Question

Inverse of \( \begin{bmatrix} 1 & 2 2 & 4 \end{bmatrix} \) ____.

• A. Exist
• B. Does not exist
• C. Unit matrix
• D. None

Hint:

A quick visual check: if one row (or column) is a multiple of another, the determinant is always zero. Here, the second row $[2, 4]$ is exactly double the first row $[1, 2]$.

Answer 1

The Correct Option is B

Step 1: Suppose, for contradiction, that the inverse \( A^{-1} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) does exist for \( A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \), so that \( A \cdot A^{-1} = I \).

Step 2: Multiplying out the first column of this product gives two equations from the two rows of \( A \): \( a + 2c = 1 \) (from row 1) and \( 2a + 4c = 0 \) (from row 2).

Step 3: But the second equation simplifies to \( a + 2c = 0 \) after dividing by 2, directly contradicting the first equation, which requires \( a + 2c = 1 \). Since \( a + 2c \) cannot equal both \( 0 \) and \( 1 \) at once, no such \( a, b, c, d \) can exist.

This contradiction confirms that no inverse can exist for this matrix: \[ \boxed{\text{Inverse does not exist}} \]