Missing Matrix Members

we could use $MM^{-1}= I$ and $M^{-1}M=I$ to get a total of 18 different equations, the strategy to solve this problem would be to optimally pick 8 that makes this easier.

let us first define some notation, let $_i A$ denote the ith row of $A$ , and $A_i$ denote the ith column of $A$ . also, notice by definition of the identity matrix $_i A A^{-1}_j = \delta_{ij}$ .

Notice that $_2 M$ and $M_2$ are independent of our variables, meaning products involving them will only be linear in each variable. furthermore notice some columns and rows of $M^{-1}$ contains only one variable. use this to your advantage to get the following: $\begin{array}{c}_2M M^{-1}_1 &&=&& \begin{pmatrix} 2 && 2 && 3 \end{pmatrix}\begin{pmatrix} -5\\ F \\ 2 \end{pmatrix}&&=&&-10 +2F+6 &&=&&0 \to && \boxed{F=2}\\ _2M M^{-1}_3 &&=&& \begin{pmatrix} 2 && 2 && 3 \end{pmatrix}\begin{pmatrix} -11\\ G \\ 4 \end{pmatrix}&&=&&-22 +2G+12 &&=&&0 \to && \boxed{G=5}\\ _1 M^{-1} M_2 &&=&& \begin{pmatrix} -5 && E && -11 \end{pmatrix}\begin{pmatrix} 1\\ 2 \\ 5 \end{pmatrix}&&=&&-5 +2E-55 &&=&&0 \to && \boxed{E=30}\\ _3 M^{-1} M_2 &&=&& \begin{pmatrix} 2 && H && 4 \end{pmatrix}\begin{pmatrix} 1\\ 2 \\ 5 \end{pmatrix}&&=&&2 +2H+20 &&=&&0 \to && \boxed{H=-11} \end{array}$ from this we have solved for $M^{-1}$ , i.e $M^{-1} = \begin{pmatrix} -5 & 30 & -11 \\ 2 & -13 & 5 \\ 2 & -11 & 4 \end{pmatrix}$ Now we solve for $A,B,C,D$ , which can be done as follows(note that other orders of equation would also work, there are many optimal solutions to this part): $\begin{array}{c}_1M M^{-1}_1 &&=&& \begin{pmatrix} A && 1 && B \end{pmatrix}\begin{pmatrix} -5\\ 2 \\ 2 \end{pmatrix}&&=&&-5A+2+2B &&=&& 1\to&& \\ _1M M^{-1}_2 &&=&& \begin{pmatrix} A && 1 && B \end{pmatrix}\begin{pmatrix} 30\\ -13 \\ -11 \end{pmatrix}&&=&&30A -13-11B &&=&&0 \to && \boxed{A=3,B=7}\\ _1 M^{-1} M_1 &&=&& \begin{pmatrix} -5 && 30 && -11 \end{pmatrix}\begin{pmatrix} 3\\ 2 \\ C \end{pmatrix}&&=&&-15 +60-11C &&=&&1 \to && \boxed{C=4}\\ _1 M^{-1} M_3 &&=&& \begin{pmatrix} -5 && 30 && -11 \end{pmatrix}\begin{pmatrix} 7\\ 3 \\ D \end{pmatrix}&&=&&-35 +90-11D &&=&&0 \to && \boxed{D=5} \end{array}$ of course, multiply the two matrices found to confirm that they are indeed inverses, as we skipped over some eqns.

We note that $MM^{-1} = I$ , where $I$ is the identity matrix. Let the elements of $I$ be $a_{ij}$ . Then we have:

$a_{21} = 2(-5) + 2F + 3(2) = 0 \implies F = 2 \\ a_{23} = 2(-11) + 2G + 3(4) = 0 \implies G = 5 \\ \begin{cases} a_{11} = -5A + 2(1) + 2B = 1 \\ a_{13} = -11A + 5(1) + 4B = 0 \end{cases} \implies \begin{cases} A = 3 \\ B = 7 \end{cases} \\ \begin{cases} a_{31} = -5C + 2(5) + 2D = 0 \\ a_{33} = -11C + 5(5) + 4D = 1 \end{cases} \implies \begin{cases} C = 4 \\ D = 5 \end{cases} \\ \begin{cases} a_{12} = 3E + 1(-13) + 7H = 0 \\ a_{32} = 4E + 5(-13) + 5H = 0 \end{cases} \implies \begin{cases} E = 30 \\ H = -11 \end{cases}$

Therefore, $A+B+C+D+E+F+G+H = 3+7+4+5+30+2+5-11 = \boxed{45}$ .

$a+b+c+d+e+f+g+h\text{/.}\, \text{Solve}\left[\left( \begin{array}{ccc} a & 1 & b \\ 2 & 2 & 3 \\ c & 5 & d \\ \end{array} \right).\left( \begin{array}{ccc} -5 & e & -11 \\ f & -13 & g \\ 2 & h & 4 \\ \end{array} \right)=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\right]\Rightarrow 45$