A Suppose A is an n x n matrix with real entries and R is it

A. Suppose A is an n x n matrix with real entries and R is its row reduced echelon form. Using infomation on elementary matrices, explain the connection between det(A) and det(R). Note: you may use the fact that if M,N are two square matrices os the same size then det (MN) = det(M) det(N).

B. Suppose A is an invertible nxn matrix.

- Prove: If x is an n x 1 matrix such that Ax = 0 then x = 0.

- prove: Then A^-1 is also invertible and (A^-1)^-1 = A.

NOTE: Indicate property of matrix multication used.

Solution

During reducing a matrix A to its reduced row echelon form (RREF) R, three types of elementary row   operations are performed:

During the process of reduction of a matrix A to its RREF   R, after an elementary row operation, the matrix obtained is called an equivalent matrix, say B. We know that:

a. If B results from the interchange of two rows of A, then, det B = - det A

b. If B results from multiplying any row (or column) of A by a scalar k, then, det B = k det A.

c. If B results from adding a multiple of one row of A to another row, then, det B = det A.

Since the process of changing the n x n matrix A to its RREF R may involve some or all of the above mentioned elementary row operations, det R = - det A or, k det A or det A or – k detA depending upon the row operations used in the process.

B. Let A be an n x n invertible matrix and let x be an n x 1 matrix such that Ax = 0. Then on multiplying to the left by A^-1 , we get A^-1 (Ax) = A^-1 0 = 0 or, (A^-1 A) x = 0 or, Ix = 0 so that x = 0.

If A is invertible, we know that there exists an invertible A^-1 such that AA^-1=A^-1 A=I. We also know that (AB)^-1 =B^-1A^-1.

Now, (AA^-1 )^-1 = I^-1 = I so that (A^-1 )^-1 A^-1 = I Then, on right multiplying by A, we have [(A^-1 )^-1 A^-1 ] A = I A or,

(A^-1)^-1 (A^-1 A) = A or, (A^-1)^-1 I = A or, (A^-1)^-1 = A.

The indicative property of matrix multiplication is Associativity of matrix multiplication.