Matrix Transformation from Identity Matrix using Elementary Row Transformations

To determine which matrix can be obtained by performing elementary row transformations on the $3 \times 3$ identity matrix $I_3$, we recall the properties of elementary row operations:

1. Performing an elementary row operation on a matrix $A$ is equivalent to premultiplying $A$ by an elementary matrix $E$.
2. Every elementary matrix $E$ is non-singular (i.e., invertible), meaning $\det(E) \neq 0$.
3. If a $3 \times 3$ matrix $M$ is obtained from $I_3$ via a sequence of elementary row transformations, then: $M = E_k E_{k-1} \cdots E_1 I_3$ where each $E_i$ is an elementary matrix.
4. Taking the determinant on both sides: $\det(M) = \det(E_k) \cdot \det(E_{k-1}) \cdots \det(E_1) \cdot \det(I_3)$ Since $\det(E_i) \neq 0$ for all $i$ and $\det(I_3) = 1$, it follows that $\det(M) \neq 0$.

Conversely, any non-singular $3 \times 3$ matrix can be transformed into $I_3$ (and vice-versa) using elementary row transformations. Therefore, a matrix can be obtained from $I_3$ by elementary row transformations if and only if its determinant is non-zero.

Now, we compute the determinants of the given matrices:

• Option A: $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ Since all three rows are identical, the rows are linearly dependent, so: $\det(A) = 0$

• Option B: $B = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{bmatrix}$ Evaluating the determinant of $B$: $\det(B) = 1(3 \cdot 1 - 4 \cdot 2) - 1(2 \cdot 1 - 4 \cdot 1) + 1(2 \cdot 2 - 3 \cdot 1)$ $\det(B) = 1(3 - 8) - 1(2 - 4) + 1(4 - 3) = -5 + 2 + 1 = -2 \neq 0$

• Option C: $C = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 2 & 5 & 8 \end{bmatrix}$ Performing the row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 2R_1$: $\det(C) = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 3 & 6 \end{vmatrix} = 1(1 \cdot 6 - 2 \cdot 3) = 0$

• Option D: $D = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 2 \\ 0 & 2 & 3 \end{bmatrix}$ Observing that $R_3 = R_1 + R_2$ since $(1, 1, 1) + (-1, 1, 2) = (0, 2, 3)$, the rows are linearly dependent. Hence: $\det(D) = 0$

Since only the matrix in Option B has a non-zero determinant, it is the only matrix that can be obtained from the $3 \times 3$ identity matrix using elementary row transformations.

Correct Option: B