How do I go about solving this for z?
$$ \begin{pmatrix} a & b & c \\ 1 & 2 & 3 \\ d & e & f \\ \end{pmatrix} ^{-1}= \begin{pmatrix} 1 & 2 & 3 \\ x & y & z \\ 4 & 5 & 6 \\ \end{pmatrix} $$
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The long way is to write out the equations that give you the inverse of either matrix and then set about solving them. – abiessu Apr 20 '21 at 23:10
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Have you tried taking the augmented matrix $(A|I)$? – Anik Bhowmick Apr 20 '21 at 23:10
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How would I go about taking the augmented matrix? – coolkid3127 Apr 20 '21 at 23:15
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what about multiplying on matrix with abc to get identity matrix on the left? It should lead to $2z + 3 + 18 = 0$, thus $z=21/2=10.5$ – guest Apr 20 '21 at 23:29
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As : $$\underbrace{\begin{pmatrix} a & b & c \\ \color{red}{1} & \color{red}{2} & \color{red}{3} \\ d & e & f \\ \end{pmatrix}}_A \underbrace{\begin{pmatrix} 1 & 2 & \color{red}{3} \\ x & y & \color{red}{z} \\ 4 & 5 & \color{red}{6} \\ \end{pmatrix}}_B= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & \color{red}{0} \\ 0 & 0 & 1 \\ \end{pmatrix}$$
multiplication of the second row of $A$ by the third column of $B$ will give $\color{red}{0}$, i.e.,
$$21+2z=0 \implies z=-21/2$$
Besides, it isn't evident a priori that a global solution exists with these constraints. I have checked using a Computer Algebra System that there is indeed a unique solution with this data:
$$a=-3/2, b=-1, c=-1, d=1/6, e=-1, f=-5/3, x=-13/2, y=-8, z=-21/2$$
Jean Marie
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The equation $A^{-1} = B$ implies that $AB = I_3$ where $I$ is the identity matrix, so there are two ways you could go about this:
- Take the second equation I wrote and multiply it out, i.e. start with
$\left[\begin{matrix} a & b & c \\ 1 & 2 & 3 \\ d & e & f \end{matrix}\right] \left[\begin{matrix} 1 & 2 & 3 \\ x & y & z \\ 4 & 5 & 6 \end{matrix}\right] = \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$
and then do the full multiplication to get a set of equations, which you can then use to find an equation (or possibly multiple) for $z$ to solve.
- Use the standard method for manually finding matrix inverses - take the augmented matrix and row-reduce it:
$\left[\begin{array}{c c c | c c c} a & b & c & 1 & 0 & 0 \\ 1 & 2 & 3 & 0 & 1 & 0 \\ d & e & f & 0 & 0 & 1\end{array}\right] \rightarrow \left[\begin{array}{c c c | c c c} 1 & 0 & 0 & ? & ? & ? \\ 0 & 1 & 0 & ? & ? & ? \\ 0 & 0 & 1 & ? & ? & ? \end{array}\right]$
You should find that the question marks match up with the second matrix in your question, and hence you can find the value of $z$ in there.
ConMan
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