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3 years ago

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Given the following matrices A and B, find an invertible matrix U such that UA = B:

1 answer:

Naddik [55]3 years ago

3 0

<span>A and B must be invertible, we have UA=B, since A is invertible. A^-1 exists, by multiplying with A^-1, we have UA A^-1 =B A^-1. But AA^-1 = I (identity matrix) and XI=X, for all matrix X, we find UI= B A^-1, and U= B A^-1.</span>

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1. True
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Read 2 more answers

Maths functions question

yarga [219]

Answer:

a) OA = 1 unit

b) OB = 3 units

c) AB = √10 units

Step-by-step explanation:

<u>Given function</u>:

$g(x)=2^x$

<h3><u>Part (a)</u></h3>

Point A is the y-intercept of the exponential curve (so when x = 0).

To find the y-value of Point A, substitute x = 0 into the function:

$\implies g(0)=2^0=1$

Therefore, A (0, 1) so OA = 1 unit.

<h3><u>Part (b)</u></h3>

If BC = 8 units then the y-value of Point C is 8.

The find the x-value of Point C, set the function to 8 and solve for x:

$\begin{aligned}f(x) & = 8 \\\implies 2^x & = 8\\2^x & = 2^3\\\implies x &= 3\end{aligned}$

Therefore, C (3, 8) so Point B is (3, 0). Therefore, OB = 3 units.

<h3><u>Part (c)</u></h3>

From parts (a) and (b):

A = (0, 1)
B = (3, 0)

To find the length of AB, use the distance between two points formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points.}$

Therefore:

$\implies \sf AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

$\implies \sf AB=\sqrt{(3-0)^2+(0-1)^2}$

$\implies \sf AB=\sqrt{(3)^2+(-1)^2}$

$\implies \sf AB=\sqrt{9+1}$

$\implies \sf AB=\sqrt{10}\:\:units$

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