Question

Challenge For Mod and Brainly star ​

Answer 1

Given,

$\longrightarrow AB=rI$

Taking the determinant,

$\longrightarrow |AB|=|rI|$

Since A and B are of order 3 each,

$\longrightarrow AB=\left[\begin{array}{ccc}r&0&0\\0&r&0\\0&0&r\end{array}\right]$

$\longrightarrow |AB|=r^3$

$\longrightarrow |A||B|=r^3\quad\quad\dots(1)$

Then,

$\longrightarrow |\mathrm{adj}(AB)|=|AB|^2$

[because $|\mathrm{adj}(A)|=|A|^{n-1}$ if $n$ is order of square matrix A.]

$\longrightarrow |\mathrm{adj}(AB)|=r^6\quad\quad\dots(2)$

Let $(ab)_{ij}$ be the element of AB in i'th row and j'th column.

The element $(ab)_{12}$ is given by,

$\longrightarrow a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32}=(ab)_{12 }$

$\longrightarrow \alpha b_{12}+b_{22}-3r =0$

$\longrightarrow \alpha b_{12}+b_{22}=3r\quad\quad\dots(i)$

The element $(ab)_{22}$ is given by,

$\longrightarrow a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32}=(ab)_{22 }$

$\longrightarrow b_{12}+2b_{22}-\dfrac{9}{2}\,r =r$

$\longrightarrow b_{12}+2b_{22}=\dfrac{11}{2}\,r\quad\quad\dots(ii )$

The element $(ab)_{32}$ is given by,

$\longrightarrow a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32}=(ab)_{32 }$

$\longrightarrow 3b_{12}+b_{22}-\dfrac{3}{2}\,r=0$

$\longrightarrow 3b_{12}+b_{22}=\dfrac{3}{2}\,r\quad\quad\dots(iii )$

Solving (ii) and (iii) we get,

• $b_{12}=-\dfrac{r}{2 }$

• $b_{22}=3r$

Then from (i), since $r\neq0,$ we get,

$\longrightarrow \alpha=0$

So,

$\longrightarrow A=\left[\begin{array}{ccc}0&1&2\\1&2&3\\3&1&1\end{array}\right]$

$\longrightarrow |A|=(3\cdot3-1\cdot1)+2(1\cdot1-2\cdot3)$

$\longrightarrow |A|=-2$

Given,

$\longrightarrow |\mathrm{adj}(B) |=r^5$

$\longrightarrow |B|^2 =r^5$

$\longrightarrow |B|=r^{\frac{5}{2}}$

Then (1) becomes,

$\longrightarrow -2r^{\frac{5}{2}}=r^3$

$\longrightarrow \left( -2r^{\frac{5}{2}} \right)^2=\left(r^3\right)^2$

$\longrightarrow 4r^5=r^6$

Since $r\neq0,$

$\longrightarrow r= 4$

And then (2) becomes,

$\longrightarrow |\mathrm{adj}(AB)| =4 ^6$

$\longrightarrow\underline{\underline{|\mathrm{adj}(AB)|=2^{12}}}$

Hence (2) is the answer.