Exercise $\newline$ Write the given sum as a single column matrix $\newline$ $1$ . $3 t\left(\begin{array}{c}2 \\ t \\ -1\end{array}\right)+(t-1)\left(\begin{array}{c}-1 \\ -t \\ 3\end{array}\right)-2\left(\begin{array}{c}3 t \\ 4 \\ -5 t\end{array}\right)$ $\newline$ $2$ . $\left(\begin{array}{ccc}1 & -3 & 4 \\ 2 & 5 & -1 \\ 0 & -4 & -2\end{array}\right)\left(\begin{array}{c}t \\ 2 t-1 \\ -t\end{array}\right)+\left(\begin{array}{c}-t \\ 1 \\ 4\end{array}\right)-\left(\begin{array}{c}2 \\ 8 \\ -6\end{array}\right)$

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Algebra 2

Write a quadratic function from its x-intercepts and another point

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Q. Exercise

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Write the given sum as a single column matrix

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1

3 t\left(\begin{array}{c}2 \\ t \\ -1\end{array}\right)+(t-1)\left(\begin{array}{c}-1 \\ -t \\ 3\end{array}\right)-2\left(\begin{array}{c}3 t \\ 4 \\ -5 t\end{array}\right)

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2

\left(\begin{array}{ccc}1 & -3 & 4 \\ 2 & 5 & -1 \\ 0 & -4 & -2\end{array}\right)\left(\begin{array}{c}t \\ 2 t-1 \\ -t\end{array}\right)+\left(\begin{array}{c}-t \\ 1 \\ 4\end{array}\right)-\left(\begin{array}{c}2 \\ 8 \\ -6\end{array}\right)

Distribute scalar to first column: First, let's distribute the scalar $3t$ to the first column matrix. $3t \times \left(\begin{array}{c} 2 \ t \ -1 \end{array}\right) = \left(\begin{array}{c} 6t \ 3t^2 \ -3t \end{array}\right)$
Distribute scalar to second column: Now, distribute the scalar $(t-1)$ to the second column matrix. $(t-1) \times \left[\begin{array}{c} -1 \ -t \ 3 \end{array}\right] = \left[\begin{array}{c} -t+1 \ -t^2+t \ 3t-3 \end{array}\right]$
Distribute scalar to third column: Next, distribute the scalar $-2$ to the third column matrix. $\newline$ \(-2 \times \left(\begin{array}{c} $3$ t \ $4$ \ $-5$ t \end{array}\right) = \left(\begin{array}{c} $-6$ t \ $-8$ \ $10$ t \end{array}\right)
Add resulting column matrices: Now, let's add the three resulting column matrices together. \left(\begin{array}{c} \(6t \ $3$ t^ $2$ \ $-3$ t \end{array}\right) + \left(\begin{array}{c} -t+ $1$ \ -t^ $2$ +t \ $3$ t $-3$ \end{array}\right) + \left(\begin{array}{c} $-6$ t \ $-8$ \ $10$ t \end{array}\right) = \left(\begin{array}{c} $6$ t-t+ $1$ $-6$ t \ $3$ t^ $2$ -t^ $2$ +t $-8$ \ $-3$ t+ $3$ t $-3$ + $10$ t \end{array}\right)
Simplify resulting column matrix: Simplify the resulting column matrix. $\newline$ $[6t-t+1-6t], [3t^2-t^2+t-8], [-3t+3t-3+10t]$ = $([1], [2t^2+t-8], [7t-3])$
Multiply $3$ x $3$ matrix by column matrix: Now, let's do the same for the second set of matrices. $\newline$ First, multiply the $3$ x $3$ matrix by the column matrix. $\newline$ \left[\begin{array}{ccc}$\newline1 & -3 & 4 (\newline$2 & 5 & -1 (\newline\)0 & -4 & -2 $\newline$ \end{array}\right]\) * \left[\begin{array}{c}$\newlinet (\newline$2t-1 (\newline\)-t $\newline$ \end{array}\right]\) = \left[\begin{array}{c}$\newline1\cdot t + (-3)\cdot(2t-1) + 4\cdot(-t) (\newline$2\cdot t + 5\cdot(2t-1) + (-1)\cdot(-t) (\newline\)0\cdot t + (-4)\cdot(2t-1) + (-2)\cdot(-t) $\newline$ \end{array}\right]\)
Simplify resulting column matrix: Simplify the resulting column matrix. $\newline$ $\left(\begin{array}{c} t - 6t + 3 - 4t \ 2t + 10t - 5 + t \ 0 - 8t + 4 + 2t \end{array}\right) = \left(\begin{array}{c} -9t + 3 \ 13t - 5 \ -6t + 4 \end{array}\right)$
Add column matrix to result: Now, add the column matrix $([-t],[1],[4])$ to the result.\left(\begin{array}{c} [\(-9t + $3$ ] + [-t] \ [ $13$ t - $5$ ] + [ $1$ ] \ [ $-6$ t + $4$ ] + [ $4$ ] \end{array}\right) = \left(\begin{array}{c} [ $-9$ t + $3$ - t] \ [ $13$ t - $5$ + $1$ ] \ [ $-6$ t + $4$ + $4$ ] \end{array}\right)
Simplify resulting column matrix: Simplify the resulting column matrix. $\newline$ [[\(-9t + $3$ - t], [ $13$ t - $5$ + $1$ ], [ $-6$ t + $4$ + $4$ ]) = ([ $-10$ t + $3$ ], [ $13$ t - $4$ ], [ $-6$ t + $8$ ])\]
Subtract column matrix from result: Finally, subtract the column matrix $\left(\begin{array}{c} 2 \ 8 \ -6 \end{array}\right)$ from the result.\left(\begin{array}{c} \(-10t + $3$ \ $13$ t - $4$ \ $-6$ t + $8$ \end{array}\right) - \left(\begin{array}{c} $2$ \ $8$ \ $-6$ \end{array}\right) = \left(\begin{array}{c} $-10$ t + $3$ - $2$ \ $13$ t - $4$ - $8$ \ $-6$ t + $8$ + $6$ \end{array}\right)
Simplify final column matrix: Simplify the resulting column matrix to get the final answer. $\newline$ $[[-10t + 3 - 2], [13t - 4 - 8], [-6t + 8 + 6]$ = $[[-10t + 1], [13t - 12], [-6t + 14]$