Answer:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/jotewv27mhitkarour37pioe9p5jjmaeb8.png)
Explanation:
Let's find the answer.
Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.
Equations:
Eq. 1 : x1 + 2x2 + 5x3 = 5
Eq. 2 : x1 + x2 + x3 = 6
E1. 3 : 4x1 + 6x2 + 5x3 = 7
Coefficients for x1 ; x2 ; x3
From eq. 1 : 1 ; 2 ; 5
From eq. 2 : 1 ; 1 ; 1
From eq. 3 : 4 ; 6 ; 5
So matrix A is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/ztlkkjc0kwy1goxm6jing4q35o1yuzuhdl.png)
And the vector of vriables (X) is:
![\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/upus65k7rzgz5o8a9l0h2wf8q0ygbqwy1a.png)
Now we can find the resulting vector (B) using the 'resulting values' from each equation:
![\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/j2ay51ntr80hsr7juhv73965d2qbbrx6xn.png)
In conclusion, AX=B is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/jotewv27mhitkarour37pioe9p5jjmaeb8.png)