Only square matrices have a determinant
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I assume since you're asking if 2x2 invertible matrices are a "subspace" that you are considering the set of all 2x2 matrices as a vector space (which it certainly is). In order for the set of 2x2 invertible matrices to be a subspace of the set of all 2x2 matrices, it must be closed under addition and scalar multiplication. A 2x2 matrix is invertible if and only if its determinant is nonzero. When multiplied by a scalar (let's call it c), the determinant of a 2x2 matrix will be multiplied by c^2 since the determinant is linear in each row (two rows -> two factors of c). If the determinant was nonzero to begin with c^2 times the determinant will be nonzero, so an invertible matrix multiplied by a scalar will remain invertible. Therefore the set of all 2x2 invertible matrices is closed under scalar multiplication. However, this set is not closed under addition. Consider the matrices {[1 0], [0 1]} and {[-1 0], [0 -1]}. Both are invertible (in this case, they are both their own inverses). However, their sum is {[0 0], [0 0]}, which is not invertible because its determinant is 0. In conclusion, the set of invertible 2x2 matrices is not a subspace of the set of all 2x2 matrices because it is not closed under addition.
It isn't clear what you want to solve for. If you want to find the matrix, there is not a unique solution - there are infinitely many matrices with the same determinant.
actually MATRICES is the plural of matrix which means the array of numbers in groups and columns in a rectangular table... and determinant is used to calculate the magnitude of a matrix....
No. Determinants are only defined for square matrices.No. Determinants are only defined for square matrices.
That's a special calculation done on square matrices - for example, on a 2 x 2 matrix, or on a 3 x 3 matrix. For details, see the Wikipedia article on "Determinant".