Solving Matrix Equation Ax=B: Understanding Determinants and Inverses

[T Niemistoinen]

Homework Statement

You are given matrices A and B:

1 -2 1
A = 0 0 0
0 1 1


1
B = 0
2

Solve the equation Ax=B

Homework Equations

x=A^(-1)*B

The Attempt at a Solution

I carried out exactly the steps required for inverting a matrix. When got to th part where the determinant is calculated, I obtained a zero determinant. I thought this meant that the inverse matrix could not be calculated. This is also the result that an online matrix calculator gave for A. Hence I anwered "unsolvable" in the exam. Yet I was only given 1 out of 3 points.

What should I have done differently?

Thanks for your help!

 

[Office_Shredder]

The fact that A has no inverse means that there exist some vectors B such that
Ax = B
is not solvable. However as long as B is in the image of A it is solvable. In particular for this problem notice that A always forces the second coordinate of Ax to be equal to zero - so if B had a nonzero second coordinate the problem would be unsolvable. Luckily B has zero for its second coordinate, so it will turn out that Ax=B is solvable in this case.

As far as figuring out what x can be equal to, you can write down a system of equations from Ax=B which you should be able to solve. Try that and show us what you get!

 

[Ray Vickson]

Homework Statement

You are given matrices A and B:

1 -2 1
A = 0 0 0
0 1 1


1
B = 0
2

Solve the equation Ax=B

Homework Equations

x=A^(-1)*B

The Attempt at a Solution

I carried out exactly the steps required for inverting a matrix. When got to th part where the determinant is calculated, I obtained a zero determinant. I thought this meant that the inverse matrix could not be calculated. This is also the result that an online matrix calculator gave for A. Hence I anwered "unsolvable" in the exam. Yet I was only given 1 out of 3 points.

What should I have done differently?

Thanks for your help!

For any nxn linear system, exactly one of the following statements is true: Ax = b has (i) exactly one solution; (ii) no solutions; or (iii) infinitely many solutions. You chose (ii), but without checking the details. The easiest way is to just write out the equations:
1x -2y + 1x = 1\\<br /> 0x+0y+0z=0\\<br /> 0x+1y+1z = 2
The second equation is redundant, so we just have two equations in three unknowns:
\begin{array}{rc} x - 2y + z &amp; = 1\\<br /> y + z &amp;= 2 \end{array}<br />

 

[T Niemistoinen]

Thanks a lot guys! That's actually all familiar stuff to me and it all makes sense now, but for some reason I just can't think straight when it comes to matrices. Thanks again!