We are given that,

\( \begin{bmatrix}9&-1 &4 \\[0.3em]-2 & 1 &3\\[0.3em]\end{bmatrix}\) = A +\( \begin{bmatrix}1&2 &-1 \\[0.3em]0 & 4 &9\\[0.3em]\end{bmatrix}\)

We need to find the matrix A. 

In order to find A, 

Shift the matrix in addition with A to left hand side of the equation. 

Just like in algebraic property, 

X = A + Y 

⇒ A = X – Y 

Similarly,

\( \begin{bmatrix}9&-1 &4 \\[0.3em]-2 & 1 &3\\[0.3em]\end{bmatrix}\) = A +\( \begin{bmatrix}1&2 &-1 \\[0.3em]0 & 4 &9\\[0.3em]\end{bmatrix}\)

⇒ A = \( \begin{bmatrix}9&-1 &4 \\[0.3em]-2 & 1 &3\\[0.3em]\end{bmatrix}\) - \( \begin{bmatrix}1&2 &-1 \\[0.3em]0 & 4 &9\\[0.3em]\end{bmatrix}\)

Subtraction in matrices is done by subtraction of corresponding elements in the matrices.

⇒ A = \( \begin{bmatrix}9-1&-1-2 &4-(-1) \\[0.3em]-2-0 & 1-4 &3-9\\[0.3em]\end{bmatrix}\)

⇒ A = \( \begin{bmatrix}8&-3 &5 \\[0.3em]-2 & -3 &-6\\[0.3em]\end{bmatrix}\)

Thus, 

We get  A = \( \begin{bmatrix}8&-3 &5 \\[0.3em]-2 & -3 &-6\\[0.3em]\end{bmatrix}\).