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Find the values of
x and
y in the following scalar multiplication.

{:[-(1)/(3)*[[x],[-9]]=[[2],[y]]],[x=],[y=]:}

Find the values of $x$ and $y$ in the following scalar multiplication. $\newline$ $\begin{array}{l} -\frac{1}{3} \cdot\left[\begin{array}{c} x \\ -9 \end{array}\right]=\left[\begin{array}{l} 2 \\ y \end{array}\right] \\ x=\square \\ y=\square \end{array}$

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Even and odd functions

Full solution

Q. Find the values of

x

and

y

in the following scalar multiplication.

\newline

\begin{array}{l} -\frac{1}{3} \cdot\left[\begin{array}{c} x \\ -9 \end{array}\right]=\left[\begin{array}{l} 2 \\ y \end{array}\right] \\ x=\square \\ y=\square \end{array}

Given equation: We are given the scalar multiplication equation $-\frac{1}{3} \times \left[\begin{array}{c} x \ -9 \end{array}\right] = \left[\begin{array}{c} 2 \ y \end{array}\right]$ . To find the values of $x$ and $y$ , we need to multiply each element of the matrix $\left[\begin{array}{c} x \ -9 \end{array}\right]$ by the scalar $-\frac{1}{3}$ and then equate the resulting matrix to $\left[\begin{array}{c} 2 \ y \end{array}\right]$ .
Multiplying by scalar: First, let's multiply the scalar $-\frac{1}{3}$ by the element $x$ in the matrix. This gives us $-\frac{1}{3} \times x = -\frac{x}{3}$ .
Resulting matrix: Next, we multiply the scalar $-\frac{1}{3}$ by the element $-9$ in the matrix. This gives us $-\frac{1}{3} \times -9 = 3$ .
Equating matrices: Now we have the resulting matrix from the scalar multiplication: \left[\begin{array}{c}-\frac{x}{3}\3\end{array}\right]. We can set this equal to the matrix \left[\begin{array}{c}2\y\end{array}\right] to find the values of $x$ and $y$ .
Solving for x: By equating the first elements of the matrices, we get $-\frac{x}{3} = 2$ . To solve for $x$ , we multiply both sides of the equation by $-3$ , which gives us $x = -6$ .
Finding $y$ : By equating the second elements of the matrices, we get $3 = y$ . This directly gives us the value of $y$ , which is $y = 3$ .

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