Question 1196276
Certainly, let's evaluate the determinant of matrix 'm' and solve the equation 'mx = (1, 11, 21)' when K = Z.

**1. Evaluate the Determinant of Matrix 'm'**

* **Expand along the first row:**

   det(m) = 1 * det( 
                       7   3
                       12 -2 
                    ) 
           - (-1) * det( 
                           4   3
                           -1 -2 
                       ) 
           + K * det( 
                           4   7
                           -1  12 
                       ) 

   det(m) = 1 * (7 * (-2) - 3 * 12) + 1 * (4 * (-2) - 3 * (-1)) + K * (4 * 12 - 7 * (-1))
   det(m) = 1 * (-14 - 36) + 1 * (-8 + 3) + K * (48 + 7)
   det(m) = -50 - 5 + 55K
   det(m) = 55K - 55

**2. Solve the Equation 'mx = (1, 11, 21)' when K = Z**

* **Substitute K = Z into matrix 'm':**

   m = ( 
         1  -1   Z
         4   7   3
         -1  12 -2 
      )

* **Set up the augmented matrix:**

   [ 
      1  -1   Z  |  1
      4   7   3  |  11
      -1  12 -2  |  21
   ]

* **Perform row operations to get the matrix in row-echelon form:**

   1. **R2 = R2 - 4R1** 
      [ 
         1  -1   Z  |  1
         0  11  3-4Z  |  7
         -1  12 -2  |  21
      ]

   2. **R3 = R3 + R1** 
      [ 
         1  -1   Z  |  1
         0  11  3-4Z  |  7
         0  11  Z-2  |  22
      ]

   3. **R3 = R3 - R2** 
      [ 
         1  -1   Z  |  1
         0  11  3-4Z  |  7
         0   0  Z+1  |  15
      ]

* **Solve for Z:**

   * Z + 1 = 15
   * Z = 14

* **Back-substitute to find Y:**

   * 11Y + 3 - 4Z = 7
   * 11Y + 3 - 4(14) = 7
   * 11Y - 53 = 7
   * 11Y = 60
   * Y = 60/11

* **Back-substitute to find X:**

   * X - Y + Z = 1
   * X - (60/11) + 14 = 1
   * X = 1 + (60/11) - 14
   * X = -3 + (60/11)
   * X = 27/11

**Therefore, when K = Z, the solution to the equation 'mx = (1, 11, 21)' is:**

* **X = 27/11**
* **Y = 60/11**
* **Z = 14** 

I hope this comprehensive solution is helpful!