# Find the $3 \times 3$ matrix

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.

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

Can you do now?

• Thank you very much :) I did :)
– user874333
Jan 15 '21 at 11:29
• @Deniz you're welcome; you can show your appreciation by accepting the answer because it was the first ;) Jan 16 '21 at 7:04

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

• Thank you very much :) I did :)
– user874333
Jan 15 '21 at 11:30

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