Reduction of Second-Degree Equation to Canonical Form

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2017)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~22 min
• Difficulty mix: medium 1
• Section: B | Dominant type: derivation

Why This Chapter Matters

This topic appeared once in 2017 as a 15-mark Section B derivation, making it a low-frequency but high-marks-per-appearance atom. UPSC asks you to reduce a given second-degree equation in three variables to canonical form by diagonalising the associated symmetric matrix — the question tests whether you can carry out eigenvalue decomposition in a coordinate-geometry setting and correctly identify the resulting quadric surface. Even as a T4 atom, the technique (matrix diagonalisation + completing the square for the linear part) is a direct application of Linear Algebra skills that T1–T2 atoms also test, so the preparation cost is shared.

Minimum Theory

The General Second-Degree Equation

A general second-degree (quadric) equation in three variables is

$F(x,y,z) = ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0.$

It can be written compactly as $\mathbf{X}^T A \mathbf{X} + 2\mathbf{b}^T \mathbf{X} + d = 0$, where

$A = \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} u \\ v \\ w \end{pmatrix}, \quad \mathbf{X} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}.$

$A$ is real symmetric, so all its eigenvalues $\lambda_1, \lambda_2, \lambda_3$ are real and it is orthogonally diagonalisable.

Step 1 — Diagonalise the Quadratic Part

Find eigenvalues of $A$ from $\det(A - \lambda I) = 0$. For each eigenvalue find the normalised eigenvector. Arrange eigenvectors as columns of an orthogonal matrix $P$ (the rotation matrix).

The substitution $\mathbf{X} = P\mathbf{Y}$ (with $\mathbf{Y} = (X, Y, Z)^T$) eliminates all cross terms; the quadratic part becomes

$\lambda_1 X^2 + \lambda_2 Y^2 + \lambda_3 Z^2.$

Step 2 — Absorb the Linear Part (Translation)

After the rotation the full equation reads

$\lambda_1 X^2 + \lambda_2 Y^2 + \lambda_3 Z^2 + 2u'X + 2v'Y + 2w'Z + d' = 0,$

where $(u', v', w')^T = P^T \mathbf{b}$ and $d' = d$. Complete the square in each variable (for any nonzero $\lambda_i$):

$\lambda_i \!\left(X_i + \frac{u_i'}{\lambda_i}\right)^2 - \frac{(u_i')^2}{\lambda_i}.$

Translate: $X_i' = X_i + u_i'/\lambda_i$. Collect the constant terms on the right side to obtain the canonical form

$\lambda_1 X'^2 + \lambda_2 Y'^2 + \lambda_3 Z'^2 = k$

for some constant $k$ (or a more specialised form if some $\lambda_i = 0$).

Classification of Quadric Surfaces

Key Determinant Invariants

Let $\Delta = \det\begin{pmatrix}A & \mathbf{b}\\ \mathbf{b}^T & d\end{pmatrix}$ (the $4\times4$ matrix of the full equation) and $\delta = \det A$. These are rotation-invariant and help classify the surface after finding canonical form.

Question Archetypes

full-reduction (1 question; 2017)

Recognition Cues

• The equation has three variables with both squared and cross terms (e.g., $xy$, $yz$, $zx$ terms present)
• The question uses words like “reduce to canonical form” or “identify the quadric surface”
• 15 marks are allocated — full eigenvalue + translation working is expected
• The matrix $A$ is likely to have a repeated eigenvalue (to keep arithmetic manageable in exam conditions)

Solution Template

1. Write down the symmetric matrix $A$ by reading off coefficients; write $\mathbf{b}$ and $d$.
2. Compute $\det(A - \lambda I) = 0$ and solve for eigenvalues $\lambda_1, \lambda_2, \lambda_3$.
3. For each eigenvalue find the eigenvector; normalise and (if repeated) apply Gram–Schmidt to get an orthonormal set.
4. Form $P = [\hat{e}_1 | \hat{e}_2 | \hat{e}_3]$ (orthogonal matrix).
5. Compute $(u', v', w')^T = P^T \mathbf{b}$.
6. Complete the square: the constant accumulated is $k = -d' + \sum_{\lambda_i \neq 0} (u_i')^2/\lambda_i$.
7. Write canonical form; divide through by $k$ (if $k \neq 0$) and identify surface type.

Worked Example

2017 Paper 1, 2017-P1-Q5b (15 marks)

Reduce the quadric $2x^2 + 2y^2 + 2z^2 - 2yz - 2zx - 2xy = 3$ to canonical form and identify the surface.

Step 1. Write the matrix.

The quadratic form has $a = b = c = 2$, $f = g = h = -1$ (since coefficient of $yz$ is $2f = -2$, etc.).

$A = \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}.$

There are no linear terms, so $\mathbf{b} = \mathbf{0}$ and $d = -3$ (moving 3 to the left gives $\cdots - 3 = 0$, but we keep the equation as stated).

Step 2. Find eigenvalues.

$\det(A - \lambda I) = \det\begin{pmatrix} 2-\lambda & -1 & -1 \\ -1 & 2-\lambda & -1 \\ -1 & -1 & 2-\lambda \end{pmatrix}.$

Use the fact that $A = 3I - J$ where $J$ is the all-ones matrix. The eigenvalues of $J$ (for $3\times3$) are $3$ (once) and $0$ (twice). Hence eigenvalues of $A = 3I - J$ are $3 - 3 = 0$ (once) and $3 - 0 = 3$ (twice).

So $\lambda_1 = 0$, $\lambda_2 = \lambda_3 = 3$.

Step 3. Find eigenvectors.

For $\lambda = 0$: $(A)\mathbf{v} = 0$. All rows of $A$ sum to $0$, so $(1,1,1)^T$ is an eigenvector. Normalised: $\hat{e}_1 = \tfrac{1}{\sqrt{3}}(1,1,1)^T$.

For $\lambda = 3$: $(A - 3I)\mathbf{v} = 0$, i.e., $(-I - J + 3I)\mathbf{v} = 0$ gives $-J\mathbf{v} = 0$, meaning $v_1 + v_2 + v_3 = 0$. Two independent solutions: take $\mathbf{v}_2 = (1,-1,0)^T$ and $\mathbf{v}_3 = (1,0,-1)^T$. Apply Gram–Schmidt:

$\hat{e}_2 = \frac{1}{\sqrt{2}}(1,-1,0)^T.$

$\mathbf{v}_3' = (1,0,-1)^T - \frac{(1,0,-1)\cdot(1,-1,0)}{2}(1,-1,0)^T = (1,0,-1)^T - \frac{1}{2}(1,-1,0)^T = \left(\frac{1}{2},\frac{1}{2},-1\right)^T.$

$\hat{e}_3 = \frac{1}{\sqrt{6}}(1,1,-2)^T.$

Step 4. Rotate.

The substitution $\mathbf{X} = P\mathbf{Y}$ with $P = [\hat{e}_1|\hat{e}_2|\hat{e}_3]$ converts the equation to

$0 \cdot X^2 + 3Y^2 + 3Z^2 = 3.$

(The $\mathbf{b} = 0$ so no linear terms arise.)

Step 5. Canonical form.

$Y^2 + Z^2 = 1.$

Step 6. Identify the surface.

This is a right circular cylinder with axis along the $X$-direction (the $\lambda = 0$ eigenvector direction, i.e., the direction $(1,1,1)^T$) and radius 1.

$\boxed{Y^2 + Z^2 = 1 \quad \text{(right circular cylinder)}}$

Common Traps

• Sign of coefficients: The coefficient of $yz$ in $F$ is $2f$, not $f$ — always halve the coefficient when reading off the matrix entry.
• Forgetting to move the constant: If the equation is $\cdots = k$ (not $= 0$), the right-hand side after rotation is still $k$; do not lose it.
• Gram–Schmidt omitted for repeated eigenvalues: With a repeated eigenvalue the eigenspace is two-dimensional; any two orthogonal vectors in it work, but they must be normalised and mutually orthogonal — skipping this gives a non-orthogonal $P$.
• Misidentifying the surface: A zero eigenvalue means the corresponding direction is a cylindrical axis; if two eigenvalues are zero the surface is a pair of planes or a parabolic cylinder.

Marks-Aware Writing

A 15-mark derivation rewards systematic layout. Allocate roughly:

• 3 marks — writing $A$, $\mathbf{b}$, $d$ correctly and stating the plan.
• 4 marks — characteristic equation solved correctly, eigenvalues stated.
• 4 marks — eigenvectors found, normalised, Gram–Schmidt applied (if needed), $P$ stated.
• 2 marks — rotation applied, new equation written (with or without linear terms).
• 2 marks — canonical form stated and surface identified by name.

Write each step as a numbered heading in your answer script. Do not skip the classification step — it is explicitly asked for and carries marks.

Practice Set

Only one historical question on this atom (shown above).