We wish to solve the following system of linear equations by using Cramer's Rule (the Method of Determinants). $$ \left\{\begin{array}{l} 2 x-y=-3 \ -7 x+3 y+z=0 \ -7 x+5 z=1 \end{array}\right. $$ Begin by calculating these four determinants: (i) $$ \operatorname{det}\left[\begin{array}{ccc}-\mathbf{3} & -\mathbf{1} & \mathbf{0} \ 0 & 3 & 1 \ 1 & 0 & 5\end{array}\right]= $$ $$ \square $$ (ii) $$ \operatorname{det}\left[\begin{array}{ccc}2 & -3 & 0 \ -7 & 0 & 1 \ -7 & 1 & 5\end{array}\right]= $$ $$ \square $$ (iii) $$ \operatorname{det}\left[\begin{array}{ccc}2 & -1 & -3 \ -7 & 3 & 0 \ -7 & 0 & 1\end{array}\right]= $$ $$ \square $$ (iv) $$ \operatorname{det}\left[\begin{array}{ccc}2 & -1 & 0 \ -7 & 3 & 1 \ -7 & 0 & 5\end{array}\right]= $$ $$ \square $$ Now divide these determinants appropriately to find $$ x, y $$, and $$ z $$, rounding to two decimal places where necessary, $$ \begin{array}{l} x=\square \ y=\square \ z=\square \end{array} $$ Submit Ouestion

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UpStudy AI · Accepted Answer

**(i)**  
We are given the matrix  
$$
\begin{pmatrix}
-3 & -1 & 0 \
0 & 3 & 1 \
1 & 0 & 5
\end{pmatrix}.
$$  
Its determinant is  
$$
\begin{aligned}
\det & = -3\Bigl(3\cdot5-1\cdot0\Bigr)
-(-1)\Bigl(0\cdot5-1\cdot1\Bigr)
+0\Bigl(0\cdot0-3\cdot1\Bigr)\[1mm]
& = -3(15) +1(-1)+0\[1mm]
& = -45-1\[1mm]
& = -46.
\end{aligned}
$$

**(ii)**  
We are given the matrix  
$$
\begin{pmatrix}
2 & -3 & 0 \
-7 & 0 & 1 \
-7 & 1 & 5
\end{pmatrix}.
$$  
Its determinant is  
$$
\begin{aligned}
\det & = 2\Bigl(0\cdot5-1\cdot1\Bigr)
-(-3)\Bigl((-7)\cdot5-1\cdot(-7)\Bigr)
+0\Bigl((-7)\cdot1-0\cdot(-7)\Bigr)\[1mm]
& = 2(0-1) +3\Bigl(-35+7\Bigr) +0\[1mm]
& = 2(-1) +3(-28)\[1mm]
& = -2 -84\[1mm]
& = -86.
\end{aligned}
$$

**(iii)**  
We are given the matrix  
$$
\begin{pmatrix}
2 & -1 & -3 \
-7 & 3 & 0 \
-7 & 0 & 1
\end{pmatrix}.
$$  
Its determinant is  
$$
\begin{aligned}
\det & = 2\Bigl(3\cdot1-0\cdot0\Bigr)
-(-1)\Bigl((-7)\cdot1-0\cdot(-7)\Bigr)
+(-3)\Bigl((-7)\cdot0-3\cdot(-7)\Bigr)\[1mm]
& = 2(3) -(-1)(-7) +(-3)(0+21)\[1mm]
& = 6 -7 -63\[1mm]
& = -64.
\end{aligned}
$$

**(iv)**  
We are given the matrix  
$$
\begin{pmatrix}
2 & -1 & 0 \
-7 & 3 & 1 \
-7 & 0 & 5
\end{pmatrix}.
$$  
Its determinant is  
$$
\begin{aligned}
\det & = 2\Bigl(3\cdot5-1\cdot0\Bigr)
-(-1)\Bigl((-7)\cdot5-1\cdot(-7)\Bigr)
+0\Bigl((-7)\cdot0-3\cdot(-7)\Bigr)\[1mm]
& = 2(15) +1\Bigl(-35+7\Bigr) +0\[1mm]
& = 30 -28\[1mm]
& = 2.
\end{aligned}
$$

Now, the original system of equations is  
$$
\begin{cases}
2x-y= -3,\[1mm]
-7x+3y+z=0,\[1mm]
-7x+5z=1.
\end{cases}
$$  
Its coefficient matrix is  
$$
A=\begin{pmatrix}
2 & -1 & 0\[1mm]
-7 & 3 & 1\[1mm]
-7 & 0 & 5
\end{pmatrix},
$$  
with determinant $$(iv)=2$$.

Using Cramer’s Rule, we construct and use the following determinants:
- Replace the first column with the constants $$(-3,0,1)$$ to obtain  
  $$
  D_x = -46.
  $$
- Replace the second column with the constants to obtain  
  $$
  D_y = -86.
  $$
- Replace the third column with the constants to obtain  
  $$
  D_z = -64.
  $$

Then, the solutions are given by  
$$
x=\frac{D_x}{D}=\frac{-46}{2}=-23,
$$
$$
y=\frac{D_y}{D}=\frac{-86}{2}=-43,
$$
$$
z=\frac{D_z}{D}=\frac{-64}{2}=-32.
$$

Thus, the calculated values are  
$$
\text{(i) } -46,\quad \text{(ii) } -86,\quad \text{(iii) } -64,\quad \text{(iv) } 2,
$$  
and  
$$
x=-23,\quad y=-43,\quad z=-32.
$$
**(i)** Determinant = -46 **(ii)** Determinant = -86 **(iii)** Determinant = -64 **(iv)** Determinant = 2 Solutions: $$ x = -23,\quad y = -43,\quad z = -32 $$

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