Question

Let Ab e an invertiblen×nmatrix, and let B be ann×pmatrix. Show thatthe equationAX=Bhas a unique solutionA−1B. Hint: You need to show thatA−1Bis a solution, and that there is no other solution. You might do this secondpart by making an argument each column ofB

Answer 1

Answer:

In fact, $A^{-1} B$  is a solution to the equation AX = B. Let me show you.

Step-by-step explanation:

Step 1 .

Remember that there's a very special $n$x$n$ matrix, called "the identity matrix", we denote it as $I_{N}$.

That matrix is special because

               

                                    $I_{N}$*(Any matrix)  = Any matrix,

So, when I say, "Any matrix", that literally means, "Any martrix", well, any $n$x$n$ matrix. Therefore;

 

                                                     $I_{N}$*A = A

Step 2.

Remember that A is invertible if there exist a matrix $A^{-1}$ such that

                                                      $A$ *$A^{-1}$ =   $I_{N}$.

$A^{-1}$  is just the name of it, the most important is the property I just mentioned.

Step 3.

Now, to show that,  we just have to show that $A^{-1} B$ is a solution of $AX = B$, in other words we have to show that,  $A*(A^{-1} B) = B$

Notice that

                     $A*(A^{-1} * B ) = (A*A^{-1} ) *B = I_{N} * B = B$

Therefore, $A^{-1} B$ is a solution of $AX = B$.