Let me drop the matrix context, but rephrase your problem like this: You want to encrpyt a clear text $R$ into three (or more) cypher texts $I_1, I_2, I_3$ (probably handed to three persons) such that
- it is "impossible" to find $R$ with only one of the cyphers
- it is possible to find $R$ by combining any two of the cyphers.
This is like a safe that can be opened only if the majority of a comittee approves.
The three persons create RSA key pairs, keeping their own secret keys $S_i$ and publishing the public keys $P_i$. Let $C_1=P_2(P_3(R))$, $C_2=P_1(P_3(R))$, $C_3=P_1(P_2(R))$. Then let $C$ be the concatenation of $C_1$, $C_2$, $C_3$. This can be published. The role of the $I_i$ is played by $C$ together with $S_i$.
I am afraid that this method is not as hard to attack as it seems. Note that the first key holder can obtain $P_3(R)$ and $P_2(P_3(R))$. Having a clear text and the corresponding capher text might help in finding $S_2$!
Again, each user creates a key pair. Hand $P_2(R)$ to the first person, $P_3(R)$ to the second, and $P_1(R)$ to the third person.
Nobody has the correct key for the cypher in his hand, but when two people cooperate, one has the key for the cypher held by the othre.
Disadvantage: If persons 1 and 2 cooperate, the second can use $S_2$ do decode 1's $P_2(R)$. However, he may lie to preson 1 about the content.
Yet another Alternative:
Create a single temporary RSA key pair (private key $S$, public key $P$), compute and publish $C=P(R)$. Assume $S$ consists of $3n$ bits. For $i=1, 2, 3$, let $I_i$ be $S$ with all bits at positions $\equiv i\pmod 3$ replaced with $0$'s. (After that, $S$ must be "trustworthily discarded"). Then any two can reconstruct $S$ (and hence $R$ by decoding $C$) by ORing their partial secrets.
Even if we assume that brute force is the only attack against RSA, the key size in my second alternative must be three times as big as is usually considered sufficient (because a single partial secret owner knows already $2n$ of the $3n$ bits).