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**1. Find the adjoint of the matrix **

The co factor of a is d, the co factor of b is -c, the co factor of c is -b and the co factor of d is a. The matrix formed by the co factors taken in order is the co factor matrix of A.

The co factor matrix of A is = Taking transpose of the co factor matrix, we get the adjoint of A.

The adjoint of

**2. State and Prove Reversal Law for Inverses .**

Statement:

If A,B are any two non-singular matrices of the same order, then AB is also non-singular and

(AB)^{-1} = B^{-1}A^{-1} ie, the inverse of a product is product of inverses taken in the reverse order.

Proof

**3. Prove that "For any non-singular matrix A, (A ^{T})^{-1} = (A^{-1})^{T}.**

Proof

Taking transpose on both sides of AA^{-1} = I

We have (AA^{-1})^{T} = I^{T}

By reversal law for transposes, we get

(A^{-1})^{T} A^{T} = I → (1)

Similarly by taking transpose on both of A^{-1}A = I, we have

(A^{T}) (A^{-1})^{T} = I → (2)

From (1) and (2)

(A^{-1})^{T} A^{T} = A^{T} (A^{-1})^{T} = I

(A^{-1})^{T} is the inverse of A^{T}

A is a non-singular matrix. Hence it is invertible. The matrix formed by the co factors is

**7. Solve by matrix inversion method x + y = 3, 2x + 3y = 8.**

The given system of equations can be written in the form of

The highest order minor of A is given by . Since the second order minor vanishes . We have to try for at least one non-zero first order minor, ie, atleast one non-zero element of A. This is possible because A has non-zero elements.

.

**10. Find the rank of the matrix .**

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