If A = [1 2 2; 2 1 -2; a 2 b] is a matrix satisfying the equation AAᵀ = 9I, where I is 3 x 3 identit

Q: Wait, why do we reject the negative root here? Or do we?

In this specific problem, we are solving a system of linear equations. We don't have a quadratic equation that leads to multiple roots, so there is no 'negative root' to reject. We are simply solving for $a$ and $b$ using linear algebra. If you encounter a quadratic later, always check if the solution satisfies the original matrix condition $AA^T = 9I$.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure mathematics problem involving matrix algebra, not a physics problem. While the logic of 'conservation' or 'invariance' is useful in physics, here we rely on the definition of matrix multiplication and the properties of the identity matrix $I$.

Q: Do we really need to calculate the entire matrix multiplication $AA^T$?

Absolutely not! That would be a waste of time. The problem only asks for the ordered pair $(a, b)$, which are located in the third row. By focusing only on the third row of the product $AA^T$, we can extract the necessary equations directly, saving precious time in the exam.

Question

If $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix}$ is a matrix satisfying the equation $AA^T = 9I$, where $I$ is $3 	imes 3$ identity matrix, then the ordered pair $(a, b)$ is equal to:

Options:

(A) $(2, 1)$

(B) $(-2, -1)$

(C) $(2, -1)$

(D) $(-2, 1)$

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Matrices

Welcome, fellow traveler on the JEE journey! Today, we are going to peel back the layers of a matrix problem that might look intimidating at first glance but is actually a beautiful exercise in structural symmetry.

We are given a matrix 
$$A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix}$$ 
and the condition $AA^T = 9I$.

This isn't just a random equation; it's a statement about the very nature of the matrix $A$. When we see $AA^T = 9I$, we should immediately think of orthogonality and scaling. It tells us that the rows of $A$ are not just vectors; they are vectors that, when dotted with themselves, yield $9$, and when dotted with each other, yield $0$.

### Phase 1: The Mirror Image

Before we can dance with the matrix, we need its partner: the transpose $A^T$. The transpose is like a mirror image where we take the rows of $A$ and turn them into columns.

The first row $[1, 2, 2]$ becomes the first column. The second row $[2, 1, -2]$ becomes the second column. The third row, which holds our mysterious variables $a$ and $b$, becomes the third column.

Thus, the transpose is:
$$A^T = \begin{bmatrix} 1 & 2 & a \ 2 & 1 & 2 \ 2 & -2 & b \end{bmatrix}$$

### Phase 2: The Dot Product Insight

Many students make the mistake of trying to multiply the entire matrix. Don't fall into that trap! We only care about the third row of $A$.

The element in the third row, first column of the resulting matrix $AA^T$ is the dot product of the third row of $A$ and the first column of $A^T$:
$$a(1) + 2(2) + b(2) = a + 4 + 2b$$

Since $9I$ has a $0$ at this position, we have our first equation:
$$a + 2b = -4$$

Similarly, the third row, second column of the product is the dot product of the third row of $A$ and the second column of $A^T$:
$$a(2) + 2(1) + b(-2) = 2a + 2 - 2b$$

This must also be $0$, giving us:
$$2a - 2b = -2$$

### Phase 3: The Algebraic Dance

Now, we have a system of two linear equations:
1) $a + 2b = -4$
2) $2a - 2b = -2$

Let's simplify the second equation by dividing by $2$, which gives $a - b = -1$. Now, look at how elegantly they align!

If we add the two equations, the $b$ terms vanish:
$$(a + 2b) + (a - b) = -4 + (-1)$$
$$2a + b = -5$$

Actually, let us subtract the simplified second equation from the first:
$$(a + 2b) - (a - b) = -4 - (-1)$$
$$3b = -3 \Rightarrow b = -1$$

Substituting $b = -1$ back into $a - b = -1$, we get $a - (-1) = -1$, which means $a + 1 = -1$, so $a = -2$. We have found our pair: **$(a, b) = (-2, -1)$**.

### Phase 4: The Final Verification

In the high-stakes environment of the JEE, never leave without checking your work. Let's verify the third diagonal element, which is the dot product of the third row with itself:
$$a^2 + 2^2 + b^2$$

Substituting $a = -2$ and $b = -1$:
$$(-2)^2 + 2^2 + (-1)^2 = 4 + 4 + 1 = 9$$

It matches perfectly! The condition $AA^T = 9I$ is satisfied. You have successfully navigated the matrix, solved the system, and verified the result.

Animated Solution for Mathematics - Matrices and Determinants: If $A = 12 a 212 2 - 2 b$ is a matrix satisfying the equation $A A^{T} = 9 I$ , where $I$ is $3 \times 3$ identity matrix, then the ordered pair $(a, b)$ is equal to:

Select Answer:

Visualized Solution

Problem Analysis for $A A^{T} = 9 I$

Finding the Transpose $A^{T}$

Matrix Multiplication $A A^{T}$ Setup

Extracting Equations for $a$ and $b$

Simplifying the Linear Equations

Solving for $a$ and $b$

Verification and Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

The Hidden Geometry of Matrices

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Problem Analysis for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AAT=9I

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

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

Extracting 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; } b

Simplifying the Linear Equations

Solving 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; } b