Solve the determinant of 𝑥, two, negative, five, and 𝑥 minus the determinant of negative five, two, three, and 𝑥 is equal to 10.
Let’s start by reminding ourselves of the formula for the determinant of a two-by-two matrix. We multiply the top left and bottom right elements and subtract the product of the top right and bottom left. In the case of our first matrice then, to find the determinant, we calculate 𝑥 multiplied by 𝑥 minus two multiplied by negative five, which gives us 𝑥 squared plus 10.
The determinant of our second matrix is negative five multiplied by 𝑥 minus two multiplied by three. That’s negative five 𝑥 minus six. Now let’s substitute this back into our original equation. Notice how I’ve placed the value of the second determinant in a bracket. That’s because it is being subtracted. Expanding that bracket, we get 𝑥 squared plus 10 plus five 𝑥 plus six is equal to 10.
At this point, we have a quadratic equation. We can only solve quadratic equations when they’re equal to zero. So we’re going to subtract 10 from both sides of the equation and then factorize. Since the coefficient of 𝑥 squared is one, this is a fairly straightforward example.
To factorize this, we get 𝑥 plus three multiplied by 𝑥 plus two is equal to zero. This means that either 𝑥 plus three must be equal to zero or 𝑥 plus two is equal to zero. We solve the first linear equation by subtracting three from both sides, and we solve the second by subtracting two from both sides.
The solution to our equation is, therefore, 𝑥 is equal to negative two or 𝑥 is equal to negative three.