Solve the system using Cramer's Rule. $$ \begin{array}{l}5 x-6 y-3 z=16 \ -2 x-4 y+6 z=-24 \ 3 x+4 y-5 z=18\end{array} $$ Find the determinant $$ D $$ (denominator). $$ D= $$ Find the determinant $$ D_{z} $$ associated with $$ x $$. $$ D_{z}= $$ Find the determinant $$ D_{z} $$ associated with $$ y $$. $$ D_{v}=\square $$ Find the determinant $$ D_{z} $$ associated with $$ z $$.

Question

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We are given the system
$$
\begin{array}{rcl}
5x-6y-3z &=& 16,\[1mm]
-2x-4y+6z &=& -24,\[1mm]
3x+4y-5z &=& 18.
\end{array}
$$

Using Cramer’s Rule, we first write the coefficient matrix
$$
A=\begin{pmatrix}
5 & -6 & -3\[1mm]
-2 & -4 & 6\[1mm]
3 & 4 & -5
\end{pmatrix}.
$$

### 1. Finding $$ D = \det(A) $$

We compute
$$
D = 
5\det\begin{pmatrix}-4 & 6\[1mm]4 & -5\end{pmatrix}
-(-6)\det\begin{pmatrix}-2 & 6\[1mm]3 & -5\end{pmatrix}
+(-3)\det\begin{pmatrix}-2 & -4\[1mm]3 & 4\end{pmatrix}\,.
$$

Calculate the $$2 \times 2$$ determinants:

- For the first minor:
  $$
  \det\begin{pmatrix}-4 & 6\[1mm]4 & -5\end{pmatrix}
  = (-4)(-5) - (6)(4) = 20 - 24 = -4\,.
  $$
- For the second minor:
  $$
  \det\begin{pmatrix}-2 & 6\[1mm]3 & -5\end{pmatrix}
  = (-2)(-5) - (6)(3) = 10 - 18 = -8\,.
  $$
- For the third minor:
  $$
  \det\begin{pmatrix}-2 & -4\[1mm]3 & 4\end{pmatrix}
  = (-2)(4) - (-4)(3) = -8 + 12 = 4\,.
  $$

Now substitute back:
$$
\begin{aligned}
D &= 5(-4) - (-6)(-8) + (-3)(4)\[1mm]
  &= -20 - 48 - 12\[1mm]
  &= -80\,.
\end{aligned}
$$
Thus,
$$
D = -80\,.
$$

---

### 2. Finding $$ D_x $$ (determinant for $$ x $$)

To find $$ D_x $$, we replace the first column of $$ A $$ by the constant vector:
$$
A_x = \begin{pmatrix}
16 & -6 & -3\[1mm]
-24 & -4 & 6\[1mm]
18 & 4 & -5
\end{pmatrix}\,.
$$

Now compute
$$
D_x = 
16\det\begin{pmatrix}-4 & 6\[1mm]4 & -5\end{pmatrix}
-(-6)\det\begin{pmatrix}-24 & 6\[1mm]18 & -5\end{pmatrix}
+(-3)\det\begin{pmatrix}-24 & -4\[1mm]18 & 4\end{pmatrix}\,.
$$

We have already computed the first minor:
$$
\det\begin{pmatrix}-4 & 6\[1mm]4 & -5\end{pmatrix} = -4\,.
$$

Next, compute the second minor:
$$
\det\begin{pmatrix}-24 & 6\[1mm]18 & -5\end{pmatrix}
= (-24)(-5) - (6)(18) = 120 - 108 = 12\,.
$$

Then, compute the third minor:
$$
\det\begin{pmatrix}-24 & -4\[1mm]18 & 4\end{pmatrix}
= (-24)(4) - (-4)(18) = -96 + 72 = -24\,.
$$

Now substitute:
$$
\begin{aligned}
D_x &= 16(-4) + 6(12) + (-3)(-24)\[1mm]
    &= -64 + 72 + 72\[1mm]
    &= 80\,.
\end{aligned}
$$
Thus,
$$
D_x = 80\,.
$$

---

### 3. Finding $$ D_y $$ (determinant for $$ y $$)

To find $$ D_y $$, we replace the second column of $$ A $$ by the constant vector:
$$
A_y = \begin{pmatrix}
5 & 16 & -3\[1mm]
-2 & -24 & 6\[1mm]
3 & 18 & -5
\end{pmatrix}\,.
$$

Compute
$$
D_y = 5\det\begin{pmatrix}-24 & 6\[1mm]18 & -5\end{pmatrix}
-16\det\begin{pmatrix}-2 & 6\[1mm]3 & -5\end{pmatrix}
+(-3)\det\begin{pmatrix}-2 & -24\[1mm]3 & 18\end{pmatrix}\,.
$$

First minor:
$$
\det\begin{pmatrix}-24 & 6\[1mm]18 & -5\end{pmatrix}
= (-24)(-5) - (6)(18) = 120 - 108 = 12\,.
$$

Second minor (already computed):
$$
\det\begin{pmatrix}-2 & 6\[1mm]3 & -5\end{pmatrix}
= 10 - 18 = -8\,.
$$

Third minor:
$$
\det\begin{pmatrix}-2 & -24\[1mm]3 & 18\end{pmatrix}
= (-2)(18) - (-24)(3) = -36 + 72 = 36\,.
$$

Now, substitute into the formula:
$$
\begin{aligned}
D_y &= 5(12) - 16(-8) -3(36)\[1mm]
    &= 60 + 128 - 108\[1mm]
    &= 80\,.
\end{aligned}
$$
Thus,
$$
D_y = 80\,.
$$

---

### 4. Finding $$ D_z $$ (determinant for $$ z $$)

To find $$ D_z $$, we replace the third column of $$ A $$ by the constant vector:
$$
A_z = \begin{pmatrix}
5 & -6 & 16\[1mm]
-2 & -4 & -24\[1mm]
3 & 4 & 18
\end{pmatrix}\,.
$$

Now compute
$$
D_z = 5\det\begin{pmatrix}-4 & -24\[1mm]4 & 18\end{pmatrix}
-(-6)\det\begin{pmatrix}-2 & -24\[1mm]3 & 18\end{pmatrix}
+16\det\begin{pmatrix}-2 & -4\[1mm]3 & 4\end{pmatrix}\,.
$$

Compute the minors one by one:

- First minor:
  $$
  \det\begin{pmatrix}-4 & -24\[1mm]4 & 18\end{pmatrix}
  = (-4)(18) - (-24)(4) = -72 + 96 = 24\,.
  $$
- Second minor:
  $$
  \det\begin{pmatrix}-2 & -24\[1mm]3 & 18\end{pmatrix}
  = (-2)(18) - (-24)(3) = -36 + 72 = 36\,.
  $$
- Third minor (previously computed):
  $$
  \det\begin{pmatrix}-2 & -4\[1mm]3 & 4\end{pmatrix}
  = -8 + 12 = 4\,.
  $$

Substitute:
$$
\begin{aligned}
D_z &= 5(24) + 6(36) + 16(4)\[1mm]
    &= 120 + 216 + 64\[1mm]
    &= 400\,.
\end{aligned}
$$
Thus,
$$
D_z = 400\,.
$$

---

### Final Answers

$$
\begin{aligned}
D &= -80,\[1mm]
D_x &= 80,\[1mm]
D_y &= 80,\[1mm]
D_z &= 400\,.
\end{aligned}
$$
$$ \begin{aligned} D &= -80,\ D_x &= 80,\ D_y &= 80,\ D_z &= 400. \end{aligned} $$

Solve the system using Cramer’s Rule.

Find the determinant (denominator).

Find the determinant associated with .

Find the determinant associated with .
Find the determinant associated with .

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Solución

1. Finding

2. Finding (determinant for )

3. Finding (determinant for )

4. Finding (determinant for )

Final Answers

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Step 1: Calculate

Step 2: Calculate , , and

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Solve the system using Cramer’s Rule. Find the determinant (denominator). Find the determinant associated with . Find the determinant associated with . Find the determinant associated with .