Inverse of Matrix: Reduce Augmented Matrix [A|I] with Row Ops

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To find the inverse of a matrix using row operations on the augmented matrix [A|I], the process involves transforming the left side into the identity matrix. The recommended approach is to first convert the matrix into upper-triangular form, followed by transforming it into the identity matrix. Key steps include making the first column of rows 2-4 zero, then the second column of rows 3 and 4, and finally the third column of row 4. A common strategy is to switch rows to avoid fractions early in the process. The participant ultimately achieved a result, though they expressed uncertainty about its correctness.
EvanQ
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Homework Statement



Find the inverse of

matrixonline.jpg


by reducing the augmented matrix [A|I] using row operations.

The Attempt at a Solution



I just can't seem to get this one out, it's the last in a series of 9 similar questions and the first I am to have a hard time with.
First few attempts involved operations like:
R1 <- R1 x 1/4
R3 <- R3 + R2
R4 <- R4 + R1
R1 <- R1 - 2R4

But I don't seem to be getting any closer to a solution with the operations that i am trying.

Any help, even just a pointer as to what row operations to start off with, would be gratefully appreciated.
 
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You want to use row operations to transform the left half of the augmented matrix into the identity matrix. Do this in two stages: (1) transform the left half of the augmented matrix into upper-triangular form, and (2) transform this into the identity matrix. You can do the scaling so that the diagonals are all ones as part of the first stage or at the very end.

Now split the first stage into substages. Use row operations to make the first column of rows 2-4 zero. Then make the second column of rows 3 and 4 zero. Finally, make the third column of row 4 zero.

You started fine. Your second step, R3 <- R3 + R2, is where you started to go awry.
 
I usually start these by switching row 3 with row 1, so I don't need to deal with fractions from step one (because you already have a 1 in slot 1,1). Note you can leave row 2 where it is and you already have a 1, so if you just kill the first two columns before going on to deal with columns three and four, you've already reduced the number of fractions you're dealing with by half
 
thanks a lot guys.
just got it out then, unsure of correctness, but I've gotten a result and that's at least worth part marks :p
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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