Question

Consider the system of linear equations given below.
\[ ax + y = b \] \[ 16x + ay = 24 \] Suppose the values of \(a\) and \(b\) are chosen such that the system of linear equations produce multiple solutions. Then the product of \(a\) and \(b\) is _____. (answer in integer)

Hint:

Two linear equations have infinitely many solutions when they are scalar multiples of each other; equate the ratios of x-coefficients, y-coefficients, and constants.

Answer 1

Correct Answer: 24

Step 1: Use the determinant view of infinitely many solutions.
Write the system as
\[ ax + y - b = 0 \]
\[ 16x + ay - 24 = 0 \]
Two lines coincide, giving infinitely many common points, exactly when the coefficient rows are proportional, which for this pair means $a \cdot a - 1 \cdot 16 = 0$.

Step 2: Solve the determinant equation.
\[ a^2 - 16 = 0 \]
\[ a^2 = 16 \]
\[ a = 4 \ \text{or} \ a = -4 \]
This matches the requirement that the two rows of coefficients be proportional.

Step 3: Pin down $b$ using the same scale factor, case $a=4$.
When $a=4$, the first equation is $4x+y=b$. For this to be a scaled copy of $16x+4y=24$, notice the second equation is exactly $4$ times $\left(4x+y=6\right)$, since $4 \times 4x = 16x$, $4\times y=4y$, and $4\times 6=24$. So the first equation must read $4x+y=6$, meaning $b=6$.

Step 4: Pin down $b$ for case $a=-4$.
When $a=-4$, the first equation is $-4x+y=b$. The second equation $16x-4y=24$ is $-4$ times $\left(-4x+y=-6\right)$, since $-4\times(-4x)=16x$, $-4\times y=-4y$, and $-4\times(-6)=24$. So $b=-6$ here.

Step 5: Multiply $a$ and $b$ in both cases.
For $a=4,\,b=6$: $ab=24$.
For $a=-4,\,b=-6$: $ab=24$.
Both scenarios land on the exact same product, so the answer does not depend on which sign we pick.

Step 6: State the final value.
\[ \boxed{ab = 24} \]