Question

Which one of the following matrices can be obtained by performing elementary row transformations on the \(3\times3\) identity matrix?

• A. \[ \begin{bmatrix} 1&1&1 1&1&1 1&1&1 \end{bmatrix} \]
• B. \[ \begin{bmatrix} 1&1&1 2&3&4 1&2&1 \end{bmatrix} \]
• C. \[ \begin{bmatrix} 1&1&1 2&3&4 2&5&8 \end{bmatrix} \]
• D. \[ \begin{bmatrix} 1&1&1 -1&1&2 0&2&3 \end{bmatrix} \]

Hint:

A matrix reachable from the identity matrix by row operations must have full rank, meaning none of its rows can be written as a combination of the other two. Before jumping to a determinant calculation, check quickly whether one row is an obvious sum or scaled combination of the other two rows; if it is, that matrix is ruled out right away.

Answer 1

The Correct Option is B

Step 1: Use the property of elementary row transformations.
A matrix obtained from identity matrix using elementary row operations must be: \[ \mathrm{Non\text{-}singular} \] Thus determinant must satisfy: \[ \det(A)\ne 0 \]

Step 2: Check Option (A).
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}=0 \] All rows are identical. \[ \Rightarrow \mathrm{Singular} \] Therefore: \[ \Rightarrow \mathrm{Option\ (A)\ is\ Incorrect} \]

Step 3: Check Option (B).
\[ \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} \] Expanding: \[ =1(3-8)-1(2-4)+1(4-3) \] \[ =-5+2+1 \] \[ =-2 \] Since determinant is non-zero: \[ \Rightarrow \mathrm{Non\text{-}singular} \] Thus: \[ \Rightarrow \mathrm{Option\ (B)\ can\ be\ obtained} \]

Step 4: Check Option (C).
\[ \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 2 & 5 & 8 \end{vmatrix} \] Expanding: \[ =1(24-20)-1(16-8)+1(10-6) \] \[ =4-8+4 \] \[ =0 \] Therefore: \[ \Rightarrow \mathrm{Option\ (C)\ is\ Incorrect} \]

Step 5: Check Option (D).
\[ \begin{vmatrix} 1 & 1 & 1 \\ -1 & 1 & 2 \\ 0 & 2 & 3 \end{vmatrix} \] Expanding: \[ =1(3-4)-1(-3)+1(-2) \] \[ =-1+3-2 \] \[ =0 \] Thus: \[ \Rightarrow \mathrm{Option\ (D)\ is\ Singular} \]

Step 6: Identify the correct option.
Only option \[ \boxed{\mathrm{(B)}} \] can be obtained from elementary row transformations on identity matrix.

Answer 2

Concept:

• Elementary row operations never change the rank of a matrix, and the identity matrix has full rank $3$. So a matrix can be obtained from the identity matrix by row operations only if it also has full rank $3$, that is, its rows are linearly independent. Any full-rank $3\times3$ matrix can likewise be reduced back to the identity matrix, so full rank is exactly the condition needed.
• A row is linearly dependent on the other two if it can be written as $R_3=aR_1+bR_2$ for some numbers $a,b$. This can often be spotted directly, without computing a full determinant.

Step 1: Test option (A).
All three rows are $[1,1,1]$, so $R_2=R_1$ directly. The rows are dependent, rank is less than $3$, so this matrix cannot be obtained from the identity matrix.

Step 2: Test option (C).
Rows are $R_1=[1,1,1]$, $R_2=[2,3,4]$, $R_3=[2,5,8]$. Try $R_3=aR_1+bR_2$:
$a+2b=2$ and $a+3b=5$, so subtracting gives $b=3$, then $a=2-2(3)=-4$.
Check the third entry: $a+4b=-4+12=8$, which matches the third entry of $R_3$. So $R_3=-4R_1+3R_2$, the rows are dependent, rank is less than $3$. This matrix cannot be obtained from the identity matrix.

Step 3: Test option (D).
Rows are $R_1=[1,1,1]$, $R_2=[-1,1,2]$, $R_3=[0,2,3]$. Adding the first two rows directly: $R_1+R_2=[0,2,3]$, which is exactly $R_3$. So $R_3=R_1+R_2$, the rows are dependent, rank is less than $3$. This matrix cannot be obtained from the identity matrix.

Step 4: Test option (B).
Rows are $R_1=[1,1,1]$, $R_2=[2,3,4]$, $R_3=[1,2,1]$. Try $R_3=aR_1+bR_2$:
$a+2b=1$ and $a+3b=2$, so subtracting gives $b=1$, then $a=1-2(1)=-1$.
Check the third entry: $a+4b=-1+4=3$, but the third entry of $R_3$ is $1$, not $3$. The values of $a,b$ that fit the first two entries fail the third, so no such combination exists.

Step 5: Conclude for option (B).
Since $R_3$ cannot be written as a combination of $R_1$ and $R_2$, the three rows are linearly independent, giving full rank $3$. This matrix can be obtained from the identity matrix by elementary row operations.

Final Answer: Option (B)