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Matrices and Determinants question

2022 · 29 Jul · Shift 2 · Q24
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Matrices and Determinants question

2022 · 29 Jul · Shift 2 · Q24

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Which of the following matrices can NOT be obtained from the matrix [−121−1]\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right][−11​2−1​] by a single elementary row operation ?
  1. A
    [011−1]\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right][01​1−1​]
  2. B
    [1−1−12]\left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right][1−1​−12​]
  3. C
    [−12−27]\left[\begin{array}{rr}-1 & 2 \\ -2 & 7\end{array}\right][−1−2​27​]
  4. D
    [−12−13]\left[\begin{array}{ll}-1 & 2 \\ -1 & 3\end{array}\right][−1−1​23​]
View written solutionFree

Correct answer: C

Let A=[−121−1].A=\begin{bmatrix}-1 & 2\\ 1 & -1\end{bmatrix}.A=[−11​2−1​].

We need to check which option cannot be obtained from AAA using one elementary row operation.

The three types of elementary row operations are:

  1. Interchange two rows
  2. Multiply a row by a nonzero constant
  3. Add a multiple of one row to another row

We test each option.

1. Check option A

Option A is [011−1].\begin{bmatrix}0 & 1\\ 1 & -1\end{bmatrix}.[01​1−1​].

Starting from A=[−121−1],A=\begin{bmatrix}-1 & 2\\ 1 & -1\end{bmatrix},A=[−11​2−1​], keep R2R_2R2​ unchanged and do R1→R1+R2.R_1 \to R_1+R_2.R1​→R1​+R2​.

Then R1=(−1,2)+(1,−1)=(0,1).R_1=(-1,2)+(1,-1)=(0,1).R1​=(−1,2)+(1,−1)=(0,1).

So we get [011−1].\begin{bmatrix}0 & 1\\ 1 & -1\end{bmatrix}.[01​1−1​].

Hence, A can be obtained by one elementary row operation.

2. Check option B

Option B is [1−1−12].\begin{bmatrix}1 & -1\\ -1 & 2\end{bmatrix}.[1−1​−12​].

This is obtained by interchanging the two rows of AAA: R1↔R2.R_1 \leftrightarrow R_2.R1​↔R2​.

So

\to \begin{bmatrix}1 & -1\\ -1 & 2\end{bmatrix}.$$ Hence, **B can be obtained**. --- ## 3. Check option C Option C is $$\begin{bmatrix}-1 & 2\\ -2 & 7\end{bmatrix}.$$ Here the first row is unchanged, so if this comes from one row operation, the likely possibility is: $$R_2 \to R_2+kR_1$$ for some constant $k$. Compute: $$R_2+kR_1=(1,-1)+k(-1,2)=(1-k,-1+2k).$$ We want $$(1-k,-1+2k)=(-2,7).$$ From first entry: $$1-k=-2 \implies k=3.$$ Then second entry becomes $$-1+2(3)=5 \neq 7.$$ So this is not possible. Could it come from scaling $R_2$? If $$R_2 \to cR_2,$$ then $$(1,-1) \to (c,-c).$$ This can never become $(-2,7)$, since the two entries would have to be negatives of each other. Could it come from row interchange? No, because row 1 remains $(-1,2)$. Could it come from scaling $R_1$? That would change row 1, not row 2. So **C cannot be obtained** by a single elementary row operation. --- ## 4. Check option D Option D is $$\begin{bmatrix}-1 & 2\\ -1 & 3\end{bmatrix}.$$ Again row 1 is unchanged, so try $$R_2 \to R_2+kR_1.$$ Then $$R_2+kR_1=(1,-1)+k(-1,2)=(1-k,-1+2k).$$ Set this equal to $(-1,3)$: From first entry, $$1-k=-1 \implies k=2.$$ Then second entry: $$-1+2(2)=3,$$ which matches. So option D is obtained by $$R_2 \to R_2+2R_1.$$ Hence, **D can be obtained**. --- ## Final conclusion The only matrix that **cannot** be obtained from the given matrix by a single elementary row operation is: $$\boxed{C}$$
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