JEE Main 2019 (11 January)

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Animated Solution for Mathematics - Matrices and Determinants: Let $A = 0 p p 2 q q - q r - r r$ . If $A A^{T} = I_{3}$ , then $∣ p ∣$ is :

Q: Why do we need the absolute value $|p|$? Can't we just find $p$?

Great observation! When we solve $p^2 = \frac{1}{2}$, we get $p = \pm \frac{1}{\sqrt{2}}$. Since the question asks for $|p|$, we are looking for the magnitude, which is always positive. The sign of $p$ doesn't affect the validity of the orthogonal condition in this specific setup, so $|p|$ is the most precise answer.

Select Answer:

Visualized Solution

The Given Matrix and Condition

Given matrix $A = 0 p p 2 q q - q r - r r$
Condition: $A A^{T} = I_{3}$
Objective: Find the value of $∣ p ∣$

Orthogonal Matrix Properties

A matrix satisfying $A A^{T} = I$ is called an orthogonal matrix.
Property 1: The sum of squares of elements in any row is $1$ (Rows are unit vectors).
Property 2: The dot product of any two distinct rows is $0$ (Rows are mutually perpendicular).

Row 2 as a Unit Vector

Let's look at Row 2: $(p, q, - r)$
Sum of squares of its elements must be $1$ .
$p^{2} + q^{2} + (- r)^{2} = 1$
$p^{2} + q^{2} + r^{2} = 1$ --- (Equation 1)

Dot Product of Row 2 and Row 3

Row 2: $R_{2} = (p, q, - r)$
Row 3: $R_{3} = (p, - q, r)$
Since rows are mutually perpendicular, their dot product is zero: $R_{2} \cdot R_{3} = 0$

Calculating the Dot Product

Multiply corresponding elements and add:
$(p) (p) + (q) (- q) + (- r) (r) = 0$
$p^{2} - q^{2} - r^{2} = 0$

Expressing $p^{2}$ in terms of $q$ and $r$

From $p^{2} - q^{2} - r^{2} = 0$ , we can isolate $p^{2}$ .
Move $q^{2}$ and $r^{2}$ to the right side.
$p^{2} = q^{2} + r^{2}$ --- (Equation 2)

Substituting Equation 2 into Equation 1

Recall Equation 1: $p^{2} + (q^{2} + r^{2}) = 1$
Substitute $q^{2} + r^{2}$ with $p^{2}$ from Equation 2.
$p^{2} + p^{2} = 1$

Simplifying the Expression

Adding the terms: $2 p^{2} = 1$
Divide by 2: $p^{2} = \frac{1}{2}$

Finding $∣ p ∣$

Take the square root on both sides: $p^{2} = \frac{1}{2}$
Since $p^{2} = ∣ p ∣$ , we get:
$∣ p ∣ = \frac{1}{2}$
This matches the first option.

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The Sigma Insight: Types of Matrices

Analyzing the Setup

We are given the matrix:

A = 0 p p 2 q q - q r - r r

The condition

A A^{T} = I_{3}

is the hallmark of an orthogonal matrix. In linear algebra, an orthogonal matrix represents a transformation that preserves lengths and angles.

This implies that the rows of the matrix are unit vectors that are mutually perpendicular to each other. This geometric property is our secret weapon to avoid tedious matrix multiplication.

Utilizing Row Properties

Let us focus on the second row,

R_{2} = (p, q, - r)

. Since it is a unit vector, the sum of the squares of its components must equal

1

This gives us our first equation:

p^{2} + q^{2} + (- r)^{2} = 1

Which simplifies to:

p^{2} + q^{2} + r^{2} = 1

The Orthogonality Condition

Next, we consider the third row,

R_{3} = (p, - q, r)

. Because the rows are mutually perpendicular, their dot product must be zero:

R_{2} \cdot R_{3} = 0

Calculating this dot product, we get:

(p) (p) + (q) (- q) + (- r) (r) = 0

This simplifies to:

p^{2} - q^{2} - r^{2} = 0

Final Calculation

We can rewrite the previous result as

p^{2} = q^{2} + r^{2}

. Now, we substitute this into our first equation:

p^{2} + (q^{2} + r^{2}) = 1

Since

q^{2} + r^{2} = p^{2}

, we substitute directly to obtain:

p^{2} + p^{2} = 1

2 p^{2} = 1

Solving for

p

, we find

p^{2} = \frac{1}{2}

, which leads to the final result:

∣ p ∣ = \frac{1}{2}

Common Traps & FAQs

Wait, why can't we just multiply the matrix

A

by its transpose

A^{T}

and solve the resulting system of equations?

You absolutely could! However, that would involve calculating nine different entries in the resulting matrix, leading to a system of equations that is much more tedious and prone to algebraic errors. Recognizing the properties of an orthogonal matrix allows you to bypass the brute force and solve the problem in just a few elegant steps.

Why does

A A^{T} = I

imply that the rows are unit vectors and mutually perpendicular?

Think about the definition of matrix multiplication. When you multiply

A

A^{T}

, the element in the

i

-th row and

j

-th column of the result is the dot product of the

i

-th row of

A

and the

j

-th column of

A^{T}

(which is the

j

-th row of

A

). If the result is the identity matrix

I

, then the diagonal elements (where

i = j

) must be

1

(the dot product of a vector with itself is its magnitude squared, so magnitude is

1

), and the off-diagonal elements (where

i \neq = j

) must be

0

(the dot product of perpendicular vectors).

Why do we need the absolute value

∣ p ∣

? Can't we just find

p

Great observation! When we solve

p^{2} = \frac{1}{2}

, we get

p = \pm \frac{1}{2}

. Since the question asks for

∣ p ∣

, we are looking for the magnitude, which is always positive. The sign of

p

doesn't affect the validity of the orthogonal condition in this specific setup, so

∣ p ∣

is the most precise answer.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 0 p p 2 q q - q r - r r$ . If $A A^{T} = I_{3}$ , then $∣ p ∣$ is :

Select Answer:

Visualized Solution

The Given Matrix and Condition

Orthogonal Matrix Properties

Row 2 as a Unit Vector

Dot Product of Row 2 and Row 3

Calculating the Dot Product

Expressing $p^{2}$ in terms of $q$ and $r$

Substituting Equation 2 into Equation 1

Simplifying the Expression

Finding $∣ p ∣$

The Sigma Insight: Types of Matrices

Analyzing the Setup

Utilizing Row Properties

The Orthogonality Condition

Final Calculation

Similar Questions

Let $a, b, c \in R$ be all non-zero and satisfy $a^{3} + b^{3} + c^{3} = 2$ . If the matrix $A = a b c b c a c a b$ satisfies $A A^{T} = I$ , then a value of $ab c$ can be :

Comprehension Passage

The number of $A$ in $T_{p}$ such that $A$ is either symmetric or skew-symmetric or both, and $det (A)$ divisible by $p$ is

The number of $A$ in $T_{p}$ such that the trace of $A$ is not divisible by $p$ but $det (A)$ is divisible by $p$ is

The number of $A$ in $T_{p}$ such that $det (A)$ is not divisible by $p$ is

Let $A = [2 a 30], a \in R$ be written as $P + Q$ where $P$ is a symmetric matrix and $Q$ is a skew symmetric matrix. If $det (Q) = 9$ , then the modulus of the sum of all possible values of determinant of $P$ is equal to :

Let $A$ be a square matrix such that $A A^{T} = I$ . Then $\frac{1}{2} A [(A + A^{T})^{2} + (A - A^{T})^{2}]$ is equal to

If $A = \frac{1}{2} [1 - 3 31]$ , then :

Let $A$ be a symmetric matrix such that $∣ A ∣ = 2$ and $[23 1 \frac{3}{2}] A = [1 α 2 β]$ . If the sum of the diagonal elements of $A$ is $s$ , then $\frac{β s}{α ^{2}}$ is equal to ______.

Let $A = [α 6 - 1 β]$ , $α > 0$ , such that $det (A) = 0$ and $α + β = 1$ . If $I$ denotes $2 \times 2$ identity matrix, then the matrix $(I + A)^{8}$ is:

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; } Let A=​0pp​2qq−q​r−rr​​. If AAT=I3​, then ∣p∣ is :

Select Answer:

Visualized Solution

The Given Matrix and Condition

Orthogonal Matrix Properties

Row 2 as a Unit Vector

Dot Product of Row 2 and Row 3

Calculating the Dot Product

Substituting Equation 2 into Equation 1

Simplifying the Expression

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣p∣

The Sigma Insight: Types of Matrices

Analyzing the Setup

Utilizing Row Properties

The Orthogonality Condition

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c∈R be all non-zero and satisfy a3+b3+c3=2. If the matrix A=​abc​bca​cab​​ satisfies AAT=I, then a value of abc can be :

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of A in Tp​ such that A is either symmetric or skew-symmetric or both, and det(A) divisible by p is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of A in Tp​ such that the trace of A is not divisible by p but det(A) is divisible by p is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of A in Tp​ such that det(A) is not divisible by p is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a square matrix such that AAT=I. Then 21​A[(A+AT)2+(A−AT)2] is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=21​[1−3​​3​1​], then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a symmetric matrix such that ∣A∣=2 and [23​123​​]A=[1α​2β​]. If the sum of the diagonal elements of A is s, then α2βs​ is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=[α6​−1β​], α>0, such that det(A)=0 and α+β=1. If I denotes 2×2 identity matrix, then the matrix (I+A)8 is:

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 0 p p 2 q q - q r - r r$ . If $A A^{T} = I_{3}$ , then $∣ p ∣$ is :

Finding $∣ p ∣$

Let $a, b, c \in R$ be all non-zero and satisfy $a^{3} + b^{3} + c^{3} = 2$ . If the matrix $A = a b c b c a c a b$ satisfies $A A^{T} = I$ , then a value of $ab c$ can be :

The number of $A$ in $T_{p}$ such that $A$ is either symmetric or skew-symmetric or both, and $det (A)$ divisible by $p$ is

The number of $A$ in $T_{p}$ such that the trace of $A$ is not divisible by $p$ but $det (A)$ is divisible by $p$ is

The number of $A$ in $T_{p}$ such that $det (A)$ is not divisible by $p$ is

Let $A$ be a square matrix such that $A A^{T} = I$ . Then $\frac{1}{2} A [(A + A^{T})^{2} + (A - A^{T})^{2}]$ is equal to

If $A = \frac{1}{2} [1 - 3 31]$ , then :

Let $A$ be a symmetric matrix such that $∣ A ∣ = 2$ and $[23 1 \frac{3}{2}] A = [1 α 2 β]$ . If the sum of the diagonal elements of $A$ is $s$ , then $\frac{β s}{α ^{2}}$ is equal to ______.

Let $A = [α 6 - 1 β]$ , $α > 0$ , such that $det (A) = 0$ and $α + β = 1$ . If $I$ denotes $2 \times 2$ identity matrix, then the matrix $(I + A)^{8}$ is: