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One value to avoid

Linear algebraEasy

Let $A = 121 2 k 1 132$ . The system $A x = b$ has exactly one solution for every vector $b$ unless $k$ takes one particular value. Find that value of $k$ .

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Hint

A unique solution for every right-hand side requires an invertible matrix.

Solution

Expanding along row $1$ gives $det (A) = 1 \times (2 k - 3) - 2 \times (4 - 3) + 1 \times (2 - k)$ .
This simplifies to $2 k - 3 - 2 + 2 - k = k - 3$ .
The determinant is zero, and uniqueness fails, exactly when $k = 3$ .

Answer: $3$

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