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Sava's question via email about matrix multiplication

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Given the matrix $\displaystyle \begin{align*} A = \left[ \begin{matrix} 3 & 0 & -4 \\ 4 & 0 & \phantom{-}3 \\ 0 & 5 & \phantom{-}0 \end{matrix}\right] \end{align*}$ evaluate $\displaystyle \begin{align*} A\,A^T \end{align*}$ and $\displaystyle \begin{align*} A^T\,A \end{align*}$. Hence write down $\displaystyle \begin{align*} A^{-1} \end{align*}$.
$\displaystyle \begin{align*} A\,A^T &= \left[\begin{matrix} 3 & 0 & -4 \\ 4 & 0 & \phantom{-}3 \\ 0 & 5 & \phantom{-}0 \end{matrix}\right]\left[ \begin{matrix} \phantom{-}3 & 4 & 0 \\ \phantom{-}0 & 0 & 5 \\ -4 & 3 & 0 \end{matrix}\right] \\ &= \left[ \begin{matrix} 3\cdot 3 + 0 \cdot 0 + \left( -4 \right) \cdot \left( -4 \right) & 3 \cdot 4 + 0 \cdot 0 + \left( -4 \right) \cdot 3 & 3 \cdot 0 + 0 \cdot 5 + \left( -4 \right) \cdot 0 \\ 4 \cdot 3 + 0 \cdot 0 + 3 \cdot \left( -4 \right) & 4 \cdot 4 + 0 \cdot 0 + 3 \cdot 3 & 4\cdot 0 + 0 \cdot 5 + 3 \cdot 0 \\ 0 \cdot 3 + 5 \cdot 0 + 0 \cdot \left( -4 \right) & 0 \cdot 4 + 5 \cdot 0 + 0 \cdot 3 & 0 \cdot 0 + 5 \cdot 5 + 0 \cdot 0 \end{matrix} \right] \\ &= \left[ \begin{matrix} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{matrix} \right] \\ &= 25\,I \end{align*}$

Since post-multiplying $\displaystyle \begin{align*} A \end{align*}$ by $\displaystyle \begin{align*} A^T \end{align*}$ gave $\displaystyle \begin{align*} 25\,I \end{align*}$, it suggests that $\displaystyle \begin{align*} A^{-1} = \frac{1}{25}\,A^T \end{align*}$.

Of course, we must also check that $\displaystyle \begin{align*} A^T\,A = 25\,I \end{align*}$ as well. This can be left to the OP/Reader. IF this is the case, then $\displaystyle \begin{align*} A^{-1} = \frac{1}{25}\,A^T \end{align*}$.