Let A = bmatrix a_ij bmatrix = bmatrix _5 128 & _4 5 _5 8 & _4 25 bmatrix .

Question

Let $ A = \begin{bmatrix} a_{ij} \end{bmatrix} = \begin{bmatrix} \log_5 128 & \log_4 5 \ \log_5 8 & \log_4 25 \end{bmatrix} $. If $ A_{ij} $ is the cofactor of $ a_{ij} $, $ C_{ij} = \sum\limits_{k=1}^{2} a_{ik} A_{jk} , 1 \leq i, j \leq 2 $, and $ C=[C_{ij}] $, then $ 8|C| $ is equal to :

Accepted Answer

To solve the problem, we need to determine the determinant of matrix $ A $:

$  A = \begin{bmatrix}  \log_5 128 & \log_4 5 \  \log_5 8 & \log_4 25  \end{bmatrix}  $

The determinant of $ A $, denoted as $|A|$, is calculated as:

$  |A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8)  $

Evaluating each component:

$\log_5 128$ can be simplified using change of base formula:

$\log_5 128 = \frac{\log_{10}128}{\log_{10}5}$

$\log_4 5$ using change of base:

$\log_4 5 = \frac{\log_{10}5}{\log_{10}4}$

$\log_5 8$:

$\log_5 8 = \frac{\log_{10}8}{\log_{10}5}$

$\log_4 25$:

$\log_4 25 = \frac{\log_{10}25}{\log_{10}4}$

Now, substitute these values into $|A|$:

$ |A| = \left(\frac{\log_{10}128}{\log_{10}5}ight) \left(\frac{\log_{10}25}{\log_{10}4}ight) - \left(\frac{\log_{10}5}{\log_{10}4}ight) \left(\frac{\log_{10}8}{\log_{10}5}ight) $

Next, we find the cofactors of matrix $ A $:

$ A_{11} = \log_4 25 $

$ A_{12} = -\log_5 8 $

$ A_{21} = -\log_4 5 $

$ A_{22} = \log_5 128 $

Then, calculate matrix $ C $ whose elements are given by $ C_{ij} = \sum\limits_{k=1}^{2} a_{ik} A_{jk} $:

$ C_{11} = a_{11}A_{11} + a_{12}A_{12} = |A| = \frac{11}{2} $

$ C_{12} = a_{11}A_{21} + a_{12}A_{22} = 0 $

$ C_{21} = a_{21}A_{11} + a_{22}A_{12} = 0 $

$ C_{22} = a_{21}A_{21} + a_{22}A_{22} = |A| = \frac{11}{2} $

Thus, matrix $ C $ is:

$ C = \begin{bmatrix}  \frac{11}{2} & 0 \  0 & \frac{11}{2}  \end{bmatrix}  $

To find $|C|$, we compute:

$  |C| = \left(\frac{11}{2}ight)\left(\frac{11}{2}ight) = \frac{121}{4}  $

Finally, calculate $ 8|C| $:

$  8|C| = 8 	imes \frac{121}{4} = 242  $

Solutions · 1