Vishal K. answered 08/10/25
Love tutoring others and helping them understand difficult math!
To solve this problem, it might be easier to think of our matrix as a system of equations:
8x - 7y = 3 -- Equation 1
24x + hy = 9 -- Equation 2
Where x and y are variables. We want to find h.
A general rule of thumb: If your system has more linearly independent equations than variables, and none of the coefficients in your equations are 0, then that means that your system has infinite solutions. A trivial example:
2x + 3y = 4
This is just one line (one equation), but because there are two variables, there are an infinite number of solutions; (2, 0), and (0, 4/3), just to name a few.
Now, how do we make it so that our current system of equations has infinite solutions? We simply set h such that our two equations are no longer linearly independent. This makes one equation redundant; it's effectively as if our system only has one equation, with two variables!
In order to make this system linearly independent, one equation has to be a multiple of another (since there are only two equations, that's the only way to make one a linear combination of another). So we want to find a constant a such that, if you multiply both sides of Equation 1 by a, you get Equation 2. Since the coefficients of the variables have to match, this means that:
a * 8 = 24 (The coefficients of x must match)
a * (-7) = h (The coefficients of y must match)
a * 3 = 9
Based on the first equation and the third equation, we can see that a must be 3. This means that 3 * (-7) = h, so h = -21.
As a bonus: let's check our new system:
8x - 7y = 3 (Equation 1)
24x - 21y = 9 (Equation 2)
We see that if we multiply both sides of Equation 1 by 3, we get Equation 2. So our system is not linearly independent, and there is only one "real" equation.