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Bezzdna [24]
3 years ago
10

A= [1, 3; 2, 1], B=[3, 6; -1, 1]. Find AB & BA if possible

Mathematics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}

Step-by-step explanation:

For two matrix P and Q, the product, say PQ is defined when:

The number of columns of P = The number of rows of Q

Since A is a 2×2 matrix and B is also a 2×2 matrix

Thus both AB and BA are possible.

So AB is:

AB\Rightarrow\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\\AB\Rightarrow\quad \begin{bmatrix}3\times 1+3\times (-1) & 6\times 1+3\times 1\\3\times 2+1\times (-1) & 6\times 2+1\times 1\end{bmatrix}\\AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}

BA is:

BA\Rightarrow\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\\BA\Rightarrow\quad \begin{bmatrix}3\times 1+6\times 2 & 3\times 3+6\times 1\\(-1)\times 1+1\times 2 & (-1)\times 3+1\times 1\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}

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Answer:

144

Step-by-step explanation:

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Step-by-step explanation:

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Tatiana [17]
So,

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3 years ago
(ab^2 x -b^4a^3)^2<br><br> What is the answers
Umnica [9.8K]

Answer:

\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8

Step-by-step explanation:

Given the expression

\left(ab^2\:x\:-b^4a^3\right)^2

solving the expression

\left(ab^2\:x\:-b^4a^3\right)^2

\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a-b\right)^2=a^2-2ab+b^2

a=ab^2x,\:\:b=b^4a^3

so the expression becomes

=\left(ab^2x\right)^2-2ab^2xb^4a^3+\left(b^4a^3\right)^2

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Thus,

\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8

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2 years ago
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klemol [59]

Answer:

4

Step-by-step explanation:

And let the missing number be x

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x = 44/11

x = 4

For verification:

4(11) - 10 = 34

44 - 10 = 34

34 = 34

Hence, Proved

3 0
2 years ago
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