Step 1 (Formula / Governing Law): (1/2 Mark)

By the definition of the inverse of a matrix, if $A$ is a matrix and $B$ is its inverse, then $AB = I$, where $I$ is the identity matrix of the same order. Let:

$$A = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 3 & 3 \\ 1 & \lambda & 3 \\ 1 & 3 & 4 \end{bmatrix}$$

The condition $AB = I$ implies:

$$\begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 3 \\ 1 & \lambda & 3 \\ 1 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Step 2 (Step-by-Step Working): (1/4 Mark)

To find the value of $\lambda$, we can compute the element at the second row and second column ($c_{22}$) of the product matrix $AB$. According to matrix multiplication:

$$c_{22} = (\text{Row 2 of } A) \cdot (\text{Column 2 of } B)$$ $$c_{22} = (-1 \times 3) + (1 \times \lambda) + (0 \times 3)$$

Since $AB = I$, the element $c_{22}$ must be equal to the element at the second row and second column of the identity matrix, which is $1$.

$$-3 + \lambda + 0 = 1$$ $$\lambda = 1 + 3$$ $$\lambda = 4$$

Step 3 (Final Answer): (1/4 Mark)

Comparing the calculated value with the given options:

• (A) -4
• (B) 1
• (C) 3
• (D) 4

The correct option is (D). The value of $\lambda$ is 4.