomposition Because every row is a multiple of the first, we can decompose matrix $A$ into the product of a column vector $C$ and a row vector $C^T$. Let us define: $$C = \begin{bmatrix} \sqrt{2} \\ 2 \\ 2\sqrt{2} \end{bmatrix}$$ Performing the multiplication $CC^T$ reconstructs matrix $A$. To find $A^2$, we use the property: $$A^2 = (CC^T)(CC^T) = C(C^TC)C^T$$ Here, $C^TC$ is a scalar value calculated as follows: $$C^TC = (\sqrt{2})^2 + 2^2 + (2\sqrt{2})^2 = 2 + 4 + 8 = 14$$ ### The Final Calculation Given $A^2 = 14(CC^T)$, it follows that $A^2 = 14A$. Therefore, the third row of $A^2$ is simply $14$ times the third row of $A$. The third row of $A$ is $[4, 4\sqrt{2}, 8]$. The sum of these elements is: $$4 + 4\sqrt{2} + 8 = 12 + 4\sqrt{2}$$ To find the sum of the elements in the third row of $A^2$, we multiply this sum by $14$: $$14(12 + 4\sqrt{2}) = 168 + 56\sqrt{2}$$ Comparing this to the form $\alpha + \beta\sqrt{2}$, we identify $\alpha = 168$ and $\beta = 56$. The final result is: $$\alpha + \beta = 168 + 56 = \mathbf{224}$$

JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [a_{ij}]$ be a matrix of order $3 \times 3$ , with $a_{ij} = (2)^{i + j}$ . If the sum of all the elements in the third row of $A^{2}$ is $α + β 2, α, β \in Z$ , then $α + β$ is equal to :

Select Answer:

Visualized Solution

Defining the Matrix Element Rule

Given matrix $A = [a_{ij}]_{3 \times 3}$
Rule for elements: $a_{ij} = (2)^{i + j}$
Our goal is to find the sum of elements in the third row of $A^{2}$

Constructing Matrix $A$

$a_{11} = (2)^{1 + 1} = 2$ , $a_{12} = (2)^{1 + 2} = 22$ , $a_{13} = 4$
Similarly, $a_{21} = 22$ , $a_{22} = 4$ , $a_{23} = 42$
And $a_{31} = 4$ , $a_{32} = 42$ , $a_{33} = 8$

Observing the Rank-1 Structure

Observe that $A$ is a rank-1 matrix because all rows are multiples of the first row.
We can write $A = C C^{T}$ where $C$ is a column vector.
Let $C = 2222$

Verifying the Decomposition $A = C C^{T}$

Verify: $C C^{T} = 2222 [2222]$
$= 2224224424428 = A$

Expressing $A^{2}$ using $C$

$A^{2} = (C C^{T}) (C C^{T})$
Using associativity: $A^{2} = C (C^{T} C) C^{T}$
Since $C^{T} C$ is a scalar, let's evaluate it.

Calculating the Scalar $C^{T} C$

$C^{T} C = [2222] 2222$
Sum of squares: $(2)^{2} + 2^{2} + (22)^{2}$
$C^{T} C = 2 + 4 + 8 = 14$

The Relation $A^{2} = 14 A$

Substitute $C^{T} C = 14$ back into the equation for $A^{2}$ .
$A^{2} = C (14) C^{T} = 14 (C C^{T})$
Since $C C^{T} = A$ , we get $A^{2} = 14 A$

Sum of Elements in the Third Row of $A$

Since $A^{2} = 14 A$ , the third row of $A^{2}$ is $14 \times$ (third row of $A$ ).
Third row of $A$ : $[4, 42, 8]$
Sum of elements in the third row of $A = 4 + 42 + 8 = 12 + 42$

Finding the Sum for $A^{2}$

Sum of elements in the third row of $A^{2} = 14 (12 + 42)$
$= 14 \times 12 + 14 \times 42$
$= 168 + 562$

Identifying $α$ and $β$

Given sum $= α + β 2$
Comparing with $168 + 562$ :
$α = 168$ , $β = 56$

Final Calculation: $α + β$

Calculate $α + β$ :
$168 + 56 = 224$
The final value is $224$ .

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Matrices

Solution Diagram

Analyzing the Setup

We are given a

3 \times 3

matrix

A

where the elements are defined by

a_{ij} = (2)^{i + j}

. Let us construct the matrix explicitly:

For the first row (

i = 1

a_{11} = (2)^{1 + 1} = 2

a_{12} = (2)^{1 + 2} = 22

, and

a_{13} = (2)^{1 + 3} = 4

For the second row (

i = 2

a_{21} = 22

a_{22} = 4

, and

a_{23} = 42

For the third row (

i = 3

a_{31} = 4

a_{32} = 42

, and

a_{33} = 8

Observing these rows, we see that every row is a scalar multiple of the first row. This confirms that

A

is a rank-1 matrix.

The Power of Rank-1 Decomposition

Because every row is a multiple of the first, we can decompose matrix

A

into the product of a column vector

C

and a row vector

C^{T}

. Let us define:

C = 2222

Performing the multiplication

C C^{T}

reconstructs matrix

A

. To find

A^{2}

, we use the property:

A^{2} = (C C^{T}) (C C^{T}) = C (C^{T} C) C^{T}

Here,

C^{T} C

is a scalar value calculated as follows:

C^{T} C = (2)^{2} + 2^{2} + (22)^{2} = 2 + 4 + 8 = 14

The Final Calculation

Given

A^{2} = 14 (C C^{T})

, it follows that

A^{2} = 14 A

. Therefore, the third row of

A^{2}

is simply

14

times the third row of

A

The third row of

A

[4, 42, 8]

. The sum of these elements is:

4 + 42 + 8 = 12 + 42

To find the sum of the elements in the third row of

A^{2}

, we multiply this sum by

14

14 (12 + 42) = 168 + 562

Comparing this to the form

α + β 2

, we identify

α = 168

and

β = 56

. The final result is:

α + β = 168 + 56 = 224

Common Traps & FAQs

Why can't we just multiply the matrix

A

by itself directly to find

A^{2}

You absolutely can! However, in JEE Advanced, time is your most precious resource. Calculating

A^{2}

by direct multiplication is tedious and prone to arithmetic errors. Recognizing the rank-1 structure allows you to solve the problem in seconds using the property

A^{2} = k A

, which is a much more elegant and robust method.

Is it always possible to write a matrix as

C C^{T}

Only if the matrix is of rank 1. A matrix has rank 1 if all its rows are linearly dependent—meaning they are all scalar multiples of a single row. In this problem, you can see that row 2 is

2

times row 1, and row 3 is

2

times row 1. This confirms the rank-1 structure.

Wait, why did we calculate

C^{T} C

as a scalar? I thought it would be a matrix.

That is a common point of confusion! Remember the dimensions:

C

is a

3 \times 1

column vector, and

C^{T}

is a

1 \times 3

row vector. When you multiply a row vector (

1 \times 3

) by a column vector (

3 \times 1

), the result is a

1 \times 1

matrix, which is just a scalar. It is the dot product of the vector with itself!

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [a_{ij}]$ be a matrix of order $3 \times 3$ , with $a_{ij} = (2)^{i + j}$ . If the sum of all the elements in the third row of $A^{2}$ is $α + β 2, α, β \in Z$ , then $α + β$ is equal to :

Select Answer:

Visualized Solution

Defining the Matrix Element Rule

Constructing Matrix $A$

Observing the Rank-1 Structure

Verifying the Decomposition $A = C C^{T}$

Expressing $A^{2}$ using $C$

Calculating the Scalar $C^{T} C$

The Relation $A^{2} = 14 A$

Sum of Elements in the Third Row of $A$

Finding the Sum for $A^{2}$

Identifying $α$ and $β$

Final Calculation: $α + β$

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Power of Rank-1 Decomposition

The Final Calculation

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Select Answer:

Visualized Solution

Defining the Matrix Element Rule

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

Observing the Rank-1 Structure

Verifying the Decomposition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=CCT

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

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The Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2=14A

Sum of Elements in the Third Row of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding the Sum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+β

The Sigma Insight: Algebraic Operations on Matrices

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The Power of Rank-1 Decomposition

The Final Calculation

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Constructing Matrix $A$

Verifying the Decomposition $A = C C^{T}$

Expressing $A^{2}$ using $C$

Calculating the Scalar $C^{T} C$

The Relation $A^{2} = 14 A$

Sum of Elements in the Third Row of $A$

Finding the Sum for $A^{2}$

Identifying $α$ and $β$

Final Calculation: $α + β$

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