The determinant |a b aα+b; b c bα+c; aα+b bα+c 0| is equal to zero, if (a) a, b, c are in A. P. (b)

Q: Why did we choose the operation $C_3 \to C_3 - \alpha C_1 - C_2$? How do I know that's the right operation?

Great question! In determinant problems, we look for patterns. Notice the elements in the third column: $a\alpha + b$ and $b\alpha + c$. If you multiply the first column by $\alpha$ and add the second column, you get exactly these expressions. By subtracting them from the third column, we force those elements to become zero, which makes the expansion trivial. Always look for linear dependencies between columns!

Q: Wait, why is the third row element $-(a\alpha^2 + 2b\alpha + c)$? I thought it was just zero.

Remember, when you perform a column operation on a determinant, you must apply it to every element in that column, including the one in the third row. The original element in the third row, third column was $0$. Applying the operation $C_3 \to C_3 - \alpha C_1 - C_2$ to this element gives $0 - \alpha(a\alpha + b) - (b\alpha + c)$, which simplifies to $-(a\alpha^2 + 2b\alpha + c)$. Don't forget to update all elements!

Q: Why is option 4 ($ax^2 + bx + c = 0$) incorrect? It looks so similar to the result.

That is a classic trap! Our result from the determinant expansion is $a\alpha^2 + 2b\alpha + c = 0$. If you compare this to the quadratic $ax^2 + bx + c = 0$, you'll see the middle term is different. The determinant leads to $2b\alpha$, not $b\alpha$. Always check the coefficients carefully; the factor of $2$ is crucial here.

Question

The determinant $\begin{vmatrix} a & b & a\alpha + b \ b & c & b\alpha + c \ a\alpha + b & b\alpha + c & 0 \end{vmatrix}$ is equal to zero, if

Options:

(A) $a, b, c$ are in A. P.

(B) $a, b, c$ are in G. P.

(C) $a, b, c$ are in H. P.

(D) $\alpha$ is a root of the equation $ax^2 + bx + c = 0$

(E) $(x - \alpha)$ is a factor of $ax^2 + 2bx + c$.

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Determinants

Welcome, future engineer. Today, we are going to peel back the layers of a problem that, at first glance, looks like a standard algebraic exercise.

As we dive deeper, you will see that it is actually a beautiful dance of linear algebra and polynomial theory. We are presented with a $3 	imes 3$ determinant:

$$
\begin{vmatrix}
a & b & a\alpha + b \
b & c & b\alpha + c \
a\alpha + b & b\alpha + c & 0
\end{vmatrix} = 0
$$

### The Art of Observation

Most students see this and immediately reach for the expansion formula. Stop! Before you start multiplying terms, look at the structure.

Look at the third column. The elements $a\alpha + b$ and $b\alpha + c$ are not random. They are linear combinations of the first two columns.

Specifically, if you take the first column and multiply it by $\alpha$, then add the second column, you get exactly the first two elements of the third column. This is the 'spark' of the problem, indicating that the third column is not independent.

### The Power of Column Operations

We want to simplify this by creating zeros, as they make determinant expansion trivial. Let's apply the column operation: $C_3 	o C_3 - \alpha C_1 - C_2$.

For the first row: $(a\alpha + b) - \alpha(a) - b = 0$.

For the second row: $(b\alpha + c) - \alpha(b) - c = 0$.

The first two elements of the third column vanish! However, we must apply this operation to the third row as well.

The original element was $0$. The new element becomes $0 - \alpha(a\alpha + b) - (b\alpha + c)$. Expanding this, we get $-a\alpha^2 - b\alpha - b\alpha - c$, which simplifies to $-(a\alpha^2 + 2b\alpha + c)$.

### The Elegance of Expansion

Now, look at our transformed determinant:

$$
\begin{vmatrix}
a & b & 0 \
b & c & 0 \
a\alpha + b & b\alpha + c & -(a\alpha^2 + 2b\alpha + c)
\end{vmatrix} = 0
$$

Expanding along the third column is now a breeze. We only have one non-zero term:

$$
-(a\alpha^2 + 2b\alpha + c) 	imes \begin{vmatrix} a & b \ b & c \end{vmatrix} = 0
$$

The minor is simply $ac - b^2$. So, our equation becomes:

$$
-(a\alpha^2 + 2b\alpha + c)(ac - b^2) = 0
$$

### The Final Revelation

For this product to be zero, one of the two factors must be zero.

**Case 1:** $ac - b^2 = 0$, which implies $b^2 = ac$. This is the classic condition for $a, b, c$ to be in a **Geometric Progression (G.P.)**.

**Case 2:** $a\alpha^2 + 2b\alpha + c = 0$. This tells us that $\alpha$ is a root of the quadratic equation $ax^2 + 2bx + c = 0$.

By the Factor Theorem, this means $(x - \alpha)$ is a factor of the polynomial $ax^2 + 2bx + c$. You have successfully uncovered the hidden relationships between geometric progressions and polynomial roots.

Animated Solution for Mathematics - Matrices and Determinants: The determinant $a b a α + b b c b α + c a α + b b α + c 0$ is equal to zero, if

Select Answer:

Visualized Solution

Visualizing the Determinant

Formulating the Column Operation

Simplifying the First Two Rows

Simplifying the Third Row Element

The Transformed Determinant

Expanding Along the Third Column

Splitting into Two Conditions

Case 1: $a, b, c$ in Geometric Progression

Case 2: Root of the Quadratic Equation

Applying the Factor Theorem

Summary of Correct Options

The Sigma Insight: Properties of Determinants

The Hidden Geometry of Determinants

The Art of Observation

The Power of Column Operations

The Elegance of Expansion

The Final Revelation

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