Question

Let \(A=[a_{ij}]\), \(\det(A)\neq 0\), and \(B=[b_{ij}]\) be two \(3\times 3\) matrices. If \[ b_{ij}=3^{\,i-j}\,a_{ij}\quad \text{for all } i,j=1,2,3, \] then:

• A. \(3\det(A)=\det(B)\)
• B. \(27\det(A)=\det(B)\)
• C. \(\det(A)=\det(B)\)
• D. \(\det(A)=27\det(B)\)

Hint:

When each element is multiplied by \(k^{i-j}\), separate the effect into \textbf{row scaling} and \textbf{column scaling}. If total row and column powers cancel, the determinant remains unchanged.

Answer 1

The Correct Option is C

Step 1: Interpret the given transformation Given: \[ b_{ij}=3^{i-j}a_{ij}=3^i\cdot 3^{-j}\cdot a_{ij} \] This means: \begin{itemize} \item Each row \(i\) of matrix \(A\) is multiplied by \(3^i\). \item Each column \(j\) is multiplied by \(3^{-j}\). \end{itemize} Step 2: Effect on determinant For determinants: \begin{itemize} \item Multiplying row \(i\) by \(k\) multiplies determinant by \(k\). \item Multiplying column \(j\) by \(k\) multiplies determinant by \(k\). \end{itemize} Hence, total multiplying factor is: \[ \frac{(3^1\cdot 3^2\cdot 3^3)}{(3^1\cdot 3^2\cdot 3^3)} =3^{(1+2+3)-(1+2+3)}=3^0=1 \] Step 3: Final result \[ \det(B)=\det(A) \] \[ \boxed{\det(A)=\det(B)} \]