Determinants of Matrices?

What I'm given is...Let:

[[a b c]

[d e f] = 4

[g h i]]

so the whole matrix is = to 4. I need to find,

#1

[[4a -b 3c]

[4d -e 3f]

[4g -h 3i]]

#2

[[a+g b+h c+i]

[ d e f ]

[ g h i ]]

#3

[[ a b c ]

[3d-2g 3e-2h 3f-2i ]

[ g h i ]]

Have no idea what to do or how to start, can someone please help me!

1 Answer

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  • 8 years ago
    Favorite Answer

    The determinant is:

    1.      aei + bfg + cdh - ceg - bdi - afh

    And you know this is equal to 4, so:

    2.      aei + bfg + cdh - ceg - bdi - afh = 4

    But you want to solve #1. If you look at "aei" with the matrix in #1, you see that the new a is 4a, the new e is -e and the new i is 3i. So this means 4⋅(-1)⋅3 = -12. If you now examine each of the rest of the terms showing in (1) or (2) above, and look at your new matrix values, you will see that each of them is multiplied by the same value, -12. So this means you can just divide everything by -12 and get back the same left side of (2) above. But doing that also divides 4, too, by -12. So the new determinant must be -⁴⁄₁₂ or -⅓.

    Try out the same idea, but now with #2 and #3. Just keep track of terms, perform and distribute terms, look for cancellations or ways to get back your left side so that it looks like the left side of equation (2) above. That should help a lot.

    For example, in #2 you have (after substitution):

    3.      det = (a+g)ei + (b+h)fg + (c+i)dh - (c+i)eg - (b+h)di - (a+g)fh

    4.      det = aei + bfg + cdh - ceg - bdi - afh + gei + hfg + idh - ieg - hdi - gfh

    5.      det = (aei + bfg + cdh - ceg - bdi - afh) + (gei + hfg + idh - ieg - hdi - gfh)

    6.      det = 4 + gei + hfg + idh - ieg - hdi - gfh

    7.      det = 4 + hfg + idh - hdi - gfh

    8.      det = 4 + idh - hdi

    9.      det = 4

    And there is that answer.

    You can do #3? I think so.

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