Let A=[a_ij]=pmatrix_5128 & _45 _58 & _425pmatrix.

Question

Let $A=[a_{ij}]=\begin{pmatrix}\log_5128 & \log_45\ \log_58 & \log_425\end{pmatrix}$. If $A_{ij}$ is the cofactor of $a_{ij}$, and $C_{ij}=\sum_{k=1}^{2}a_{ik}A_{jk}$, $1\le i,j\le2$, and $C=[C_{ij}]$, then $8|C|$ is equal to:

Accepted Answer

Using the change of base formula, $\log_5128=\dfrac{\log128}{\log5}$, $\log_45=\dfrac{\log5}{\log4}$, $\log_58=\dfrac{\log8}{\log5}$, $\log_425=\dfrac{\log25}{\log4}$.

$$|A|=(\log_5128)(\log_425)-(\log_45)(\log_58)=\frac{\log128\cdot\log25-\log5\cdot\log8}{\log5\cdot\log4}.$$

The cofactors are $A_{11}=\log_425,\ A_{12}=-\log_58,\ A_{21}=-\log_45,\ A_{22}=\log_5128$.

$C_{11}=a_{11}A_{11}+a_{12}A_{12}=|A|$, $C_{22}=a_{21}A_{21}+a_{22}A_{22}=|A|$, and $C_{12}=C_{21}=0$ (standard identity $\sum a_{ik}A_{jk}=|A|\delta_{ij}$).

Numerically $|A|=\dfrac{11}2$, so $C=\begin{pmatrix}\frac{11}2 & 0\0 & \frac{11}2\end{pmatrix}$, and $|C|=\left(\dfrac{11}2\right)^2=\dfrac{121}4$.

$8|C|=8\times\dfrac{121}4=242$.

Solutions · 1