Questions: Can Cramer's rule be used to solve this system of equations? 15x₁-4x₂=6 -41x₁+11x₂=-3 Yes No If Yes, write the determinants used to compute x₂. Compute x₂=

Can Cramer's rule be used to solve this system of equations?

15x₁-4x₂=6
-41x₁+11x₂=-3

Yes
No

If Yes, write the determinants used to compute x₂.

Compute

x₂=

Transcript text: Can Cramer's rule be used to solve this system of equations? \[ \left\{\begin{array}{l} 15 x_{1}-4 x_{2}=6 \\ -41 x_{1}+11 x_{2}=-3 \end{array}\right. \] Yes No If Yes, write the determinants used to compute $\mathrm{x}_{2}$. Compute \[ x_{2}= \] $\square$

Solution

Solution Steps

Step 1: Check if Cramer's rule is applicable

To apply Cramer's rule, the determinant of the coefficient matrix, $D$, must not be zero.

Given the system of equations: \[ \left\{\begin{array}{l} a_1 x_{1} + b_1 x_{2} = c_1 \\ a_2 x_{1} + b_2 x_{2} = c_2 \end{array}\right. \] The determinant $D = a_1b_2 - a_2b_1 = 15 \times 11 + 41 \times -4 = 1$.

Since $D \neq 0$, Cramer's rule can be applied to find the unique solution for $x_1$ and $x_2$.

Step 2: Compute the determinants $D_{x_1}$ and $D_{x_2}$

$D_{x_1} = c_1b_2 - c_2b_1 = 6 \times 11 + 3 \times -4 = 54$ $D_{x_2} = a_1c_2 - a_2c_1 = 15 \times -3 + 41 \times 6 = 201$

Step 3: Solve for $x_1$ and $x_2$

The solutions are obtained by dividing the determinants by $D$: $x_1 = \frac{D_{x_1}}{D} = \frac{54}{1} = 54$ $x_2 = \frac{D_{x_2}}{D} = \frac{201}{1} = 201$