1. **State the problem:** We need to find all solutions $(x,y,z)$ to the system of linear equations: $$ \begin{cases} 96x + 25y + 15z = 13 \\ -26x - 7y - 4z = 3 \\ 7x + 2y + z = 12 \end{cases} $$ 2. **Write the augmented matrix:** $$ B = \left[\begin{array}{ccc|c} 96 & 25 & 15 & 13 \\ -26 & -7 & -4 & 3 \\ 7 & 2 & 1 & 12 \end{array}\right] $$ 3. **Apply row operations to find the RREF:** - Start with $R_1$ as is. - Replace $R_2$ by $R_2 + \frac{26}{96} R_1$ to eliminate $x$ in $R_2$. - Replace $R_3$ by $R_3 - \frac{7}{96} R_1$ to eliminate $x$ in $R_3$. 4. **Calculate:** $$ R_2 = \left(-26, -7, -4, 3\right) + \frac{26}{96} \times \left(96, 25, 15, 13\right) = \left(0, -7 + \frac{26}{96} \times 25, -4 + \frac{26}{96} \times 15, 3 + \frac{26}{96} \times 13\right) $$ Calculate each term: $$ -7 + \frac{26}{96} \times 25 = -7 + \frac{650}{96} = -7 + 6.7708 = -0.2292 $$ $$ -4 + \frac{26}{96} \times 15 = -4 + \frac{390}{96} = -4 + 4.0625 = 0.0625 $$ $$ 3 + \frac{26}{96} \times 13 = 3 + \frac{338}{96} = 3 + 3.5208 = 6.5208 $$ So, $$ R_2 = (0, -0.2292, 0.0625, 6.5208) $$ Similarly for $R_3$: $$ R_3 = (7, 2, 1, 12) - \frac{7}{96} \times (96, 25, 15, 13) = (7 - 7, 2 - \frac{175}{96}, 1 - \frac{105}{96}, 12 - \frac{91}{96}) $$ Calculate each term: $$ 2 - \frac{175}{96} = 2 - 1.8229 = 0.1771 $$ $$ 1 - \frac{105}{96} = 1 - 1.09375 = -0.09375 $$ $$ 12 - \frac{91}{96} = 12 - 0.9479 = 11.0521 $$ So, $$ R_3 = (0, 0.1771, -0.09375, 11.0521) $$ 5. **Next, eliminate $y$ in $R_3$ using $R_2$:** Multiply $R_2$ by $\frac{0.1771}{-0.2292} \approx -0.7729$ and add to $R_3$: $$ R_3 = R_3 + (-0.7729) \times R_2 $$ Calculate: $$ 0.1771 + (-0.7729)(-0.2292) = 0.1771 + 0.1771 = 0 $$ $$ -0.09375 + (-0.7729)(0.0625) = -0.09375 - 0.0483 = -0.14205 $$ $$ 11.0521 + (-0.7729)(6.5208) = 11.0521 - 5.038 = 6.0141 $$ So, $$ R_3 = (0, 0, -0.14205, 6.0141) $$ 6. **Normalize pivots:** Divide $R_2$ by $-0.2292$: $$ R_2 = \left(0, 1, \frac{0.0625}{-0.2292}, \frac{6.5208}{-0.2292}\right) = (0, 1, -0.2727, -28.46) $$ Divide $R_3$ by $-0.14205$: $$ R_3 = (0, 0, 1, -42.33) $$ 7. **Back substitution:** From $R_3$: $z = -42.33$ From $R_2$: $y - 0.2727 z = -28.46 \Rightarrow y = -28.46 + 0.2727 \times (-42.33) = -28.46 - 11.54 = -40$ From $R_1$: $96x + 25y + 15z = 13$ Substitute $y$ and $z$: $$ 96x + 25(-40) + 15(-42.33) = 13 $$ $$ 96x - 1000 - 635 = 13 $$ $$ 96x - 1635 = 13 $$ $$ 96x = 1648 $$ $$ x = \frac{1648}{96} = 17.17 $$ **Final solution:** $$ \boxed{\left(x,y,z\right) = \left(17.17, -40, -42.33\right)} $$