Question

See question in attached photo.​

Answer 1

9514 1404 393

Answer:

  (a, b) = (-2, -1)

Step-by-step explanation:

The transpose of the given matrix is ...

  $A^T=\left[\begin{array}{ccc}1&2&a\\2&1&2\\2&-2&b\end{array}\right]$

Then the [3,1] term of the product is ...

  $(A\cdot A^T)_{31}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}1&2&2\end{array}\right]=a+2b+4$

and the [3,2] term is ...

  $(A\cdot A^T)_{32}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}2&1&-2\end{array}\right]=2a-2b+2$

Both of these terms in the product matrix are 0. We can solve the system of equations by adding these two terms:

  (a +2b +4) +(2a -2b +2) = (0) +(0)

  3a +6 = 0

  a = -2

Substituting for 'a' in term [3,1] gives ...

  -2 +2b +4 = 0

  b = -1

The ordered pair (a, b) is (-2, -1).