Number and Algebra
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Geometry and Trigonometry
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Calculus
Compute 3A3A3A where
Compute −2B-2B−2B where
Evaluate 5(0−321)−3(1204).5\begin{pmatrix}0 & -3\\2 & 1\end{pmatrix} - 3\begin{pmatrix}1 & 2\\0 & 4\end{pmatrix}.5(02​−31​)−3(10​24​).
Calculate 4(2135)+2(10−12).4\begin{pmatrix}2 & 1\\3 & 5\end{pmatrix} + 2\begin{pmatrix}1 & 0\\-1 & 2\end{pmatrix}.4(23​15​)+2(1−1​02​).
Let C=(−12031−2),D=(2−3405−1).C = \begin{pmatrix}-1 & 2 & 0\\3 & 1 & -2\end{pmatrix},\quad D = \begin{pmatrix}2 & -3 & 4\\0 & 5 & -1\end{pmatrix}.C=(−13​21​0−2​),D=(20​−35​4−1​). Compute C−2DC - 2DC−2D.
Given A=(1−234),B=(20−15),A = \begin{pmatrix}1 & -2\\3 & 4\end{pmatrix},\quad B = \begin{pmatrix}2 & 0\\-1 & 5\end{pmatrix},A=(13​−24​),B=(2−1​05​), find 2A+3B2A + 3B2A+3B.
Simplify 3A+2B−A3A + 2B - A3A+2B−A for A=(2103),B=(−1452).A = \begin{pmatrix}2 & 1\\0 & 3\end{pmatrix},\quad B = \begin{pmatrix}-1 & 4\\5 & 2\end{pmatrix}.A=(20​13​),B=(−15​42​).
Compute −3C+4D-3C + 4D−3C+4D using the C,DC,DC,D from the previous question.
Simplify 2(A+B)−(A−B)2(A+B) - (A - B)2(A+B)−(A−B) where A=(1230),B=(0−145).A = \begin{pmatrix}1 & 2\\3 & 0\end{pmatrix},\quad B = \begin{pmatrix}0 & -1\\4 & 5\end{pmatrix}.A=(13​20​),B=(04​−15​).
Verify the distributive property k(A+B)=kA+kBk(A+B)=kA+kBk(A+B)=kA+kB by computing both sides for k=2,A=(123−1),B=(04−25).k=2,\quad A=\begin{pmatrix}1 & 2\\3 & -1\end{pmatrix},\quad B=\begin{pmatrix}0 & 4\\-2 & 5\end{pmatrix}.k=2,A=(13​2−1​),B=(0−2​45​).
If X+2A=3BX + 2A = 3BX+2A=3B and A=(10−12),B=(2310),A = \begin{pmatrix}1 & 0\\-1 & 2\end{pmatrix},\quad B = \begin{pmatrix}2 & 3\\1 & 0\end{pmatrix},A=(1−1​02​),B=(21​30​), find XXX.
Compute 3E−F3E - F3E−F for E=(1−1203−2410),F=(20−13−11024).E = \begin{pmatrix}1 & -1 & 2\\0 & 3 & -2\\4 & 1 & 0\end{pmatrix},\quad F = \begin{pmatrix}2 & 0 & -1\\3 & -1 & 1\\0 & 2 & 4\end{pmatrix}.E=​104​−131​2−20​​,F=​230​0−12​−114​​.
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Question Type 2: Determining validity of matrix operations