Practice Exam 2 M314 1 (6 points) Let A be an n x n matrix. If the equation Ax = 0 has only the trivial solution, do the columns of A span R n?Why or why not? Answer: To say that the columns of A span R n is the same as saying that Ax = b has a solution for every b in R n. Let's define what its determinant is. So let me construct a 3 by 3 matrix here. Let's say my matrix A is equal to- let me just write its entries- first row, first column, first row, second column, first row, third column. Then you have a2 1, a2 2, a2 3. Then you have a3 1, third row first column, a3 2, and then a3 3. That is a 3 by 3 matrix. Place the results of the previous step into a new matrix of cofactors by aligning each minor matrix determinant with the corresponding position in the original matrix. Thus, the determinant that you calculated from item (1,1) of the original matrix goes in position (1,1).
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How to find all the minors and cofactors of the matrix #A=((1, -2, 3), ( 6, 7, -1 ), (-3, 1, 4))#?
1 Answer
Motrix 1 1 3 X 25
The minors of a matrix are the determinants of the smaller matrices you get when you delete one row and one column of the original matrix. The cofactors of a matrix are the matrices you get when you multiply the minor by the right sign (positive or negative).
X 2 1 X 2 4
Explanation:
There is a minor and a cofactor for every entry in the matrix -- so that's 9 altogether! I'll just go through a few of them.
The minor for the first row and column is called #M_(1,1)#. We calculate it by removing the first row and column of the matrix, so we are left with the matrix:
#((7,-1), (1,4))#
Then we take the determinant of this matrix, which is #(7*4) - (-1*1) = 28 + 1 = 29#. So #M_(1,1) = 29#.
To find the corresponding cofactor, called #A_(1,1)#, we look at the subscripts on the name of the minor -- in our case, #(1,1)#, and we add them together to get 2. Then we raise #-1# to this value, so we get #(-1)^2 = 1#, and we multiply this by the minor. So that gives us #A_(1,1)= 29*1 = 29# as the cofactor.
Let's do one more. To find the minor #M_(2,3)# we delete the second row and third column of the matrix, so we are left with:
Then we take the determinant of this matrix, which is #(1*1)-(-3*-2) = 1 - 6 = -5#. So #M_(2,3) = -5#.
3 X 2 X 1.5
To find the cofactor #A_(2,3)#, we add the subscripts #(2,3)# to get 5, and raise #-1# to this value, which gets us #(-1)^5 = -1#. So we multiply the minor #-5# by #-1# to get #A_(2,3) = 5#. Compressor powerful encoding delivered 4 4 5.