IIT-JEE 2009 · Paper 1 · Mathematics Q34

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Paragraph for Question Nos. 33 to 35

Let $\mathcal A$ be the set of all $3\times3$ symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.

The number of matrices $A$ in $\mathcal A$ for which the system of linear equations
$A\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$
has a unique solution, is

less than 4

at least 4 but less than 7

at least 7 but less than 10

at least 10

Answer and worked solution

Official answer: B

Unique solution iff $\det A\ne0$ .

$d=3$ , $p=1$ : e.g. $\begin{bmatrix}1&1&0\\1&1&0\\0&0&1\end{bmatrix}$ has two equal rows: all 3 are singular.
$d=1$ : say $a_{11}=1$ . With pairs {12, 13} the 2nd and 3rd rows are both (1, 0, 0): singular. With pairs {12, 23} or {13, 23} the determinant is −1.

So 2 of the 3 choices for each diagonal position work: $3\times2=6$ matrices.

(B) at least 4 but less than 7

Topics:Systems of linear equations Determinants and properties

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IIT-JEE 2009 · Paper 1, Mathematics question 34