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?
- \(\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}\)
- \(\begin{bmatrix}1 &-1 \\-1 & 2\end{bmatrix}\)
- \(\begin{bmatrix}-1 &2 \\-2 & 7\end{bmatrix}\)
- \(\begin{bmatrix}-1 & 2\\-1 &3\end{bmatrix}\)
The Correct Option is C
Approach Solution - 1
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:
- Swapping two rows.
- Multiplying a row by a non-zero scalar.
- Adding or subtracting a multiple of one row to another row.
Let's examine the options one by one:
- \(\begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}\)
To achieve this, we can add the first row to the second row:
| \(R_1 + R_2 \rightarrow R_1\) |
| \(\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} \rightarrow \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}\) |
- \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\)
To obtain this matrix, multiply the first row by \(-1\) and the second row by \(-1\):
| \(-R_1 \rightarrow R_1\) and \(-R_2 \rightarrow R_2\) |
| \(\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\) |
- \(\begin{bmatrix} -1 & 2 \\ -2 & 7 \end{bmatrix}\)
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.
- \(\begin{bmatrix} -1 & 2 \\ -1 & 3 \end{bmatrix}\)
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 \(\begin{bmatrix} -1 & 2 \\ -2 & 7 \end{bmatrix}\).
Approach Solution -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): \(\begin{bmatrix}-1&2\\-2&7\end{bmatrix}\)
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Concepts Used:
Matrices
Matrix:
A matrix is a rectangular array of numbers, variables, symbols, or expressions that are defined for the operations like subtraction, addition, and multiplications. The size of a matrix is determined by the number of rows and columns in the matrix.
The basic operations that can be performed on matrices are:
- Addition of Matrices - The addition of matrices addition can only be possible if the number of rows and columns of both the matrices are the same.
- Subtraction of Matrices - Matrices subtraction is also possible only if the number of rows and columns of both the matrices are the same.
- Scalar Multiplication - The product of a matrix A with any number 'c' is obtained by multiplying every entry of the matrix A by c, is called scalar multiplication.
- Multiplication of Matrices - Matrices multiplication is defined only if the number of columns in the first matrix and rows in the second matrix are equal.
- Transpose of Matrices - Interchanging of rows and columns is known as the transpose of matrices.