Determine the non-singular matrices P & Q such that PAQ is in the normal form for A. A = [2 1 -3 -6; 3 -3 1 2; 1 1 1 2]. Hence find the rank of A.

Steps to Solve

1. Identify the matrix A

   We start with the matrix $$ A = \begin{bmatrix} 2 & 1 & -3 & -6 \ 3 & -3 & 1 & 2 \ 1 & 1 & 1 & 2 \end{bmatrix} $$

2. Find the rank of matrix A

   The rank of a matrix is the maximum number of linearly independent row or column vectors. We can use row reduction to echelon form (REF) to find the rank.

   Row Operations:

   1. Swap Row 1 and Row 2: $$ \begin{bmatrix} 3 & -3 & 1 & 2 \ 2 & 1 & -3 & -6 \ 1 & 1 & 1 & 2 \end{bmatrix} $$
   2. Replace Row 2 and Row 3 to eliminate the leading coefficients:
   • Row 2: ( R_2 - \frac{2}{3}R_1 )
   • Row 3: ( R_3 - \frac{1}{3}R_1 )

   The updated matrix now looks like: $$ \begin{bmatrix} 3 & -3 & 1 & 2 \ 0 & 3 & -7 & -8 \ 0 & 2 & \frac{2}{3} & \frac{5}{3} \end{bmatrix} $$

3. Continue with row reduction

   From the current matrix, continue to simplify:

   • Make Row 2 the leading row by dividing by 3: $$ \begin{bmatrix} 3 & -3 & 1 & 2 \ 0 & 1 & -\frac{7}{3} & -\frac{8}{3} \ 0 & 2 & \frac{2}{3} & \frac{5}{3} \end{bmatrix} $$
   • Now eliminate the leading coefficient in Row 3:
   • Row 3: ( R_3 - 2R_2 )

   The resulting matrix becomes: $$ \begin{bmatrix} 3 & -3 & 1 & 2 \ 0 & 1 & -\frac{7}{3} & -\frac{8}{3} \ 0 & 0 & 6 & 8 \end{bmatrix} $$

4. Determine the rank

   The resulting echelon form shows three non-zero rows. Therefore, the rank of matrix A is:

   $$ \text{Rank}(A) = 3 $$

5. Find matrices P and Q

   To find non-singular matrices ( P ) and ( Q ) such that ( PAQ ) is in normal form, we can select:

   $$ P = I $$ (identity matrix) and $$ Q = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix} $$

   This will effectively keep the result as ( A ).