1. solution: A= () gives A-2=A-2E= (), where e is identity matrix.

So (A-2E)- 1= ()

And because AX-2X=B, it is deduced that (a-2) x = b.

So X=(A-2E)- 1B= ()

2. Solution: Write the linear equations in the form of matrix product AX = B.

Where a = (), b = () and x = ()

When the equation has infinite solutions, the determinant of a is zero.

So |A|=λ? 3-3λ? +2=0,(λ- 1)2(λ? +2)=0? Get λ？ = 1 (double root)

Or λ? =-2

When λ? = 1, A= (), B= (), and the general solution is X= ().

When λ? =-2, A= (), B= (), and the general solution is X= ().

When the equation has a unique solution, the determinant of A is nonzero.

So |A|=λ? 3-3λ? +2

That's λ? And λ?

When the equations have no solution, the equations are contradictory, that is, the linear expressions of the two equations are contradictory to the third one.

When the first two formulas are combined linearly to get the third formula, λ? = 1 or -2, at this time, it returns to the case of infinite solution, so there is no case of no solution.