A. ![\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](/tpl/images/0684/5477/936c0.png)
Step-by-step explanation:
The question is asking us to find the product of the matrices. The key difference is the second A has a little T in the exponent. This T means transpose. You multiply A by the transpose of A. To find the transpose, you turn the rows into columns.
![A^T=\left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right]](/tpl/images/0684/5477/9a043.png)
Now that we have our transpose, we can multiply the matrices.
![\left[\begin{array}{ccc}5&2\\3&-1\\\end{array}\right] \left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right] =\left[\begin{array}{ccc}5*5+2*2&5*3+2(-1)\\3*5+2(-1)&3*3+(-1)(-1)\\\end{array}\right] =\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](/tpl/images/0684/5477/39da3.png)