I had these four problems that I was having trouble with. The goal is to determine whether the matricies below are invertible or not without using row operations

Problem 1 $$ \begin{bmatrix} -1 & 4 & 3 & 7 & -5 \\ 7 & 12 & 9 & -4 & 3 \\ 21 & -8 & -6 & 7 & 13 \\ 6 & 0 & 0 & -4 & 1\\ 0 & -4 & -3 & 2 & 1 \end{bmatrix} $$

Problem 2 $$ \begin{bmatrix} 1 & -2 & 3 & -4 & -1 \\ 7 & 6 & 5 & -3 & -4 \\ 0 & 3 & 0 & -6 & 5 \\ 4 & 5 & 8 & -2 & -1\\ -2 & 4 & -6 & 8 & 2 \end{bmatrix} $$

Problem 3

Let A be the standard matrix of a linear transformation from $R^2$ to $R^2$ that rotates the plane about the origin by $\theta$ radians clockwise

Problem 4

Let A be the standard matrix of a linear transformation from $R^3$ to $R^3$ that projects $R^3$ onto the $x_1 x_2$-plane

Now standard practice for me to determine whether a matrix is invertible or not is to either do Gaussian elimination (row operations) or find the determinate. However, at this point, determinates will be covered later on and are assumed to not really be of use in this situation.

To solve most of them, I've been looking to the invertible matrix theorem. Here

For example, for the first one, it looks like you can take non zero multiples of each of the rows. Thus you would lose a pivot point and it would no longer be invertible. However, I have my doubts on that one.

Any help would be useful! Thank you in advanced!


Note that in matrix $2$, the fifth row is exactly $-2$ times the first row. Hence the determinant is zero.

In matrix $1$, the second column and the third column are $4$ times and $3$ times the column vector $[1,3,-2,0,-1]^T$ respectively. By multilinearity, these get removed, so the determinant here is zero too.

The third matrix is invertible, since its inverse can be explicitly written down as the reverse rotation matrix. That is, there exists a matrix which you can right and left compose this matrix with to get the identity, which is the reverse rotation matrix.

The fourth matrix is of rank $2$ : it is essentially projecting onto a two dimensional space, therefore is not invertible by the fifteenth(!) point of the rank theorem.

In general, if people ask you to inspect matrices and tell whether or not the determinant is zero, you will have to look for patterns of the kind I looked for. Moreover, the operations determined by these matrices (such as rotation / projection / scaling) carry a certain geometric meaning to them, which helps you propose candidates for a possible inverse.

The key, of course, is the IMT, but we have some shockers in store as well.

There are some interesting examples of matrices that look very complicated but are invertible. These include special Vandermonde matrices, Hankel matrices, and the (not so well known) strictly diagonally dominant matrices, which actually are more helpful than we would think they are. So I would also look for these properties in a question of the kind that you were given.

  • $\begingroup$ Thank you for the acceptance, and +1 to your question. $\endgroup$ Mar 16 '18 at 1:19

For the first problem, notice that $3$ times the second column is equal to $4$ times the third column. Since the columns are not linearly independent, the matrix is not invertible.

Similarly, for the second problem, the last row is equal to $-2$ times the first row, so the matrix is not invertible.

The matrix in the third problem is invertible. This is because rotation by an angle of $\theta$ has an inverse: rotation by an angle of $-\theta$. Thus the inverse of the matrix in question is the matrix representing the rotation by $-\theta$.

Similarly, the matrix in the fourth problem is not invertible since the operation which it represents is not invertible.

  • $\begingroup$ Thank you for your reply! What do you mean that the operation it represents is not invertible? How does it violate the IMT? $\endgroup$ Mar 15 '18 at 0:48

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