Let A=[gathered a_ij gathered]=[array|c|c| _5128 & _45 _58 & _425 array].

Question

Let $A=[\begin{gathered} {a}_{ij} \end{gathered}]=[\begin{array}{|c|c|} {\log }_{5}128 & {\log }_{4}5 \ {\log }_{5}8 & {\log }_{4}25 \end{array}]$. If ${A}_{ij}$ is the cofactor of ${a}_{ij}$, ${C}_{ij}=\sum _{k=1}^{2}{a}_{ik}{A}_{jk},1\le i,j\le 2$, and $C=[{C}_{ij}]$, then $8|C|$ is equal to :

Accepted Answer

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

$A=[\begin{gathered} {\log }_{5}128 {\log }_{4}5 \ {\log }_{5}8 {\log }_{4}25 ]$

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 $|A|$:

$|A|=(\frac{{\log }_{10}128}{{\log }_{10}5})(\frac{{\log }_{10}25}{{\log }_{10}4})-(\frac{{\log }_{10}5}{{\log }_{10}4})(\frac{{\log }_{10}8}{{\log }_{10}5})$

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 ${C}_{ij}=\sum _{k=1}^{2}{a}_{ik}{A}_{jk}$:

${C}_{11}={a}_{11}{A}_{11}+{a}_{12}{A}_{12}=|A|=\frac{11}{2}$

${C}_{12}={a}_{11}{A}_{21}+{a}_{12}{A}_{22}=0$

${C}_{21}={a}_{21}{A}_{11}+{a}_{22}{A}_{12}=0$

${C}_{22}={a}_{21}{A}_{21}+{a}_{22}{A}_{22}=|A|=\frac{11}{2}$

Thus, matrix $C$ is:

$C=[ \frac{11}{2} 0 \ 0 \frac{11}{2} \end{gathered}]$

To find $|C|$, we compute:

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

Finally, calculate $8|C|$:

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

Solutions · 1