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Math Help - Test question

  1. #1
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    Test question

    I had this question on a test, and I am wondering if I did it correctly.

    "If A is a 3x3 matrix, AX = B has at least one solution for every B. Explain how to construct a 3x3 matrix C such that AC = I." (3 marks)

    I basically said that since A and C are both square matrices, AC = I = CA, and thus C is just the inverse of A. So you could use the matrix inversion algorithm to find the inverse of A, and that would give you C.

    Is that correct?
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  2. #2
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    Quote Originally Posted by BrownianMan View Post
    I had this question on a test, and I am wondering if I did it correctly.

    "If A is a 3x3 matrix, AX = B has at least one solution for every B. Explain how to construct a 3x3 matrix C such that AC = I." (3 marks)

    I basically said that since A and C are both square matrices, AC = I = CA, and thus C is just the inverse of A. So you could use the matrix inversion algorithm to find the inverse of A, and that would give you C.

    Is that correct?

    The question is, at best, unclear: is it "if A is a 3x3 matrix such that Ax=b has at least one solution for...etc.", or is it "If A is a non-singular matrix then Ax=b has at least one solution for every...etc."??
    As it is it makes no much sense and it's hard to guess what were you actually asked, but I'll give it a try: I'm guessing they wanted you to show them a way to calculate the inverse matrix of a given invertible matrix A (Invertible because they told us it is or because we had to deduce it. This I can't say), but if this is true then you didn't answered correctly the question: You had to ACTUALLY show the matrix inversion algorithm and, perhaps, show them how it is doen and why it works, not just merelity quoting it.

    Tonio
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  3. #3
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    You are given the fact that AX=B has at least one solution for every B. This means that you may select the three "usual" basis vectors, e_1, e_2, and e_3 (e_i having a single 1 in the i_th position) as right hand side (B) and obtain three solutions x_1, x_2 and x_3. Note that the given info implies that there is only one solution for each B, not more. The solutions x_1, x_2 and x_3 are the columns of the inverse, C.
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    Quote Originally Posted by tonio View Post
    The question is, at best, unclear: is it "if A is a 3x3 matrix such that Ax=b has at least one solution for...etc.", or is it "If A is a non-singular matrix then Ax=b has at least one solution for every...etc."??
    As it is it makes no much sense and it's hard to guess what were you actually asked, but I'll give it a try: I'm guessing they wanted you to show them a way to calculate the inverse matrix of a given invertible matrix A (Invertible because they told us it is or because we had to deduce it. This I can't say), but if this is true then you didn't answered correctly the question: You had to ACTUALLY show the matrix inversion algorithm and, perhaps, show them how it is doen and why it works, not just merelity quoting it.

    Tonio
    I drew the diagram [A I] --> [I C], showing that as A is brought to I, I is brought to C. So C would be the inverse of A.
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    Quote Originally Posted by signaldoc View Post
    You are given the fact that AX=B has at least one solution for every B. This means that you may select the three "usual" basis vectors, e_1, e_2, and e_3 (e_i having a single 1 in the i_th position) as right hand side (B) and obtain three solutions x_1, x_2 and x_3. Note that the given info implies that there is only one solution for each B, not more. The solutions x_1, x_2 and x_3 are the columns of the inverse, C.
    Hm - but C would still be the inverse of A, no? So inverting A to get C would be right?
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  6. #6
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    Also, another question that I was unsure of:

    Does A^2 = A + I, imply that A is invertible?
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  7. #7
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    For the first question, I figure "AX=B has at least one solution" was thrown in there to show that A is invertible, since "AX=B has at least one solution" implies A^-1 exists.
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    Quote Originally Posted by BrownianMan View Post
    For the first question, I figure "AX=B has at least one solution" was thrown in there to show that A is invertible, since "AX=B has at least one solution" implies A^-1 exists.
    Not necessarily. AX=B has at least one solution for any B means A has an inverse. The underlined clause is essential. AX=B may have multiple solutions for certain B (those in the range of A) and none for others...

    You are correct in that inverting A is the answer. How to get this inverse is possibly the root of the question, and Gauss elimination as you indicated will provide it.
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    Quote Originally Posted by signaldoc View Post
    Not necessarily. AX=B has at least one solution for any B means A has an inverse. The underlined clause is essential. AX=B may have multiple solutions for certain B (those in the range of A) and none for others...

    You are correct in that inverting A is the answer. How to get this inverse is possibly the root of the question, and Gauss elimination as you indicated will provide it.
    That's right! That's what the question said in the OP and on the test ("AX=B has at least one solution for any B") - I simply made an error when copying and pasting from the OP.

    Anyways, thanks! I feel that I should get the full 3 marks for this question.
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  10. #10
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    Any thoughts on this:

    Does A^2 = A + I, imply that A is invertible?
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    Quote Originally Posted by BrownianMan View Post
    Any thoughts on this:

    Does A^2 = A + I, imply that A is invertible?

    Yes. Just bring A in the right side to the left side and factor out...

    Tonio
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