Question

The value of the determinant \(\begin{vmatrix} 4&4^2&4^3\\3&3^2&3^3\\2&2^2&2^3 \end{vmatrix}\)is

• A. 52
• B. -24
• C. 24
• D. 48
• E. -48

Answer 1

Answer: (E)

We are asked to find the value of the determinant:

\(\begin{vmatrix} 4 & 4^2 & 4^3 \\ 3 & 3^2 & 3^3 \\ 2 & 2^2 & 2^3 \end{vmatrix}\)

This is a 3x3 matrix where the elements are powers of 4, 3, and 2. We can calculate the determinant by expanding along the first row:

\(\text{det} = 4 \cdot \begin{vmatrix} 3^2 & 3^3 \\ 2^2 & 2^3 \end{vmatrix} - 4^2 \cdot \begin{vmatrix} 3 & 3^3 \\ 2 & 2^3 \end{vmatrix} + 4^3 \cdot \begin{vmatrix} 3 & 3^2 \\ 2 & 2^2 \end{vmatrix}\) 

Now, calculate the 2x2 determinants:

\(\begin{vmatrix} 3^2 & 3^3 \\ 2^2 & 2^3 \end{vmatrix} = \begin{vmatrix} 9 & 27 \\ 4 & 8 \end{vmatrix} = (9 \times 8) - (27 \times 4) = 72 - 108 = -36\)

\(\begin{vmatrix} 3 & 3^3 \\ 2 & 2^3 \end{vmatrix} = \begin{vmatrix} 3 & 27 \\ 2 & 8 \end{vmatrix} = (3 \times 8) - (27 \times 2) = 24 - 54 = -30\)

\(\begin{vmatrix} 3 & 3^2 \\ 2 & 2^2 \end{vmatrix} = \begin{vmatrix} 3 & 9 \\ 2 & 4 \end{vmatrix} = (3 \times 4) - (9 \times 2) = 12 - 18 = -6\)

Substitute these values back into the determinant expansion:

\(\text{det} = 4 \cdot (-36) - 16 \cdot (-30) + 64 \cdot (-6)\)

\(\text{det} = -144 + 480 - 384 = -48\)

The value of the determinant is \( -48 \).