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 simplifies to 7(sin3θ+2cos2θ−2)=0, so sin3θ+2cos2θ=2. Using identities, this reduces to sinθ(−4sin2θ−4sinθ+3)=0, giving θ=nπ or θ=nπ+(−1)n6π for integer n.
When a determinant equals zero, it means the rows (or columns) are linearly dependent. But here, we're not looking for a dependency relationship — we're solving for a specific variable θ that makes the determinant vanish. The direct approach is to expand the determinant and simplify the resulting trigonometric equation.
The determinant is 3×3, so expansion is straightforward. The key is to handle the trigonometric terms carefully and use identities to reduce everything to a single trigonometric function.
Step 1: Expand the determinant
Let
Δ=1−4713−7sin3θcos2θ−2
Expanding along the first row (or any row/column — choose what's simplest):
Second minor (note the minus sign in front of the cofactor):
−47cos2θ−2=(−4)(−2)−(cos2θ)(7)=8−7cos2θ
Third minor:
−473−7=(−4)(−7)−(3)(7)=28−21=7
Step 3: Assemble the expansion
Δ=1⋅(−6+7cos2θ)−1⋅(8−7cos2θ)+sin3θ⋅7
Simplify:
Δ=−6+7cos2θ−8+7cos2θ+7sin3θ
Δ=−14+14cos2θ+7sin3θ
Factor 7:
Δ=7(−2+2cos2θ+sin3θ)
Set Δ=0:
7(−2+2cos2θ+sin3θ)=0⟹−2+2cos2θ+sin3θ=0
So:
sin3θ+2cos2θ=2
Watch out
A common mistake is to forget the factor of 2 on cos2θ after expansion. Double-check the arithmetic: the two cos2θ terms add to 14cos2θ, which becomes 2cos2θ after factoring 7.
Step 4: Use trigonometric identities to reduce to one function
Method: Solving for an Unknown Angle from a Determinant Equation
When a determinant containing trigonometric functions of an unknown angle is set equal to zero, the method is to expand the determinant into a plain trigonometric equation first, then solve that equation using standard multiple-angle identities.
Steps
Step 1: Expand the determinant using cofactor expansion
Pick the row or column with the simplest (constant) entries and expand along it. This converts the determinant condition into a linear combination of the trigonometric quantities appearing in the matrix (here, sin of a multiple angle and cos of another multiple angle), multiplied by fixed numeric coefficients from the 2×2 minors.
Step 2: Set the expanded expression equal to zero and simplify
Collect like terms and divide through by any common numeric factor to reach the cleanest possible trigonometric equation, something of the form psin(mθ)+qcos(nθ)=r.
Step 3: Rewrite every term in terms of a single trig function …
Mistake 1: Dropping the alternating sign in cofactor expansion
Expanding a 3×3 determinant along a row, the middle term always carries a − sign. Missing it here turns the correctly-expanded −14+14cos2θ+7sin3θ into a wrong expression and every step after it is built on a false equation.
Mistake 2: Misremembering the triple-angle identity
Using sin3θ=3sinθ−sin3θ instead of the correct sin3θ=3sinθ−4sin3θ corrupts the cubic equation in sinθ and leads to entirely wrong roots. …