Simple Matrix Equation.

Question

$$(2\begin{bmatrix}2 &1 \\-1 &3\end{bmatrix} - 5A^{-1})^{T} = (4A^{T})^{-1}$$

I have approached this question by inverting the transpose and inverse operation on the LHS and then distributing the transpose on the RHS on both matrices. But I tried my solution on Matlab and it seems a wrong approach. How shall I solve this equation?

Answer 1

Setting $B:=\begin{bmatrix} 2 & 1\\ -1 & 3\end{bmatrix}$, then solving for the matrix $A$, the equation $$(2B-5A^{-1})^{T}=(4A^{T})^{-1}$$ we have \begin{align*} (2B-5A^{-1})^{T}=(4A^{T})^{-1} &\iff (2B)^{T}-(5A^{-1})^{T}=(4A^{T})^{-1}\\ &\iff (4A^{T})^{-1}+(5A^{-1})^{T}=(2B)^{T}\\ &\iff (4A^{T})^{-1}+5(A^{-1})^{T}=(2B)^{T}\\ &\iff (4A^{T})^{-1}+5(A^{T})^{-1}=(2B)^{T}\\ & \iff \frac{1}{4}(A^{T})^{-1}+5(A^{T})^{-1}=(2B)^{T}\\ &\iff \left( \frac{1}{4}+5\right)(A^{T})^{-1}=(2B)^{T}\\ &\iff (A^{T})^{-1}=\frac{4}{21}(2B)^{T},\\ &\iff (A^{-1})^{T}=\left(\frac{4\cdot 2}{21}\right)B^{T}\\ &\iff ((A^{-1})^{T})^{T}=\left(\left(\frac{4\cdot 2}{21}\right)B^{T}\right)^{T}\\ &\iff A^{-1}=\left( \frac{4\cdot 2}{21} \right)B \end{align*} then the result is follows.

Remark:

• $(\lambda A)^{-1}=\frac{1}{\lambda}A^{-1}$ with $\lambda\not=0$.
• $(A^{T})^{-1}=(A^{-1})^{T}$.
• $(\lambda A)^{T}=(\lambda) A^{T}$
• $(A^{T})^{T}=A$.

with all the hypotheses well-known.

Answer 2

Since $(A^{-1})^T)=(A^{T})^{-1}$ you can easily get to the fact that $$\frac{21}{4}A^{-1} = 2 \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}$$ or $$A^{-1} = \begin{bmatrix} 16/21 & 8/21 \\ -8/21 & 8/7 \end{bmatrix}$$