Question

1. find the solution to the linear system using determinants.

\

$$\begin{cases} x + 2y = -3 \\\\ 2x - 3y = 8 \\end{cases}$$

\bigcirc\\ (-2, 1)
\bigcirc\\ (-1, 2)
\bigcirc\\ (1, -2)
\bigcirc\\ (2, -1)

1. find the solution to the linear system using determinants.

\

$$\begin{cases} x - 2y + z = 0 \\\\ x + 3y - 2z = -6 \\\\ 2x - y + 4z = 4 \\end{cases}$$

\bigcirc\\ (2, 0, -2)
\bigcirc\\ (-2, 0, 2)
\bigcirc\\ (2, 0, 2)
\bigcirc\\ (-2, 0, -2)

Explanation:

Question 3

Step1: Recall Cramer's Rule

For a system \(

$$\begin{cases}a_1x + b_1y = c_1\\a_2x + b_2y = c_2\end{cases}$$

\), the determinant \(D=

$$\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}$$

\), \(D_x=

$$\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix}$$

\), \(D_y=

$$\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}$$

\). Then \(x = \frac{D_x}{D}\), \(y=\frac{D_y}{D}\) (if \(D
eq0\)).

For the system \(

$$\begin{cases}x + 2y=-3\\2x-3y = 8\end{cases}$$

\), we have \(a_1 = 1\), \(b_1=2\), \(c_1=-3\); \(a_2 = 2\), \(b_2=-3\), \(c_2 = 8\).

Step2: Calculate \(D\)

\(D=

$$\begin{vmatrix}1&2\\2&-3\end{vmatrix}$$

=1\times(-3)-2\times2=-3 - 4=-7\)

Step3: Calculate \(D_x\)

\(D_x=

$$\begin{vmatrix}-3&2\\8&-3\end{vmatrix}$$

=(-3)\times(-3)-2\times8 = 9-16=-7\)

Step4: Calculate \(D_y\)

\(D_y=

$$\begin{vmatrix}1&-3\\2&8\end{vmatrix}$$

=1\times8-(-3)\times2=8 + 6 = 14\)

Step5: Find \(x\) and \(y\)

\(x=\frac{D_x}{D}=\frac{-7}{-7}=1\)? Wait, no, wait, I made a mistake. Wait, no, let's recalculate \(D_x\) and \(D_y\) again. Wait, no, the system is \(x + 2y=-3\) and \(2x-3y = 8\). So \(c_1=-3\), \(c_2 = 8\).

Wait, \(D_x\) should be \(

$$\begin{vmatrix}-3&2\\8&-3\end{vmatrix}$$

=(-3)\times(-3)-2\times8=9 - 16=-7\)

\(D_y=

$$\begin{vmatrix}1&-3\\2&8\end{vmatrix}$$

=1\times8-(-3)\times2=8 + 6 = 14\)

Wait, but \(D=-7\), so \(x=\frac{D_x}{D}=\frac{-7}{-7}=1\)? No, that's not matching the options. Wait, maybe I mixed up \(D_x\) and \(D_y\) definitions. Wait, \(D_x\) is replacing the \(x\)-column with constants, so \(D_x=

$$\begin{vmatrix}-3&2\\8&-3\end{vmatrix}$$

\), \(D_y=

$$\begin{vmatrix}1&-3\\2&8\end{vmatrix}$$

\). Wait, but let's check the options. Wait, the options are \((-2,1)\), \((-1,2)\), \((1,-2)\), \((2,-1)\). Wait, maybe I made a mistake in calculation.

Wait, let's solve the system by substitution. From the first equation, \(x=-3 - 2y\). Substitute into the second equation: \(2(-3 - 2y)-3y=8\)

\(-6-4y-3y = 8\)

\(-6-7y=8\)

\(-7y=14\)

\(y=-2\)

Then \(x=-3-2\times(-2)=-3 + 4 = 1\). Wait, so \(x = 1\), \(y=-2\), which is the option \((1,-2)\). Wait, my Cramer's rule calculation had an error. Let's recalculate \(D\), \(D_x\), \(D_y\) correctly.

Wait, \(D=

$$\begin{vmatrix}1&2\\2&-3\end{vmatrix}$$

=1\times(-3)-2\times2=-3-4=-7\)

\(D_x\): replace \(x\)-column with constants: \(

$$\begin{vmatrix}-3&2\\8&-3\end{vmatrix}$$

=(-3)\times(-3)-2\times8=9 - 16=-7\). So \(x=\frac{-7}{-7}=1\)

\(D_y\): replace \(y\)-column with constants: \(

$$\begin{vmatrix}1&-3\\2&8\end{vmatrix}$$

=1\times8-(-3)\times2=8 + 6 = 14\). Then \(y=\frac{14}{-7}=-2\). Ah, there we go. So \(x = 1\), \(y=-2\), so the solution is \((1,-2)\).

Step1: Recall Cramer's Rule for 3x3 system

For a system \(

$$\begin{cases}a_1x + b_1y + c_1z=d_1\\a_2x + b_2y + c_2z=d_2\\a_3x + b_3y + c_3z=d_3\end{cases}$$

\), the determinant \(D=

$$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}$$

\), \(D_x=

$$\begin{vmatrix}d_1&b_1&c_1\\d_2&b_2&c_2\\d_3&b_3&c_3\end{vmatrix}$$

\), \(D_y=

$$\begin{vmatrix}a_1&d_1&c_1\\a_2&d_2&c_2\\a_3&d_3&c_3\end{vmatrix}$$

\), \(D_z=

$$\begin{vmatrix}a_1&b_1&d_1\\a_2&b_2&d_2\\a_3&b_3&d_3\end{vmatrix}$$

\). Then \(x=\frac{D_x}{D}\), \(y=\frac{D_y}{D}\), \(z=\frac{D_z}{D}\) (if \(D
eq0\)).

For the system \(

$$\begin{cases}x-2y + z=0\\x + 3y-2z=-6\\2x-y + 4z=4\end{cases}$$

\), we have \(a_1 = 1\), \(b_1=-2\), \(c_1=1\), \(d_1=0\); \(a_2 = 1\), \(b_2=3\), \(c_2=-2\), \(d_2=-6\); \(a_3 = 2\), \(b_3=-1\), \(c_3=4\), \(d_3=4\).

Step2: Calculate \(D\)

\(D=

$$\begin{vmatrix}1&-2&1\\1&3&-2\\2&-1&4\end{vmatrix}$$

\)

Expand along the first row:

\(1\times

$$\begin{vmatrix}3&-2\\-1&4\end{vmatrix}$$

-(-2)\times

$$\begin{vmatrix}1&-2\\2&4\end{vmatrix}$$

+1\times

$$\begin{vmatrix}1&3\\2&-1\end{vmatrix}$$

\)

Calculate each minor:

\(

$$\begin{vmatrix}3&-2\\-1&4\end{vmatrix}$$

=3\times4-(-2)\times(-1)=12 - 2 = 10\)

\(

$$\begin{vmatrix}1&-2\\2&4\end{vmatrix}$$

=1\times4-(-2)\times2=4 + 4 = 8\)

\(

$$\begin{vmatrix}1&3\\2&-1\end{vmatrix}$$

=1\times(-1)-3\times2=-1 - 6=-7\)

So \(D=1\times10+2\times8+1\times(-7)=10 + 16-7=19\)? Wait, no, wait the sign for the second term: \(-(-2)=+2\), so \(+2\times8 = 16\), and the third term is \(+1\times(-7)=-7\). So \(10 + 16-7 = 19\)? Wait, but let's check by another method. Alternatively, use row operations.

Wait, maybe it's easier to test the options. Let's test option \((-2,0,2)\):

First equation: \(-2-2\times0 + 2=0\), which is \(0 = 0\) (good).

Second equation: \(-2+3\times0-2\times2=-2 - 4=-6\), which matches \(-6\) (good).

Third equation: \(2\times(-2)-0 + 4\times2=-4 + 8 = 4\), which matches \(4\) (good). Wait, so the solution is \((-2,0,2)\).

Wait, let's verify with Cramer's rule.

Calculate \(D_x\): replace \(x\)-column with \(d_1,d_2,d_3\) (0, -6, 4)

\(D_x=

$$\begin{vmatrix}0&-2&1\\-6&3&-2\\4&-1&4\end{vmatrix}$$

\)

Expand along the first row:

\(0\times

$$\begin{vmatrix}3&-2\\-1&4\end{vmatrix}$$

-(-2)\times

$$\begin{vmatrix}-6&-2\\4&4\end{vmatrix}$$

+1\times

$$\begin{vmatrix}-6&3\\4&-1\end{vmatrix}$$

\)

\(=0 + 2\times[(-6)\times4-(-2)\times4]+1\times[(-6)\times(-1)-3\times4]\)

\(=2\times(-24 + 8)+1\times(6 - 12)\)

\(=2\times(-16)+1\times(-6)=-32-6=-38\)

Then \(x=\frac{D_x}{D}\). Wait, but we found by testing that \(x=-2\), so \(D_x=-38\), then \(D\) should be \(\frac{D_x}{x}=\frac{-38}{-2}=19\). Let's recalculate \(D\) correctly.

\(D=

$$\begin{vmatrix}1&-2&1\\1&3&-2\\2&-1&4\end{vmatrix}$$

\)

Using row 2 = row 2 - row 1: \(

$$\begin{vmatrix}1&-2&1\\0&5&-3\\2&-1&4\end{vmatrix}$$

\)

Row 3 = row 3 - 2*row 1: \(

$$\begin{vmatrix}1&-2&1\\0&5&-3\\0&3&2\end{vmatrix}$$

\)

Now expand along the first column:

\(1\times

$$\begin{vmatrix}5&-3\\3&2\end{vmatrix}$$

=5\times2-(-3)\times3=10 + 9 = 19\) (so \(D = 19\))

Now \(D_x=-38\), so \(x=\frac{-38}{19}=-2\)

Calculate \(D_y\): replace \(y\)-column with \(0, -6, 4\)

\(D_y=

$$\begin{vmatrix}1&0&1\\1&-6&-2\\2&4&4\end{vmatrix}$$

\)

Expand along the second column (since it has a 0):

\(-0\times

$$\begin{vmatrix}1&-2\\2&4\end{vmatrix}$$

+(-6)\times(-1)\times

$$\begin{vmatrix}1&1\\2&4\end{vmatrix}$$

-4\times

$$\begin{vmatrix}1&1\\1&-2\end{vmatrix}$$

\)

\(=0 + 6\times(4 - 2)-4\times(-2 - 1)\)

\(=6\times2-4\times(-3)=12 + 12 = 24\)? Wait, no, that can't be. Wait, no, the formula for expanding along column \(j\) is \(\sum_{i = 1}^n (-1)^{i + j}a_{ij}M_{ij}\). For column 2, \(j = 2\):

\(i=1\): \((-1)^{1+2}…

Answer:

\((1,-2)\)

Question 4