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mafiozo [28]
3 years ago
12

Let u=<-5,1>, v=<7,-4> find 9u-6v

Mathematics
2 answers:
katrin2010 [14]3 years ago
8 0
9u -6v = 9&lt;-5, 1> -6&lt;7, -4>
.. = &lt;9*-5 -6*7, 9*1 -6*-4>
.. = &lt;-87, 33>
Fantom [35]3 years ago
5 0

Answer:

9u-6v=\langle-87, 33\rangle

Step-by-step explanation:

In order to sum the vector u and the vector v , it is convenient to define some operations.

The vector addition is given by:

u \pm v = \langle u_1 \pm v_1 , u_2\pm v_2 ,..,u_n\pm v_n\rangle

And the scalar multiplication is given by:

ku=k \langle u_1 ,u_2,..., u_n\rangle =\langle k u_1, k u_2,..., k u_n \rangle

Using the previous definitions, let's solve the problem.

First, let's find 9u :

9u=9 \langle -5,1 \rangle =\langle 9*(-5), 9*(1)\rangle=\langle-45, 9\rangle

Now, let's find 6v :

6v=6\langle 7,-4 \rangle =\langle 6*(7), 6*(-4)\rangle=\langle42, -24\rangle

Finally, let's find 9u-6v:

9u-6v=\langle-45, 9\rangle - \langle42, -24\rangle =\langle-45 -42, 9-(-24)\rangle=\langle-87, 33\rangle

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Read 2 more answers
A = ???? 4 −2
irinina [24]

Answer:

1. The matrix A isn't the inverse of matrix B.

2. |B|=12, |A|=12

Step-by-step explanation:

1. We want to know if matrix A is the inverse of matrix B, this means that if you do the product between B and A you have to obtain the identity matrix.

We have:

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

and

B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]

A and B are 2×2 matrices (2 rows and 2 columns), if you multiply them you have to obtain a 2×2 matrix.

Then if A is the inverse of B:

B.A=I

Where,

I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]

Observation:

If you have two matrices:

A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A.B=\left[\begin{array}{cc}(a.e+b.g)&(a.f+b.h)\\(c.e+d.g)&(c.f+d.h)\end{array}\right]

Now:

B.A=\left[\begin{array}{cc}3&2\\1&4\end{array}\right].\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}4.3+(-2).1&4.2+(-2).4\\(-1).3+3.1&(-1).2+3.4\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}12-2&8-8\\-3+3&-2+12\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]

B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]\neq \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=I\\\\\\B.A\neq I

Then, the matrix A isn't the inverse of matrix B.

2. If you have a matrix A:

A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]

The determinant of the matrix is:

|A|=ad-bc

Then the determinant of B is:

B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]

a=3, b=2, c=1, d=4

|B|=3.4-2.1\\|B|=12-2=10

The determinant of A is:

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

a=4, b=-2, c=-1, d=3

|A|=4.3-(-2).(-1)\\|B|=12-2=10

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