If 2132A−325−3=1001 ,then the matrix A equals

let   B=2132   and c=−325−3 given   BAC=I⇒B−1(BAC)=B−1I⇒ACC−1=B−1C−1⇒AI=B−1C−1  ∴A=(B−1)(C−1)now    B−1=14−32−1−32=2−1−32C−1=19−10−3−2−5−3=32−32(B−1)(C−1)=2−1−323253=1110A=1110