Let the determinant of a square matrix A of order m be m-n, where m and n satisfy 4 m+n=22…

Question

Let the determinant of a square matrix A of order $m$ be $m-n$, where $m$ and $n$ satisfy $4 m+n=22$ and $17 m+4 n=93$. If $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A)))=3^{a} 5^{b} 6^{c}$ then $a+b+c$ is equal to :

Accepted Answer

Given that $|A|=m-n$, and let's solve the system of linear equations to find the values of $m$ and $n$ :

$4m + n = 22$ ...... (1)

$17m + 4n = 93$ ....... (2)

We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal:

$16m + 4n = 88$ ......... (3)

Now, subtract equation (3) from equation (2):

$(17m + 4n) - (16m + 4n) = 93 - 88$

$m = 5$

Now, we can find the value of $n$ by substituting $m$ into equation (1):

$4(5) + n = 22$

$20 + n = 22$

$n = 2$

Since the order of matrix A is $m$, the order is 5. Now, let's find the determinant of $n \operatorname{adj}(\operatorname{adj}(mA))$:

We know that $\operatorname{det}(A) = m-n = 3$.

$$
\begin{aligned}
& \therefore \operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A))) \\
& =|2 \operatorname{adj}(\operatorname{adj}(5 A))| \\
& =2^5|5 A|^{16} \\
& =2^5 5^{80}|A|^{16}=2^5 \cdot 3^{16} \cdot 5^{80} \\
& =3^{11} 5^{80} 6^5
\end{aligned}
$$

So, $a+b+c=96$

Solutions · 1