Question

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Answer 1

that matrix is used to find Dy

not Dx

Answer 2

notice you replaced the second column (the y column) with the right side

Answer 3

\(\large \begin{array}{ccllll}
&D_x\\
\begin{cases}
4x + 2y = 5\\
x - 3y = 7
\end{cases}\implies
\begin{bmatrix}
4&2&|&{\color{brown}{ 5}}\\
1&-3&|&{\color{brown}{ 7}}
\end{bmatrix}\qquad thus\to &
\begin{bmatrix}
{\color{brown}{ \square }}&2\\
{\color{brown}{ \square }}&3
\end{bmatrix}
\end{array}\)

Answer 4

hmmm missing a minus there so\(\large \begin{array}{ccllll}
&D_x\\
\begin{cases}
4x + 2y = 5\\
x - 3y = 7
\end{cases}\implies
\begin{bmatrix}
4&2&|&{\color{brown}{ 5}}\\
1&-3&|&{\color{brown}{ 7}}
\end{bmatrix}\qquad thus\to &
\begin{bmatrix}
{\color{brown}{ \square }}&2\\
{\color{brown}{ \square }}&-3
\end{bmatrix}
\end{array}\)

Answer 5

then surely you know how to get the determinant

Answer 6

hmm the \(\bf D_y?\)

Answer 7

\(\large \begin{array}{ccllll}
&D_y\\
\begin{cases}
4x + 2y = 5\\
x - 3y = 7
\end{cases}\implies
\begin{bmatrix}
4&2&|&{\color{brown}{ 5}}\\
1&-3&|&{\color{brown}{ 7}}
\end{bmatrix}\qquad thus\to &
\begin{bmatrix}
4&{\color{brown}{ 5 }}\\
1&{\color{brown}{ 7 }}
\end{bmatrix}
\end{array}\)

so you're asked for the determinant of that matrix then

Answer 8

yeap,.... that's the coefficient matrix for "y"
so you'd just need to get the determinant of that

Answer 9

hmm hmm

Answer 10

heheh I see I got the wrong.... value on the equal this a different system of equations but anyhow yieap
your'e correct
for 2x + 3y = 4
5x - y = 8

the \(\bf D_y\) is -4 :)

Answer 11

yw