Without graphing, is the system independent, dependent, inconsistent? {12x=3y=12 y=-4x+5

1 answer:

7 0

Okay so you out them into the form of y=mx+b. since equation 2 is already like that you need to do it to equation 1. which is y=-4x+4. graph both equations. if it has a solution (1 point where the two lines meet) it is consistantly and independent. if they are parallel lines and the solution is 0 the system is inconsistent and the lines are dependent. if it's the same line they are consistent and dependent. the line is not the same since the y intercept is different. the slope is the same though which tells us its parallel. so the system has a solution of 0 and is inconsistent and the lines are independent.

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Read 2 more answers

Check answer please

Cerrena [4.2K]

The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

$\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}$

$\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}$

$\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}$

$\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}$

Hope it helps.

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