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vlada-n [284]
3 years ago
15

Jacob solves the system of equations by forming a matrix equation.

Mathematics
1 answer:
Lyrx [107]3 years ago
6 0

Simultaneous equations can be solved using inverse matrix operation.

The complete steps of Jacob's solution are:

\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]

We have:

4x + y = 2

-2x + 4y = -22

Calculate the determinant of \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]

|A| = 4 \times 3 -1 \times -2

|A| = 12 +2

|A| = 14

So, the inverse matrix becomes

A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]

Replace the first column with \left[\begin{array}{c}2&-22\end{array}\right] to calculate the value of x

x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]

So, we have:

x = \frac{1}{14}(2 \times 3 - 1 \times -22)

x = \frac{1}{14}(6 +22)

x = \frac{1}{14}(28)

x = 2

Replace the second column with \left[\begin{array}{c}2&-22\end{array}\right] to calculate the value of y

y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]

So, we have:

y = \frac{1}{14}(4 \times -22 - 2 \times -2)

y = \frac{1}{14}(-88 +4)

y = \frac{1}{14}(-84)

y = -6

Hence, the complete process is:

\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]

Read more about matrices at:

brainly.com/question/11367104

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