Question

If $\mathrm{A}=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$, then $\mathrm{A}(\text{adj }\mathrm{A}) =$

• A. $\left[\begin{array}{ccc} -1/3 & 0 & 0 \\ 0 & -1/3 & 0 \\ 0 & 0 & -1/3 \end{array}\right]$
• B. $\left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$
• C. $\left[\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]$
• D. $\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 4 \end{array}\right]$

Hint:

Never spend time calculating all 9 cofactors to construct the adjugate matrix when a question asks for $A(\text{adj } A)$. The theorem $A(\text{adj } A) = |A|I$ ensures the answer must be a scalar diagonal matrix. Calculating the single scalar determinant $3$ points directly to option (B) in under twenty seconds!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a $3 \times 3$ square matrix $A$. We need to compute the matrix product $A(\text{adj } A)$.

Step 2: Key Formula or Approach:
According to the properties of square matrices, the product of a matrix and its adjugate matrix is always equal to the determinant of the matrix multiplied by the identity matrix ($I$): $$A(\text{adj } A) = |A| \cdot I$$ Therefore, we only need to compute the determinant $|A|$ to find the final matrix representation.

Step 3: Detailed Explanation:
Calculate the determinant of matrix $A$ by expanding along the first row: $$|A| = \begin{vmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{vmatrix}$$ $$|A| = 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} - 2 \cdot \begin{vmatrix} -1 & 2 \\ 1 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix}$$ Evaluate each $2 \times 2$ minor determinant: First minor: $(1)(4) - (2)(2) = 4 - 4 = 0$ Second minor: $(-1)(4) - (2)(1) = -4 - 2 = -6$ Third minor: $(-1)(2) - (1)(1) = -2 - 1 = -3$ Combine the components: $$|A| = 1(0) - 2(-6) + 3(-3)$$ $$|A| = 0 + 12 - 9 = 3$$ Now, substitute this value into our matrix property formula: $$A(\text{adj } A) = 3 \cdot I = 3 \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$$

Step 4: Final Answer:
The resulting matrix matches option (B).