Question

If the matrix A = \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\), then which of the following is true?

• A. The matrix is invertible
• B. The matrix is singular
• C. The matrix is diagonalizable
• D. The matrix is symmetric

Hint:

For a 3x3 matrix with elements in an arithmetic progression like this one, the determinant is always zero. Notice that the elements in each row (1,2,3), (4,5,6), (7,8,9) and each column (1,4,7), (2,5,8), (3,6,9) are in AP. A property of determinants states that if we perform the operation \(C_2 \rightarrow C_2 - C_1\) and \(C_3 \rightarrow C_3 - C_2\), the new columns will be identical, making the determinant zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a 3x3 matrix A and asked to determine its properties from the given options.
The key properties to check are whether the matrix is singular, invertible, or symmetric.

Step 2: Key Formula or Approach:
A matrix is

singular if its determinant is zero (\(\det(A) = 0\)).
A matrix is

invertible (or non-singular) if its determinant is non-zero (\(\det(A) \neq 0\)).
A matrix is

symmetric if it is equal to its transpose (\(A = A^T\)).
We will calculate the determinant of A to check if it is singular or invertible.

Step 3: Detailed Explanation:
The given matrix is:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \]
Let's calculate the determinant of A:
\[ \det(A) = 1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} \]
\[ \det(A) = 1(5 \times 9 - 6 \times 8) - 2(4 \times 9 - 6 \times 7) + 3(4 \times 8 - 5 \times 7) \]
\[ \det(A) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) \]
\[ \det(A) = 1(-3) - 2(-6) + 3(-3) \]
\[ \det(A) = -3 + 12 - 9 \]
\[ \det(A) = 0 \]
Since the determinant of A is 0, the matrix is singular.
This also means the matrix is not invertible.
Now, let's check if the matrix is symmetric.
The transpose of A is:
\[ A^T = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} \]
Since \(A \neq A^T\), the matrix is not symmetric.

Step 4: Final Answer:
The determinant of the matrix A is 0. Therefore, the matrix is singular.