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Question is $$ \text{adj} B = \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right] $$

If $\det B = −202$, then find the $3\times 3$ matrix $B$, whose adjoint is given as above.

First, I wrote the $3x3$ matrix as $$\begin{bmatrix}a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}$$. Then I found the cofactor matrix and typed its transpose. As a result, $$ \begin{cases} -fh&+ei&=-45 \\ -id&+fg&=32 \\ dh&-eg&=-4 \\ -ib&+ch&=33 \\ ia&-cg&=-10 \\ -ah&+bg&=-24 \\ bf&-ec&=-28 \\ -af&+cd&=-16 \\ ea&-bd&=2 \end{cases} $$

came from the transactions I made. Then I applied the determinant formulas and replaced what I had just found. $$\begin{cases}-45a+32b-4c=-202 \\ 33d-10e-24f=-202\\ -28g-16h+2i=-202\end{cases}$$ But I got stuck here. How can I find $a, b, c, d, e, f, g, h, i$ here? I will be grateful if you could help me.

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$B^{-1} = \frac{adj(B)} {\det B} \implies B = \frac{(adj(B))^{-1}}{\det B} $

Can you do now?

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    $\begingroup$ Thank you very much :) I did :) $\endgroup$
    – user874333
    Jan 15 '21 at 11:29
  • $\begingroup$ @Deniz you're welcome; you can show your appreciation by accepting the answer because it was the first ;) $\endgroup$
    – utkarshk5
    Jan 16 '21 at 7:04
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The adjugate satisfies the relation $$ B\operatorname{adj}B=(\det B)I\qquad(*) $$ and therefore, if $B$ is invertible, $$ B=(\det B)(\operatorname{adj}B)^{-1} $$ Your final linear system is part of the relation $(*)$, if you notice, but you forgot some equations.

With some patience, you get $$ (\operatorname{adj}B)^{-1}= \begin{bmatrix} -1/101 & 3/202 & -2/101 \\ 0 & -1/202 & -4/101 \\ -2/101 & -3/101 & -3/202 \end{bmatrix} $$

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  • $\begingroup$ Thank you very much :) I did :) $\endgroup$
    – user874333
    Jan 15 '21 at 11:30
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You don't have to solve it backward. See this solution, please. Given

$ \mathbf {detB} = −202$, and

$ \mathbf{adj} B = \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right] $,

then, $ \mathbf{B^{-1}} = \frac{1}{detB} . adjB$.

This implies that, $\mathbf {B^{-1}} = -\frac{1}{202} \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right] $.

Now take $\mathbf{B} = \left[ \begin{array}{ccc} a& b& c\\ d& e& f\\ g& h& i\\ \end{array} \right] $.

Using invertibility relation, $\mathbf{BB^{-1}} = \mathbf{B^{-1}B} = \mathbf{I}$, i.e., $$ -\frac{1}{202} \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right] \left[ \begin{array}{ccc} a& b& c\\ d& e& f\\ g& h& i\\ \end{array} \right] = \left[ \begin{array}{ccc} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ \end{array} \right] $$.

Using simple matrix multiplication, you'll get the following system of equations by using $B^{-1}$ against the first, second, and third columns of the matrix $B$ in that order.

$$ (\mathbf{1st}) \rightarrow -\frac{1}{202} \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right] \left[ \begin{array}{ccc} a\\ d\\ g\\ \end{array} \right] = \left[ \begin{array}{c} 1\\ 0\\ 0\\ \end{array} \right] $$

$$ (\mathbf{2nd}) \rightarrow -\frac{1}{202} \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right] \left[ \begin{array}{ccc} b\\ e\\ h\\ \end{array} \right] = \left[ \begin{array}{c} 0\\ 1\\ 0\\ \end{array} \right] $$

$$ (\mathbf{3rd}) \rightarrow -\frac{1}{202} \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right] \left[ \begin{array}{ccc} c\\ f\\ i\\ \end{array} \right] = \left[ \begin{array}{c} 0\\ 0\\ 1\\ \end{array} \right] $$

Now, you can multiply both sides, in each of the equations, by $-202$ to eliminate the $-\frac{1}{202}$. Then, solve the 3 equations to get the first, second, and third columns of the matrix $B$. If you solve it correctly, you'll get the matrix $B$ below $$ \mathbf (1st) \rightarrow \left[ \begin{array}{ccc} a\\ d\\ g\\ \end{array} \right] = \left[ \begin{array}{c} 2\\ 0\\ 4\\ \end{array} \right] $$ $$ \mathbf (2nd) \rightarrow \left[ \begin{array}{ccc} b\\ e\\ h\\ \end{array} \right] = \left[ \begin{array}{c} -3\\ 1\\ 6\\ \end{array} \right] $$ $$ \mathbf (3rd) \rightarrow \left[ \begin{array}{ccc} c\\ f\\ i\\ \end{array} \right] = \left[ \begin{array}{c} 4\\ 8\\ 3\\ \end{array} \right] $$

Hence, $$ \mathbf B = \left[ \begin{array}{ccc} 2& -3& 4 \\ 0& 1& 8\\ 4& 6& 3\\ \end{array} \right] $$

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