The inverse of matrix N = $\begin{vmatrix} 2 & 3 \\ 1 & 4\end{vmatrix}$ is

Step 1: Analyze the question We need the inverse of a 2×2 matrix N = [[2, 3], [1, 4]].

Step 2: The concept For a matrix [[a, b], [c, d]], the inverse is:

N⁻¹ = (1/det) × [[d, -b], [-c, a]], where det = ad - bc.

In words: swap the main diagonal entries (a and d), change the signs of the other two (b and c), and divide by the determinant.

Step 3: Solution

• a = 2, b = 3, c = 1, d = 4
• Determinant = (2)(4) - (3)(1) = 8 - 3 = 5
• Swap 2 and 4, negate 3 and 1:

N⁻¹ = (1/5) [[4, -3], [-1, 2]]

Step 4: Check Multiply N by this matrix (ignoring the 1/5 first):

• Row 1: (2×4 + 3×(-1), 2×(-3) + 3×2) = (5, 0)
• Row 2: (1×4 + 4×(-1), 1×(-3) + 4×2) = (0, 5)

This gives 5I, and dividing by 5 gives the identity matrix I. ✓

Tip: Remember "swap the diagonal, negate the others, divide by the determinant." Option A is the transpose, and C and D do not give the identity when multiplied by N. If the determinant were 0, no inverse would exist, so always compute it first.