Find the matrix A if $\left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]=A+\left[ \begin{matrix}
   1 & 2 & -1 \\
   0 & 4 & 9 \\
\end{matrix} \right]$

Hint: We will solve the given question by using the addition-subtraction rule of the matrices. We will first take the matrix on RHS to the LHS and then use the subtraction rule by simply subtracting corresponding terms in the matrices and hence find matrix A.

Complete step-by-step solution:
We have to find matrix A where we have given
$\left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]=A+\left[ \begin{matrix}
   1 & 2 & -1 \\
   0 & 4 & 9 \\
\end{matrix} \right]$
We can see that it is a problem of matrix addition and subtraction.
As we know two matrices are equal if and only if the order of the matrices is the same and the corresponding elements of the matrices are the same.
Similarly, while doing subtraction of matrices, the order of the matrices must be the same. We can subtract the corresponding elements in the matrices to get the resultant matrix.
Matrix subtraction is not commutative. Also, matrix subtraction is not associative.
We can solve it by taking variables as shown:
Let’s consider $P=\left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]$ and $Q=\left[ \begin{matrix}
   1 & 2 & -1 \\
   0 & 4 & 9 \\
\end{matrix} \right]$.
We have been given P = A + Q in the question.
Then we get A = P – Q. We can find matrix A, by subtracting two given matrices.
As we know matrix subtraction rule is $\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
\end{matrix} \right]-\left[ \begin{matrix}
   g & h & i \\
   j & k & l \\
\end{matrix} \right]=\left[ \begin{matrix}
   a-g & b-h & c-i \\
   d-j & e-k & f-l \\
\end{matrix} \right]$
Here P and Q are $3\times 2$ dimensional matrices hence their subtraction matrix also $3\times 2$dimensional.
Now we will subtract P-Q i.e.
$\left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]-\left[ \begin{matrix}
   1 & 2 & -1 \\
   0 & 4 & 9 \\
\end{matrix} \right]$
We will apply the matrix subtraction rule as mentioned above. We get,
$\begin{align}
  & \therefore \left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]-\left[ \begin{matrix}
   1 & 2 & -1 \\
   0 & 4 & 9 \\
\end{matrix} \right]=\left[ \begin{matrix}
   9-1 & -1-2 & 4-(-1) \\
   -2-0 & 1-4 & 3-9 \\
\end{matrix} \right] \\
 & \\
 & \Rightarrow \left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]-\left[ \begin{matrix}
   1 & 2 & -1 \\
   0 & 4 & 9 \\
\end{matrix} \right]=\left[ \begin{matrix}
   8 & -3 & 5 \\
   -2 & -3 & -6 \\
\end{matrix} \right] \\
\end{align}$
Here we get the solution as matrix $A=\left[ \begin{matrix}
   8 & -3 & 5 \\
   -2 & -3 & -6 \\
\end{matrix} \right]$.

Note: Use subtraction rule by simply subtracting terms in the matrices. Here given matrices are $3\times 2$ dimensional matrices hence their subtraction matrix also be $3\times 2$ dimensional i.e. matrix A. If we do not follow these, we might get confused and make silly mistakes. We can also assume matrix A as $\left[ \begin{align}
  & \begin{matrix}
   a & b & c \\
\end{matrix} \\
 & \begin{matrix}
   d & e & f \\
\end{matrix} \\
\end{align} \right]$ and then the given equation can be written as $\left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
\end{matrix} \right]+\left[ \begin{matrix}
   1 & 2 & -1 \\
   0 & 4 & 9 \\
\end{matrix} \right]$. Then, we can add them as $\left[ \begin{matrix}
   9 & -1 & 4 \\
   -2 & 1 & 3 \\
\end{matrix} \right]=\left[ \begin{matrix}
   a+1 & b+2 & c-1 \\
   d & e+4 & f+9 \\
\end{matrix} \right]$ . Now, since both matrices are equal, we can equate the corresponding elements, find the values of a, b, c, d, e and f. For example, we get 9 = a + 1, which gives a = 8. Then finally, we can substitute these and we can obtain matrix A.