Suppose A and B are square matrices of the same size. Which of the following are necessarily true?

Mathematics

1 answer:

kiruha [24]3 years ago

5 0

Answer:

c. (A + B)^2-A^2 + 2AB + B^2

Step-by-step explanation:

Given that:

(A-B)^2 = A^2 - 2AB + B^2

If and only if AB = BA

Then;

(A-B)^2 = (A -B ) (A - B)

(A-B)^2 = A^2 - AB-BA + B^2 (FALSE)

(AB)^2=A^2B^2

on true if any only if AB =BA

(AB)^2= (AB) (AB)

(A+ B)^2 = A^2 + 2AB + B^2

(A+ B)^2 = (A + B) (A+B)

(A+ B)² = A × A + A × B + B × A + B × B

(A+ B)^2 = A^2 + A*B + B*A + B^2

This is true

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Help me ...Given △ABC≅△EFG, which congruency statement is true?

MrRa [10]

Answer:

(C.) AC¯¯¯¯¯≅EG¯¯¯¯¯ is true

Step-by-step explanation:

The positions of the letters in naming the triangles in the statement of congruent tells you which sides are congruent.

△ABC≅△EFG

Sides AB and EF are congruent

△ABC≅△EFG

Sides BC and FG are congruent

△ABC≅△EFG

Sides AC and EG are congruent

Look at the choices:

(C.) AC¯¯¯¯¯≅EG¯¯¯¯¯ is true

3 0

1 year ago

1/2x + 1/3y = 7

pashok25 [27]

let's multiply both sides in each equation by the LCD of all fractions in it, thus doing away with the denominator.

$\begin{cases} \cfrac{1}{2}x+\cfrac{1}{3}y&=7\\\\ \cfrac{1}{4}x+\cfrac{2}{3}y&=6 \end{cases}\implies \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6\left( \cfrac{1}{2}x+\cfrac{1}{3}y \right)=6(7)}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{1}{4}x+\cfrac{2}{3}y\right)=12(6)} \end{cases}\implies \begin{cases} 3x+2y=42\\ 3x+8y=72 \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{using elimination}}{ \begin{array}{llll} 3x+2y=42&\times -1\implies &\begin{matrix} -3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-2y=&-42\\ 3x+8y-72 &&~~\begin{matrix} 3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+8y=&72\\ \cline{3-4}\\ &&~\hfill 6y=&30 \end{array}} \\\\\\ y=\cfrac{30}{6}\implies \blacktriangleright y=5 \blacktriangleleft \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{substituting \underline{y} on the 1st equation}~\hfill }{3x+2(5)=42\implies 3x+10=42}\implies 3x=32 \\\\\\ x=\cfrac{32}{3}\implies \blacktriangleright x=10\frac{2}{3} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left(10\frac{2}{3}~~,~~5 \right)~\hfill$