Question

(a) The n × n matrices A, B, C, and X satisfy the equation AX(B + CX) ?1 = C Write an expression for the matrix X in terms of A, B, and C. You may assume invertibility of any matrix when necessary.

(b) Suppose D is a 3 × 5 matrix, E is a 5 × c matrix, and F is a 4 × d matrix. Find the values of c and d for which the statement “det(DEF) = 1” can be valid. Explain your answer.

(c) Find all (real or complex) values of x such that the matrix GH is invertible, where G =

x^2, ?1

x , x ? 2

H = x ? 1 , ?2

1 , x + 1

Answer 1

a) Given, AX(B+CX)^-1 = C

i.e., [AX(B+CX)^-1](B+CX) = C(B+CX)

i.e., AX = C(B+CX)

i.e., AX = CB+C²X

i.e., AX-C²X = CB

i.e., (A-C²)X = CB

i.e., (A-C²)^-1[(A-C²)X] = (A-C²)^-1CB

i.e., X = (A-C²)^-1CB [Since inverses of all matrices exists]

Therefore, the required expression for matrix X is : X = (A-C²)^-1CB.

b) Given, D is a 3x5 matrix, E is a 5xc matrix and F is a 4xd matrix.

Here, the matrix DE is a 3xc matrix.

Now, matrix DEF exists when the number of columns of matrix DE and the number of rows of matrix F are equal.

Then, c = 4.

We can calculate the determinant of square matrices only.

Now, the matrix DEF will be square matrix when 3 = d.

Therefore, values of c and d are 4 and 3 respectively.

c) Given, G = and H =

Now, GH = = .

Here, the matrix GH is invertible if and only if det(GH) 0.

Then, 0

i.e., (x³-x²-1)(x²-3x-2) + (x²-2)(2x²+x+1) 0

i.e., (x⁵-4x⁴+x³+x²+3x+2)+(2x⁴+x³-3x²-2x-2) 0

i.e., x⁵-2x⁴+2x³-2x²+x 0

i.e., x(x-1)²(x²+1) 0

i.e., x 0,1, -i, i [where i denotes the complex number and i = ]

For all values of x except 0, 1, -i, i, the matrix GH is invertible.