Question

Is the sector (2,-5,3) in the subspace of R^3S panned by the vector (1,-3,2),(2,-4,-1),(1,-5,7)​

Answer 1

Concept:

A matrix is a rectangular array or table of numbers, symbols, or expressions, that are arranged in rows and columns.

Given:

We have,

The subspace of R^⁽³S⁾ panned by the vector (1, -3 ,2), (2, -4, -1), (1, -5, 7)

Find:

We are asked to find whether sector (2, -5, 3) lies in the subspace of R^⁽³S⁾ panned by the vector (1, -3, 2), (2, -4, -1), (1, -5, 7)​.

Solution:

So,

We have,

For a given sector to lie in the subspace panned by a vector, we should be able to write it in the terms of the given vector.

And the equations so formed, should be valid with solutions belonging to the set of real numbers.

⇒

$\left[\begin{array}{ccc}2\\(-5)\\3\end{array}\right] = x \left[\begin{array}{ccc}1\\(-3)\\2\end{array}\right] + y \left[\begin{array}{ccc}2\\(-4)\\1\end{array}\right] + z \left[\begin{array}{ccc}1\\(-5)\\7\end{array}\right]$

$\left[\begin{array}{ccc}2\\(-5)\\3\end{array}\right] = \left[\begin{array}{ccc}x+2y+z\\-3x -4y -z\\2x-y+7z\end{array}\right]$

Now,

Comparing both sides,

We get,

2 = x +2y + z       ------(1)

-5 = -3x -4y -5z   -------(2)

3 = 2x - y +7z       ------(3)

From equation (1),

we get,

x = 2 - 2y - z    -----(4)

Now,

Putting value of x from equation (4) in equation (3),

3 = 2(2 - 2y - z) - y + 7z

3 = 4 -4y - 2z - y + 7z

5y - 5z = 1

We get,

y - z = 1/5    ------(5)

Now,

Putting value of x from equation (4) in equation (2),

-5 = -3(2 - 2y - z) - 4y - 5z

-5 = -6 + 6y +3z - 4y - 5z

2y - 2z = -1

We get,

y - z = -1/2    ------(6)

From equation  (5) and  equation  (6),

We get,

It is not possible to have 2 different values for the same equation

∴ Given equations are not valid.

Hence, the given sector does not lie in the R³ sector panned by the vector (1,-3,2), (2,-4,-1), (1,-5,7)​.

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