Question

If \(A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\), then which of the following is not true?

• A. \(A^{-1}=A^{2}\)
• B. \(A^{2}\) is a triangular matrix.
• C. \(A\) is a non-singular matrix.
• D. \(A^{3}\) is an identity matrix.

Hint:

Compute the determinant, A squared and A cubed, then test each statement.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Check every option by computing the determinant, \(A^{2}\) and \(A^{3}\) for the given matrix.

Step 2: Non-singular
Expanding along the third row, \(|A|=1\times\begin{vmatrix}-1&1\\-1&0\end{vmatrix}=1\neq0\). So \(A\) is non-singular and option (C) is true.

Step 3: Powers of A
\(A^{2}=\begin{bmatrix}0&0&1\\0&-1&2\\1&-1&1\end{bmatrix}\) and \(A^{3}=A^{2}A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I\).
So (D) is true, and \(A^{3}=I\) gives \(A^{-1}=A^{2}\), so (A) is true.

Step 4: Triangular Check
\(A^{2}\) has non-zero entries above and below the main diagonal, so it is not a triangular matrix. Option (B) is not true.

Final Answer:
The statement that A squared is a triangular matrix is not true. \[ \boxed{\text{(B) }\text{A}^{2}\text{ is a triangular matrix.}} \]