Question

Which of the following matrices can NOT be obtained from the matrix \(\begin{bmatrix} -1 &2 \\  1 & -1 \end{bmatrix}\) by a single elementary row operation?

• A. \(\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}\)
• B. \(\begin{bmatrix}1 &-1 \\-1 & 2\end{bmatrix}\)
• C. \(\begin{bmatrix}-1 &2 \\-2 & 7\end{bmatrix}\)
• D. \(\begin{bmatrix}-1 & 2\\-1 &3\end{bmatrix}\)

Answer 1

The Correct Option is C

To determine which matrix cannot be obtained from the given matrix \(\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}\) by a single elementary row operation, let's first recall the three types of elementary row operations: 

1. Swapping two rows.
2. Multiplying a row by a non-zero scalar.
3. Adding or subtracting a multiple of one row to another row.

Let's examine the options one by one:

1. \(\begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}\)

To achieve this, we can add the first row to the second row:

\(R_1 + R_2 \rightarrow R_1\)

\(\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} \rightarrow \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}\)

1. \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\)

To obtain this matrix, multiply the first row by \(-1\) and the second row by \(-1\):

\(-R_1 \rightarrow R_1\) and \(-R_2 \rightarrow R_2\)

\(\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\)

1. \(\begin{bmatrix} -1 & 2 \\ -2 & 7 \end{bmatrix}\)

To achieve this transformation, it requires adding some scalar multiple of the first row to the second row; let's verify:

\( \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} \rightarrow \begin{bmatrix} -1 & 2 \\ (-2)(-1) + 1 & (-2)(2) + (-1) \end{bmatrix}\)

\(\Rightarrow (-2) \times R_1 + R_2 \rightarrow R_2 \)

By calculating: \(-2 \times (-1) + 1 = 2 + 1 = 3\) and \(-2 \times 2 + (-1) = -4 - 1 = -5\). Thus incorrect, it cannot be obtained with a single operation.

1. \(\begin{bmatrix} -1 & 2 \\ -1 & 3 \end{bmatrix}\)

Achieve this by subtracting the first row from the second row:

\(R_2 - R_1 \rightarrow R_2\)

\(\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} \rightarrow \begin{bmatrix} -1 & 2 \\ -1 & 3 \end{bmatrix}\)

Therefore, the matrix that cannot be obtained from the original matrix \(\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}\) by a single elementary row operation is \(\begin{bmatrix} -1 & 2 \\ -2 & 7 \end{bmatrix}\).

Answer 2

(1) By R1→R1+R2, \(\begin{bmatrix}0 &1\\1 &-1\end{bmatrix}\) is possible

(2) By R1↔R2, \(\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\) is possible

(3) This matrix can’t be obtained
(4) By R2→R2+2R1, \(\begin{bmatrix}-1&2\\-1&3\end{bmatrix}\)is possible

So, the correct option is (C): \(\begin{bmatrix}-1&2\\-2&7\end{bmatrix}\)