Discussion

If the matrix \(A = \begin{bmatrix} 1 & -1 \\ 4\lambda & 8 \end{bmatrix}\) is singular, then the value of \(\lambda\) is equal to

Answer: A

\(\lambda =\) \(-2\).

A singular matrix has determinant zero:

\[ \begin{aligned} |A| &= 1(8) - (-1)(4\lambda) = 8 + 4\lambda \\ 8 + 4\lambda &= 0 \;\Rightarrow\; \lambda = -2 \end{aligned} \]

\(\lambda = 2\) comes from taking \(-(-1)(4\lambda)\) as \(-4\lambda\), the usual sign slip. The other values leave the determinant non-zero.

Check: with \(\lambda = -2\) the second row is \((-8, 8)\), which is \(-8\) times the first row \((1, -1)\), so the rows are dependent.

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