91Ó°ÊÓ

We know that the number of pivots can be at most 3 since the matrix is \(3 \times 3\). Let's explore the possibilities. 1. One pivot: The pivot will always be on the first row of the matrix. We will have 1 non-zero row and 2 zero rows, like this: \[ \begin{bmatrix} 1 & * & * \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] 2. Two pivots: The pivots must be in separate rows and columns. We will have 2 non-zero rows and 1 zero row, like this: \[ \begin{bmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 0 \end{bmatrix} \] 3. Three pivots: All entries will be diagonal, and we will have no zero rows: \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

Now let's calculate the number of possible matrices for each structure by filling in the placeholders with 0's and 1's: 1. One pivot: \[ \begin{bmatrix} 1 & * & * \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] There are two free variables left (denoted by *). Each of them can be either 0 or 1, which gives us \(2^2 = 4\) possible matrices for this case. 2. Two pivots: \[ \begin{bmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 0 \end{bmatrix} \] Again, there are two free variables left. Each variable can be either 0 or 1, which gives us another \(2^2 = 4\) possible matrices for this case. 3. Three pivots: \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] In this case, there are no free variables, and the matrix is entirely determined. There is only one possible matrix for this case.

To find the total number of possible matrices, simply add the number of possible matrices for each structure. Total number of possible matrices = Number of matrices with 1 pivot + Number of matrices with 2 pivots + Number of matrices with 3 pivots = 4 + 4 + 1 = 9 So there are 9 unique \(3 \times 3\) matrices in row canonical form using only 0's and 1's.