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: …
The determinant equals xyz(1+x1+y1+z1); since it is 0 and xyz=0, the reciprocal sum must be −1 — option (D).
The idea
Evaluate the determinant in closed form. It factors as xyz times (1+∑1/x), so the condition Δ=0 (with none of x,y,z zero) pins the reciprocal sum immediately.
Method: Evaluating a Determinant Whose Rows Differ by a Common Additive Shift
When a determinant's entries look like "1 plus a variable" on the diagonal and plain 1's elsewhere, don't expand it term by term — use a column (or row) operation to expose a repeated factor, then reduce to a much smaller determinant before setting it equal to a given value.
Steps
Step 1: Spot the structure and choose an operation that creates a common column/row
For a matrix like
1+x1111+y1111+z,
subtracting one column from a neighbouring one (e.g. C1→C1−C2, C2→C2−C3) turns most entries into the single variable that column "owns", isolating x, y, z while leaving simple constants elsewhere. This is always safe — subtracting one column from another never changes the determinant's value.
Step 2: Expand the reduced determinant and factor
After the operation, expand along the row or column with the most zeros. You'll typically land on an expression of the form
Mistake 1: Dividing by xyz without checking it's non-zero
Why it's wrong: the step from xyz(1+x1+y1+z1)=0 to 1+x1+y1+z1=0 is only valid because the question states x,y,z=0, so xyz=0. Skipping this justification is a logic gap examiners penalise even when the final number is right. Correct approach: explicitly cite x,y,z=0⇒xyz=0 before cancelling it from both sides.
Mistake 2: Picking the "looks similar" distractor −x−y−z instead of −1 …