Given the matrix A=begin{bmatrix}0 & 0&1 0 & 3&0 1&0 & -10 end{bmatrix} and suppose that we have the following row reduction to its PREF B A=begin{bma

Question

Given the matrix

A = [\begin{matrix} 0 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & - 10 \end{matrix}]

and suppose that we have the following row reduction to its PREF B

A = [\begin{matrix} 0 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & - 10 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 0 & - 10 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 0 & - 10 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]

Write

A and A^{- 1}

as product of elementary matrices.

Answer 1

Given the matrix
$A = [\begin{matrix} 0 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & - 10 \end{matrix}]$
Write $A and A^{- 1}$ as product of elementary matrices. Step 2 Take the matrix A and exchange the first and second rows. Let $E_{1}$ be the elementary row matrix corresponding to the above row operation.
$E_{1} = [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}]$
Notice that $E_{1} A = [\begin{matrix} 1 & 0 & - 10 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{matrix}]$
Step 3 Next, take the matrix $[\begin{matrix} 1 & 0 & - 10 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{matrix}]$ and divide the second row by the scalar 3. Let $E_{2}$ be the elementary row matrix corresponding to the above row operation.
$E_{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{matrix}]$
Notice that $E_{2} [\begin{matrix} 1 & 0 & - 10 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & - 10 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
Step 4
Next, take the matrix $[\begin{matrix} 1 & 0 & - 10 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ and add 10 times the third row to the first row. Let $E_{3}$ be the elementary row matrix corresponding to the above row operation.
$E_{3} = [\begin{matrix} 1 & 0 & 10 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
Notice that $E_{3} [\begin{matrix} 1 & 0 & - 10 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I$
Step 5
Now,
$I = E_{3} [\begin{matrix} 1 & 0 & - 10 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = E_{3} E_{2} [\begin{matrix} 1 & 0 & - 10 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{matrix}] = E_{3} E_{2} E_{1} A$
and so $A^{- 1} = E_{3} E_{2} E_{1}$ , that is
$A^{- 1} = [\begin{matrix} 1 & 0 & 10 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & - 10 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}]$
Step 6
Also,
$A = E_{1}^{- 1} E_{2}^{- 1} E_{3}^{- 1}$
$=$