Sometimes a determinant is not just a number to compute — it is set equal to a given value, and that equality becomes an equation you must solve. The unknown sits inside the matrix, so you first evaluate the determinant as an expression in that unknown, then solve the resulting ordinary equation.
Note
Core idea: a determinant containing a variable is just a polynomial in disguise. "Expand the determinant, set it equal to the given value, solve" — that is the whole recipe.
The basic move
Suppose you are told
x32x=10.
Expand the left side: x⋅x−2⋅3=x2−6. Now it is an equation you already know how to handle:
x2−6=10⇒x2=16⇒x=±4.
A 2×2 gives a quadratic; a 3×3 typically gives a cubic. The number of solutions matches the degree of the polynomial you get.
Tip
Before expanding a 3×3, use row/column operations to create zeros. Fewer non-zero entries means a much shorter polynomial to solve — the value of the determinant is unchanged when you add a multiple of one row to another.
The important special case: equals zero
Most board problems set the determinant to 0:
1241xx21416=0.
Expanding gives a polynomial in x; its roots are the required values. Geometrically, a determinant being zero means the rows (or columns) are linearly dependent — the matrix is singular — so these equations often ask "for what value does the system collapse?"
A classic application: three points on one line
Three points A(x1,y1), B(x2,y2), C(x3,y3) are collinear exactly when the area of triangle ABC is zero. Since that area is 21 of a determinant, the collinearity condition is a determinant equation: …
Method: Equating Two Determinants and Solving the Resulting Equation
This method solves for an unknown that appears inside a determinant by evaluating both sides of a given determinant equation and reducing it to an ordinary algebraic equation.
Steps
Step 1: Evaluate the side with no unknown first
If one determinant is purely numeric, evaluate it completely — this becomes a fixed target value.
Step 2: Evaluate the side containing the unknown, keeping it symbolic
Apply the same ad−bc (or larger) formula to the determinant containing the variable, leaving the result as an algebraic expression in that variable.
Step 3: Set the two results equal
This converts the determinant equation into a standard algebraic equation (linear, quadratic, etc.) in the unknown.
Mistake 1: Dropping the negative root when solving x2=8
Why it's wrong: x2=8 has two solutions, x=8 and x=−8 — reporting only the positive root misses half the valid answers, since nothing in the problem restricts x to be positive. Correct approach: whenever solving x2=k for k>0, always state both x=±k.
Mistake 2: Sign error evaluating the numeric determinant on the right-hand side …