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UPSC 2026 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Solution of system of linear equations · Linear Algebra · asked 8× in 14 yrs · Read the full method →

Question

Find all the values of $a$ for which the system of homogeneous equations

has non-trivial solutions, and hence determine all the solutions for each value of $a$.

Technique

A homogeneous square system $A\mathbf{x}=\mathbf{0}$ has a non-trivial solution iff $\det A = 0$ (equivalently $\operatorname{rank}A < n$). Compute $\det A$ as a polynomial in $a$, solve $\det A=0$, then for each root reduce the system and read off the null space with its free parameter.

Solution

Step 1 — Write the coefficient matrix and state the criterion.

The system is homogeneous with as many equations as unknowns, so $\mathbf{x}=\mathbf 0$ is always a solution. A non-trivial solution exists if and only if $A$ is singular:

Step 2 — Evaluate $\det A$.

Expand along the first row:

Now $\begin{vmatrix}1 & -1\\ 1 & -a\end{vmatrix} = -a+1 = 1-a$ and $\begin{vmatrix}1 & 1\\ 1 & 1\end{vmatrix}=0$. Hence

(Cross-check by row reduction: $R_2 \to R_2-R_3$ turns the second row into $(0,\,0,\,a-1)$, and expanding along it gives $\det A = (a-1)\cdot(-1)^{2+3}\begin{vmatrix}a&1\\1&1\end{vmatrix} = -(a-1)(a-1) = -(a-1)^2$. ✓)

Step 3 — Solve $\det A = 0$.

The determinant is a perfect square, so $a=1$ is a repeated root and it is the only value of $a$; there is no second value to hunt for. For every $a \neq 1$ we have $\det A = -(a-1)^2 \neq 0$, so $A$ is invertible and the only solution is the trivial one $x=y=z=0$.

Step 4 — Confirm Step 3 without determinants (elimination check).

Subtracting the third equation from the second:

• If $a\neq 1$: $z=0$. Then equation (2) gives $x+y=0$ and equation (1) gives $ax+y=0$; subtracting, $(a-1)x=0 \Rightarrow x=0 \Rightarrow y=0$. Only the trivial solution. ✓
• If $a=1$: the identity $(a-1)z=0$ is vacuous — equations (2) and (3) coincide, so the system genuinely loses a constraint. ✓

This independently confirms $a=1$ is the unique value.

Step 5 — The complete solution set at $a=1$.

Put $a=1$. The system becomes

Equations (2) and (3) are identical, so effectively two equations remain. Adding (1) and (2): $2(x+y)=0$; subtracting (2) from (1): $2z=0$. Hence

The coefficient matrix at $a=1$ is $\begin{pmatrix}1&1&1\\1&1&-1\\1&1&-1\end{pmatrix}$, which row-reduces to $\begin{pmatrix}1&1&0\\0&0&1\\0&0&0\end{pmatrix}$: $\operatorname{rank}=2$, so by rank–nullity the null space has dimension $3-2=1$ — a one-parameter family, as found.

Taking $x=t$ as the free parameter,

So the solution space is the line $\operatorname{span}\{(1,-1,0)\}$; $t\neq 0$ gives the non-trivial solutions.

Verification by substitution ($a=1$): $x+y+z = t-t+0 = 0$ ✓; $x+y-z = t-t-0 = 0$ ✓; third equation same as second ✓.

Answer