Answer:
a.A+B can not find out
b.
=![\left[\begin{array}{ccc}13&8&11\\17&10&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D13%268%2611%5C%5C17%2610%2613%5Cend%7Barray%7D%5Cright%5D)
=![\left[\begin{array}{ccc}a_1b_1+a_2b_4+a_3b_7&a_1b_2+a_2b_5+a_3b_8&a_1b_3+a_2b_6+a_3b_9\\a_4b_1+a_5b_4+a_6b_7&a_4b_2+a_5b_5+a_6b_8&a_4b_3+a_5b_6+a_6b_9\\a_7b_1+a_8b_4+a_9b_7&a_7b_2+a_8b_2+a_8b_5+a_9b_8&a_a_7b-3+a_8b_6+a_9b_9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1b_1%2Ba_2b_4%2Ba_3b_7%26a_1b_2%2Ba_2b_5%2Ba_3b_8%26a_1b_3%2Ba_2b_6%2Ba_3b_9%5C%5Ca_4b_1%2Ba_5b_4%2Ba_6b_7%26a_4b_2%2Ba_5b_5%2Ba_6b_8%26a_4b_3%2Ba_5b_6%2Ba_6b_9%5C%5Ca_7b_1%2Ba_8b_4%2Ba_9b_7%26a_7b_2%2Ba_8b_2%2Ba_8b_5%2Ba_9b_8%26a_a_7b-3%2Ba_8b_6%2Ba_9b_9%5Cend%7Barray%7D%5Cright%5D)
=![\left[\begin{array}{ccc}a_1b_1+a_2b_3&a_1b_2+a_2b_4\\a_3b_1+a_4b_3&a_3b_2+a_4b_4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1b_1%2Ba_2b_3%26a_1b_2%2Ba_2b_4%5C%5Ca_3b_1%2Ba_4b_3%26a_3b_2%2Ba_4b_4%5Cend%7Barray%7D%5Cright%5D)
In similar way multiply two matrix of order ![4\times 4](https://tex.z-dn.net/?f=4%5Ctimes%204)
c.No,because A is not a square matrix and determinant of B is zero.
Step-by-step explanation:
We are given that two matrix
A=![\left[\begin{array}{ccc}1&3&1\\2&3&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%261%5C%5C2%263%262%5Cend%7Barray%7D%5Cright%5D)
B=![\left[\begin{array}{ccc}2&1&1\\3&2&3\\2&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%261%5C%5C3%262%263%5C%5C2%261%261%5Cend%7Barray%7D%5Cright%5D)
In matrix A , two rows and 3 columns therefore, the order of matrix ![2\times 3](https://tex.z-dn.net/?f=%202%5Ctimes%203)
In matrix B, 3 rows and 3 columns therefore, the order of matrix B is ![3\times 3](https://tex.z-dn.net/?f=%203%5Ctimes%203)
a.A+B can no find because when add two matrix then the order of two matrix should be same .
b.![A\times B](https://tex.z-dn.net/?f=%20A%5Ctimes%20B)
When we multiply on matrix to other matrix then number of columns of first matrix equals to number of rows of second matrix.
Therefore, number of columns of matrix A is equals to number of rows of matrix B.So, we can multiply
=
\times![\left[\begin{array}{ccc}2&1&1\\3&2&3\\2&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%261%5C%5C3%262%263%5C%5C2%261%261%5Cend%7Barray%7D%5Cright%5D)
=![\left[\begin{array}{ccc}13&8&11\\17&10&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D13%268%2611%5C%5C17%2610%2613%5Cend%7Barray%7D%5Cright%5D)
Formula for multiply of matrix of order ![3\times3](https://tex.z-dn.net/?f=3%5Ctimes3%20)
Let A and B are square matrix of order ![3\times 3](https://tex.z-dn.net/?f=%203%5Ctimes%203)
Let A=
and B=![\left[\begin{array}{ccc}b_1&b_2&b_3\\b_4&b_5&b_6\\b_7&b_8&b_9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db_1%26b_2%26b_3%5C%5Cb_4%26b_5%26b_6%5C%5Cb_7%26b_8%26b_9%5Cend%7Barray%7D%5Cright%5D)
=![\left[\begin{array}{ccc}a_1b_1+a_2b_4+a_3b_7&a_1b_2+a_2b_5+a_3b_8&a_1b_3+a_2b_6+a_3b_9\\a_4b_1+a_5b_4+a_6b_7&a_4b_2+a_5b_5+a_6b_8&a_4b_3+a_5b_6+a_6b_9\\a_7b_1+a_8b_4+a_9b_7&a_7b_2+a_8b_2+a_8b_5+a_9b_8&a_a_7b-3+a_8b_6+a_9b_9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1b_1%2Ba_2b_4%2Ba_3b_7%26a_1b_2%2Ba_2b_5%2Ba_3b_8%26a_1b_3%2Ba_2b_6%2Ba_3b_9%5C%5Ca_4b_1%2Ba_5b_4%2Ba_6b_7%26a_4b_2%2Ba_5b_5%2Ba_6b_8%26a_4b_3%2Ba_5b_6%2Ba_6b_9%5C%5Ca_7b_1%2Ba_8b_4%2Ba_9b_7%26a_7b_2%2Ba_8b_2%2Ba_8b_5%2Ba_9b_8%26a_a_7b-3%2Ba_8b_6%2Ba_9b_9%5Cend%7Barray%7D%5Cright%5D)
In similar way we multiply of matrix of order
and matrix multiply of order ![4\times 4](https://tex.z-dn.net/?f=4%5Ctimes%204)
Let A and B are matrix of order ![2\times 2](https://tex.z-dn.net/?f=2%5Ctimes%202)
Let![A=\left[\begin{array}{ccc}a_1&a_2\\a_3&a_4\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1%26a_2%5C%5Ca_3%26a_4%5Cend%7Barray%7D%5Cright%5D)
![B=\left[\begin{array}{ccc}b_1&b_2\\b_3&b_4\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db_1%26b_2%5C%5Cb_3%26b_4%5Cend%7Barray%7D%5Cright%5D)
=![\left[\begin{array}{ccc}a_1b_1+a_2b_3&a_1b_2+a_2b_4\\a_3b_1+a_4b_3&a_3b_2+a_4b_4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1b_1%2Ba_2b_3%26a_1b_2%2Ba_2b_4%5C%5Ca_3b_1%2Ba_4b_3%26a_3b_2%2Ba_4b_4%5Cend%7Barray%7D%5Cright%5D)
In similar way we multiply two matrix of order ![4\times 4](https://tex.z-dn.net/?f=%204%5Ctimes%204)
C.Matrix A is not a square matrix .Therefore, it is not a invertible matrix.
![\mid B\mid=\begin{vmatrix}2&1&1\\3&2&3\\2&1&1\end{vmatrix}](https://tex.z-dn.net/?f=%5Cmid%20B%5Cmid%3D%5Cbegin%7Bvmatrix%7D2%261%261%5C%5C3%262%263%5C%5C2%261%261%5Cend%7Bvmatrix%7D)
![\mid B\mid=2(2-3)-1(3-6)+1(3-4)=-2+3-1=0](https://tex.z-dn.net/?f=%5Cmid%20B%5Cmid%3D2%282-3%29-1%283-6%29%2B1%283-4%29%3D-2%2B3-1%3D0)
Therefore, the determinant of B is equal to zero therefore, inverse of matrix B does not exist.
Hence, Both matrix A and B are no invertible.