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07.42
Pertanyaan 2 dari 40
118:03
Diketahui matriks berikut :
Apabila B - A
=C^(t), dimana
C^(t)= transpose matriks
C, maka nilai
x.y=
qquad

B=[[x+y,2],[3,y]]" dan "C=[[7,2],[3,1]]
25
10
30
20
15
SEBELUMNYA
SELANJUTNYA >

$07$ . $42$ $\newline$ Pertanyaan $2$ dari $40$ $\newline$ $118$ : $03$ $\newline$ Diketahui matriks berikut : $\newline$ Apabila B - A $=C^{t}$ , dimana $C^{t}=$ transpose matriks $C$ , maka nilai $x . y=$ $\qquad$ $\newline$ $B=\left[\begin{array}{cc} x+y & 2 \\ 3 & y \end{array}\right] \text { dan } C=\left[\begin{array}{ll} 7 & 2 \\ 3 & 1 \end{array}\right]$ $\newline$ $25$ $\newline$ $10$ $\newline$ $30$ $\newline$ $20$ $\newline$ $15$ $\newline$ SEBELUMNYA $\newline$ SELANJUTNYA >

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Convert customary and metric units using proportions

Full solution

Q.

07

.

42

\newline

Pertanyaan

2

dari

40

\newline

118

:

03

\newline

Diketahui matriks berikut :

\newline

Apabila B - A

=C^{t}

, dimana

C^{t}=

transpose matriks

C

, maka nilai

x . y=

\qquad

\newline

B=\left[\begin{array}{cc} x+y & 2 \\ 3 & y \end{array}\right] \text { dan } C=\left[\begin{array}{ll} 7 & 2 \\ 3 & 1 \end{array}\right]

\newline

25

\newline

10

\newline

30

\newline

20

\newline

15

\newline

SEBELUMNYA

\newline

SELANJUTNYA >

Write Matrix A: First, let's write down matrix $A$ by comparing $B - A = C^{t}$ with the given matrices $B$ and $C$ .
B = \left[\begin{array}{cc}x+y & 2\3 & y\end{array}\right]
C^{t} = \left[\begin{array}{cc}7 & 3\2 & 1\end{array}\right]
So, $A = B - C^{t}$ .
Calculate A: Now, let's calculate A by subtracting $C^{(t)}$ from B. $\newline$ $A = [[x+y - 7, 2 - 3], [3 - 2, y - 1]]$ $\newline$ $A = [[x+y - 7, -1], [1, y - 1]]$
Set Up Equations: Since $A$ is the result of $B - C^{t}$ , and we know that $A$ must be a matrix with zero entries (because $B - A = C^{t}$ ), we can set up equations based on the entries of $A$ . $x+y - 7 = 0$ and $y - 1 = 0$ .
Solve Equations: Solve the equations for $x$ and $y$ . From $y - 1 = 0$ , we get $y = 1$ . Substitute $y = 1$ into $x+y - 7 = 0$ to get $x + 1 - 7 = 0$ .
Calculate Product: Solve for $x$ . $x - 6 = 0$ $x = 6$
Calculate Product: Solve for $x$ . $x - 6 = 0$ $x = 6$ Now we have $x = 6$ and $y = 1$ . Calculate the product $x \times y$ . $x \times y = 6 \times 1$ $x \times y = 6$

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