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: …
Expanding gives Δ=2a2+8a+44; setting it equal to 86 yields a2+4a−21=0, whose two roots sum to −4 — option (C).
The idea
A determinant with an unknown inside is just a polynomial in that unknown. Evaluate it, set it equal to the given number, and solve the resulting equation. Here that equation is a quadratic, so there are two values of a and we only need their sum.
Expand the determinant
Because the (3,1) entry is 0, expanding along the first column is quickest:
Method: Solving a "Determinant Equals a Number" Equation for a Sum of Unknowns
When a determinant containing one unknown is set equal to a given number and the question only asks for the sum of the solutions (not each individual value), expand the determinant into a polynomial equation and read the sum off its coefficients — don't solve for each root separately if you don't have to.
Steps
Step 1: Expand the determinant along the row or column with the most zeros
Pick whichever row or column has a 0 entry (or create one with a row/column operation) to shorten the cofactor expansion. Keep careful track of the alternating sign pattern for a column expansion — the cofactor of the i-th entry down a column carries sign (−1)i+1, so the middle term is subtracted, not added.
Step 2: Set the resulting polynomial equal to the given value and simplify to standard form
Move everything to one side to get a polynomial equation in the unknown, typically a quadratic pa2+qa+r=0 once you subtract the given determinant value from both sides.
Step 3: Use Vieta's formula instead of finding each root, when only the sum is needed …
Mistake 1: Using the wrong sign for the middle cofactor in a column/row expansion
Why it's wrong: expanding along a column, the cofactor signs alternate +,−,+,… down the column, not all +. Treating the middle term's cofactor as + instead of − (or vice versa) flips the sign of one term in the polynomial and produces a wrong quadratic — and hence a wrong sum of roots. Correct approach: write out the (−1)i+j sign for each term explicitly before substituting the minors.
Mistake 2: Solving the full quadratic for individual roots and then mis-adding them …