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Determine those values of λ for which the matrix   1 λ 0 3 2 0 1 2 1   is not invertible

Question

Determine those values of λ \lambda for which the matrix

(1λ0320121) \begin{pmatrix} 1 & \lambda & 0 \\ 3 & 2 & 0 \\ 1 & 2 & 1 \end{pmatrix}

is not invertible.

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Solution

A matrix is not invertible (or singular) if and only if its determinant is zero. So, we need to find the values of λ that make the determinant of the matrix zero.

The matrix is:

1 λ 0 3 2 0 1 2 1

The determinant of a 3x3 matrix

a b c d e f g h i

is given by the formula:

a(ei−fh)−b(di−fg)+c(dh−eg)

So, the determinant of our matrix is:

1*(21 - 02) - λ*(31 - 01) + 0*(32 - 11)

Simplify this to:

2 - 3λ

Set the determinant equal to zero and solve for λ:

2 - 3λ = 0 3λ = 2 λ = 2/3

So, the matrix is not invertible for λ = 2/3.

This problem has been solved

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