Question

Let \(A\) be a \(3\times3\) matrix. If \[ A \begin{bmatrix} 001 \end{bmatrix} = \begin{bmatrix} 123 \end{bmatrix}, \quad A \begin{bmatrix} 101 \end{bmatrix} = \begin{bmatrix} 10-1 \end{bmatrix}, \quad A \begin{bmatrix} 110 \end{bmatrix} = \begin{bmatrix} 110 \end{bmatrix}, \] then the rank of \((A-I)\) is

• A. \(3\)
• B. \(2\)
• C. \(1\)
• D. \(0\)

Hint:

Whenever images of vectors under a matrix are given, first determine the columns of the matrix and then evaluate the required rank or determinant.

Answer 1

The Correct Option is B

Concept: The columns of a matrix can be found using the images of standard basis vectors. Also, \[ \operatorname{rank}(A-I) = 3-\text{nullity}(A-I) \] using Rank–Nullity theorem.

Step 1: Find columns of matrix \(A\).
Given \[ A \begin{bmatrix} 001 \end{bmatrix} = \begin{bmatrix} 123 \end{bmatrix} \] Hence third column of \(A\) is \[ C_3= \begin{bmatrix} 123 \end{bmatrix} \] Now \[ A \begin{bmatrix} 101 \end{bmatrix} = C_1+C_3 = \begin{bmatrix} 10-1 \end{bmatrix} \] Thus \[ C_1 = \begin{bmatrix} 10-1 \end{bmatrix} - \begin{bmatrix} 123 \end{bmatrix} = \begin{bmatrix} 0-2-4 \end{bmatrix} \] Similarly, \[ A \begin{bmatrix} 110 \end{bmatrix} = C_1+C_2 = \begin{bmatrix} 110 \end{bmatrix} \] Hence \[ C_2 = \begin{bmatrix} 110 \end{bmatrix} - \begin{bmatrix} 0-2-4 \end{bmatrix} = \begin{bmatrix} 134 \end{bmatrix} \] Therefore \[ A= \begin{bmatrix} 0& 1& 1 -2& 3& 2 -4& 4& 3 \end{bmatrix} \]

Step 2: Find \(A-I\).
\[ A-I= \begin{bmatrix} -1& 1& 1 -2& 2& 2 -4& 4& 2 \end{bmatrix} \]

Step 3: Compute rank.
\[ R_2\to R_2-2R_1 \] \[ R_3\to R_3-4R_1 \] gives \[ \begin{bmatrix} -1& 1& 1 0& 0& 0 0& 0&-2 \end{bmatrix} \] There are two non-zero rows. Hence \[ \boxed{\operatorname{rank}(A-I)=2} \]