Discussion Forum

Let P=$[a_{ij}]$ be 3x3 matrix and let Q=$[b_{ij}]$,  where $b_{ij}=2^{i+j}a_{ij}$ for  $1 \leq i, j \leq 3$ . If the determinant of P is 2, then the determinant of the matrix Q is 

Answer:

Explanation:

Concept Involved it is a simple question on scalar  multipication  i.e, 

$\begin{bmatrix}ka_{1} & ka_{2}&ka_{3} \\b_{1} & b_{2}&b_{3}\\c_{1}&c_{2}&c_{3} \end{bmatrix}=k \begin{bmatrix}a_{1} & a_{2}&a_{3} \\b_{1} & b_{2}&b_{3}\\c_{1}&c_{2}&c_{3} \end{bmatrix}$

Description of Situation Construction of Matrix

i.e, if A=$[a_{ij}]_{3\times 3}= \begin{bmatrix}a_{11} & a_{12}&a_{13} \\a_{21} & a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{bmatrix}$

 sol. here , P=$[a_{ij}]_{3\times 3}= \begin{bmatrix}a_{11} & a_{12}&a_{13} \\a_{21} & a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{bmatrix}$

 $Q=[b_{ij}]_{3x3}= \begin{bmatrix}b_{11} & b_{12}&b_{13} \\b_{21} & b_{22}&b_{23}\\b_{31}&b_{32}&b_{33} \end{bmatrix}$

 where    $b_{ij}=2^{i+j}a_{ij}$

 $\therefore$   $|Q|= \begin{bmatrix}4a_{11} & 8a_{12}&16a_{13} \\8a_{21} &16 a_{22}&32a_{23}\\16a_{31}&32a_{32}&64a_{33} \end{bmatrix}$

   =$4 \times 8 \times 16  \begin{bmatrix}a_{11} & a_{12}&a_{13} \\2a_{21} & 2a_{22}&2a_{23}\\4a_{31}&4a_{32}&4a_{33} \end{bmatrix}$

=$2^{9} \times 2 \times 4  \begin{bmatrix}a_{11} & a_{12}&a_{13} \\a_{21} & a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{bmatrix}$

 =$2^{12}.|P|=2^{12}.2=2^{13}$