Solved on Nov 15, 2023

Find $\frac{1}{3} B+2 A$ given $A=\left[\begin{array}{ll}6 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}-15 & 6\end{array}\right]$ .

STEP 1

Assumptions1. The matrix $A$ is given as $A=\left[\begin{array}{ll} 6 &3\end{array}\right]$ . The matrix $B$ is given as $B=\left[\begin{array}{ll} -15 &6\end{array}\right]$
3. We need to find $\frac{1}{3} B+ A$

STEP 2

First, we need to find $\frac{1}{} B$ . We can do this by multiplying each element of matrix $B$ by $\frac{1}{}$ .
$\frac{1}{} B = \frac{1}{} \left[\begin{array}{ll} -15 &6\end{array}\right]$

STEP 3

Now, calculate the result of $\frac{1}{3} B$ .
$\frac{1}{3} B = \left[\begin{array}{ll} -5 &2\end{array}\right]$

STEP 4

Next, we need to find $2 A$ . We can do this by multiplying each element of matrix $A$ by $2$ .
$2 A =2 \left[\begin{array}{ll} 6 &3\end{array}\right]$

STEP 5

Now, calculate the result of $2 A$ .
$2 A = \left[\begin{array}{ll} 12 &\end{array}\right]$

STEP 6

Now that we have $\frac{1}{3} B$ and $2 A$ , we can add these two matrices to find $\frac{1}{3} B+2 A$ .
$\frac{1}{3} B+2 A = \left[\begin{array}{ll} -5 &2\end{array}\right] + \left[\begin{array}{ll} 12 &6\end{array}\right]$

STEP 7

Now, calculate the result of $\frac{1}{3} B+2 A$ .
$\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]$ So, $\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]$ .

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Find $\frac{1}{3} B+2 A$ given $A=\left[\begin{array}{ll}6 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}-15 & 6\end{array}\right]$ .

STEP 1

Assumptions1. The matrix $A$ is given as $A=\left[\begin{array}{ll} 6 &3\end{array}\right]$ . The matrix $B$ is given as $B=\left[\begin{array}{ll} -15 &6\end{array}\right]$
3. We need to find $\frac{1}{3} B+ A$

STEP 2

First, we need to find $\frac{1}{} B$ . We can do this by multiplying each element of matrix $B$ by $\frac{1}{}$ .
$\frac{1}{} B = \frac{1}{} \left[\begin{array}{ll} -15 &6\end{array}\right]$

STEP 3

Now, calculate the result of $\frac{1}{3} B$ .
$\frac{1}{3} B = \left[\begin{array}{ll} -5 &2\end{array}\right]$

STEP 4

Next, we need to find $2 A$ . We can do this by multiplying each element of matrix $A$ by $2$ .
$2 A =2 \left[\begin{array}{ll} 6 &3\end{array}\right]$

STEP 5

Now, calculate the result of $2 A$ .
$2 A = \left[\begin{array}{ll} 12 &\end{array}\right]$

STEP 6

Now that we have $\frac{1}{3} B$ and $2 A$ , we can add these two matrices to find $\frac{1}{3} B+2 A$ .
$\frac{1}{3} B+2 A = \left[\begin{array}{ll} -5 &2\end{array}\right] + \left[\begin{array}{ll} 12 &6\end{array}\right]$

STEP 7

Now, calculate the result of $\frac{1}{3} B+2 A$ .
$\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]$ So, $\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]$ .

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Find 13B+2A\frac{1}{3} B+2 A31​B+2A given A=[63]A=\left[\begin{array}{ll}6 & 3\end{array}\right]A=[6​3​] and B=[−156]B=\left[\begin{array}{ll}-15 & 6\end{array}\right]B=[−15​6​].

STEP 1

Assumptions1. The matrix AAA is given as A=[63]A=\left[\begin{array}{ll} 6 &3\end{array}\right]A=[6​3​] . The matrix BBB is given as B=[−156]B=\left[\begin{array}{ll} -15 &6\end{array}\right]B=[−15​6​]3. We need to find 13B+A\frac{1}{3} B+ A31​B+A

STEP 2

First, we need to find 1B\frac{1}{} B1​B. We can do this by multiplying each element of matrix BBB by 1\frac{1}{}1​.1B=1[−156]\frac{1}{} B = \frac{1}{} \left[\begin{array}{ll} -15 &6\end{array}\right]1​B=1​[−15​6​]

STEP 3

Now, calculate the result of 13B\frac{1}{3} B31​B.13B=[−52]\frac{1}{3} B = \left[\begin{array}{ll} -5 &2\end{array}\right]31​B=[−5​2​]

STEP 4

Next, we need to find 2A2 A2A. We can do this by multiplying each element of matrix AAA by 222.2A=2[63]2 A =2 \left[\begin{array}{ll} 6 &3\end{array}\right]2A=2[6​3​]

STEP 5

Now, calculate the result of 2A2 A2A.2A=[12]2 A = \left[\begin{array}{ll} 12 &\end{array}\right]2A=[12​​]

STEP 6

Now that we have 13B\frac{1}{3} B31​B and 2A2 A2A, we can add these two matrices to find 13B+2A\frac{1}{3} B+2 A31​B+2A.13B+2A=[−52]+[126]\frac{1}{3} B+2 A = \left[\begin{array}{ll} -5 &2\end{array}\right] + \left[\begin{array}{ll} 12 &6\end{array}\right]31​B+2A=[−5​2​]+[12​6​]

STEP 7

Now, calculate the result of 13B+2A\frac{1}{3} B+2 A31​B+2A.13B+2A=[7]\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]31​B+2A=[7​​]So, 13B+2A=[7]\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]31​B+2A=[7​​].

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Find $\frac{1}{3} B+2 A$ given $A=\left[\begin{array}{ll}6 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}-15 & 6\end{array}\right]$ .

Assumptions1. The matrix $A$ is given as $A=\left[\begin{array}{ll} 6 &3\end{array}\right]$ . The matrix $B$ is given as $B=\left[\begin{array}{ll} -15 &6\end{array}\right]$
3. We need to find $\frac{1}{3} B+ A$

First, we need to find $\frac{1}{} B$ . We can do this by multiplying each element of matrix $B$ by $\frac{1}{}$ .
$\frac{1}{} B = \frac{1}{} \left[\begin{array}{ll} -15 &6\end{array}\right]$

Now, calculate the result of $\frac{1}{3} B$ .
$\frac{1}{3} B = \left[\begin{array}{ll} -5 &2\end{array}\right]$

Next, we need to find $2 A$ . We can do this by multiplying each element of matrix $A$ by $2$ .
$2 A =2 \left[\begin{array}{ll} 6 &3\end{array}\right]$

Now, calculate the result of $2 A$ .
$2 A = \left[\begin{array}{ll} 12 &\end{array}\right]$

Now that we have $\frac{1}{3} B$ and $2 A$ , we can add these two matrices to find $\frac{1}{3} B+2 A$ .
$\frac{1}{3} B+2 A = \left[\begin{array}{ll} -5 &2\end{array}\right] + \left[\begin{array}{ll} 12 &6\end{array}\right]$

Now, calculate the result of $\frac{1}{3} B+2 A$ .
$\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]$ So, $\frac{1}{3} B+2 A = \left[\begin{array}{ll} 7 &\end{array}\right]$ .