Discussion
Which one of the following matrices can be obtained by performing elementary row transformations on the \(3 \times 3\) identity matrix ?
Answer: B
The matrix that can be obtained is \(\begin{bmatrix}1&1&1\\2&3&4\\1&2&1\end{bmatrix}\).
Every elementary row operation is reversible and multiplies the determinant by a non-zero number (swap: \(-1\); scaling a row by \(k \ne 0\): \(k\); adding a multiple of one row to another: 1). Starting from \(I\), whose determinant is 1, you can therefore reach exactly the invertible matrices, so the test is a non-zero determinant.
\[ \begin{aligned} \text{B: } &1(3 - 8) - 1(2 - 4) + 1(4 - 3) = -2 \\ \text{C: } &1(24 - 20) - 1(16 - 8) + 1(10 - 6) = 0 \\ \text{D: } &1(3 - 4) - 1(-3 - 0) + 1(-2 - 0) = 0 \end{aligned} \]
The matrix of all 1s has three equal rows, so its determinant is 0 as well. Only the second matrix is invertible, so it alone can be reached from the identity.
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