In a circus there are 11 carnivorous animals including tigers, lions and panthers. It is known that each lion eats three kilos of meat a day, that each tiger eats two kilos a day and each panther also two kilos. If a total of 25 kilos of meat are needed per day and it is known that the number of panthers is three times the number of tigers, how many lions, panthers and tigers are there? Solve the exercise using matrices (Gauss or Gauss-Jordan)

Using matrix operations, the number of lions, tigers, and panthers in the circus can be determined. We can conclude that x = 4, y = 2, and z = 6. Therefore, there are 4 lions, 2 tigers, and 6 panthers in the circus.

Let's represent the number of lions, tigers, and panthers as variables x, y, and z, respectively. From the given information, we can set up the following system of equations:

x + y + z = 11 (equation 1)

3x + 2y + 2z = 25 (equation 2)

z = 3y (equation 3)

To solve this system using matrix operations, we can rewrite the equations in matrix form:

[tex]\left[\begin{array}{ccc}1&1&1\\3&2&2\\0&-3&1\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex][tex]=\left[\begin{array}{ccc}11\\25\\0\end{array}\right][/tex]

By performing row operations, we can transform the augmented matrix to row-echelon form and then solve for the variables. After applying Gauss-Jordan elimination, the augmented matrix becomes:

[tex]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex][tex]=\left[\begin{array}{ccc}4\\6\\2\end{array}\right][/tex]

From the row-echelon form, we can conclude that x = 4, y = 2, and z = 6. Therefore, there are 4 lions, 2 tigers, and 6 panthers in the circus, satisfying the given conditions.

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