Please perform the indicated operations

\(\textbf{1)}\,\,\)\( \left[ {\begin{array}{ccc}3 & 45 & 6 \\-8 & 2 & 4 \\1 & 0 & 3 \\\end{array} } \right]\)\(+\left[ {\begin{array}{ccc}3 & 45 & 6 \\-8 & 2 & 4 \\1 & 0 & 3 \\\end{array} } \right]\)

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The answer is \( \left[ {\begin{array}{ccc}6 & 90 & 12 \\-16 & 4 & 8 \\2 & 0 & 6 \\\end{array} } \right]\)\(\)

 

\(\textbf{2)}\,\,\)\(4\cdot\left[ {\begin{array}{ccc}3 & 45 & 6 \\-8 & 2 & 4 \\1 & 0 & 3 \\\end{array} } \right]\)

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The answer is \( \left[ {\begin{array}{ccc}12 & 180 & 24 \\-32 & 8 & 16 \\4 & 0 & 12 \\\end{array} } \right]\)\(\)

 

\(\textbf{3)}\)\(\frac{2}{3}\cdot\left[ {\begin{array}{ccc}6 & 15 & 6 \\-9 & 2 & 0 \\4 & -12 & \frac{1}{5} \\\end{array} } \right]\)

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The answer is \( \left[ {\begin{array}{ccc}4 & 10 & 4 \\-6 & \frac{4}{3} & 0 \\\frac{8}{3} & -8 & \frac{2}{15} \\\end{array} } \right]\)\(\)

 

\(\textbf{4)}\)\(\left[ {\begin{array}{cc}1 & 0 \\-3 & -4 \\\end{array} } \right]\)\(-\left[ {\begin{array}{cc}2 & -3 \\5 & -5 \\\end{array} } \right]\)

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The answer is \(\left[ {\begin{array}{cc}-1 & 3 \\-8 & 1 \\\end{array} } \right]\)

 

\(\textbf{5)}\,\,\)\( \left[ {\begin{array}{ccc}1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9 \\\end{array} } \right]\)\(+\left[ {\begin{array}{ccc}8 & 7 & 6 \\5 & 4 & 3 \\2 & 1 & 0 \\\end{array} } \right]\)

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The answer is \( \left[ {\begin{array}{ccc}9 & 9 & 9 \\9 & 9 & 9 \\9 & 9 & 9 \\\end{array} } \right]\)\(\)

 

\(\textbf{6)}\,\,\)\(2\left[ {\begin{array}{cc}1 & 2 \\3 & 4 \\\end{array} } \right]\)

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The answer is \( \left[ {\begin{array}{cc}2 & 4 \\6 & 8 \\\end{array} } \right]\)

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\(\text{Step 1: Given}\)
\(\,\,\,\,\,\,2\left[ {\begin{array}{cc}1 & 2 \\3 & 4 \\\end{array} } \right]\)
\(\text{Step 2: Distribute the }2\)
\(\,\,\,\,\,\, \left[ {\begin{array}{cc}(2)(1) & (2)(2) \\(2)(3) & (2)(4) \\\end{array} } \right]\)
\(\text{Step 3: Simplify}\)
\(\,\,\,\,\,\, \left[ {\begin{array}{cc}2 & 4 \\6 & 8 \\\end{array} } \right]\)

 

\(\textbf{7)}\)\(\frac{1}{2}\left[ {\begin{array}{cc}6 & 16 \\4 & -12 \\\end{array} } \right]\)

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The answer is \( \left[ {\begin{array}{cc}3 & 8 \\2 & -6 \\\end{array} } \right]\)\(\)

 

\(\textbf{8)}\)\(\left[ {\begin{array}{cc}5 & 4 \\-3 & -4 \\\end{array} } \right]\)\(-\left[ {\begin{array}{cc}1 & -1 \\1 & -1 \\\end{array} } \right]\)

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The answer is \(\left[ {\begin{array}{cc}4 & 5 \\-4 & -3 \\\end{array} } \right]\)

 

See Related Pages\(\)

\(\bullet\text{ Intro to Matrices}\)
\(\,\,\,\,\,\,\,\,\)\(\left[ {\begin{array}{ccc}4 & -5 & 2 \\1 & 0 & 3 \\\end{array} } \right]\)

\(\bullet\text{ Matrix Operations}\)
\(\,\,\,\,\,\,\,\,\)\( \left[ {\begin{array}{ccc}3 & 45 & 6 \\-8 & 2 & 4 \\1 & 0 & 3 \\\end{array} } \right]\)\(+\left[ {\begin{array}{ccc}3 & 45 & 6 \\-8 & 2 & 4 \\1 & 0 & 3 \\\end{array} } \right]\)

\(\bullet\text{ Multiplying Matrices}\)
\(\,\,\,\,\,\,\,\,\)\(\left[{\begin{array}{cc} 1 & 2 \\ -3 & -4 \\\end{array} } \right]\)\(\left[ {\begin{array}{cc}6 & -3 \\5 & 0 \\\end{array} } \right]\)

\(\bullet\text{ Determinants}\)
\(\,\,\,\,\,\,\,\,\)\(\left|{\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right|=ad-bc\)

\(\bullet\text{ Cramer’s Rule}\)
\(\,\,\,\,\,\,\,\,\text{ax+by=e } \& \text{ cx+dy=f}…\)

\(\bullet\text{ Identity Matrix}\)
\(\,\,\,\,\,\,\,\,\)\(\left[{\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]\)

\(\bullet\text{ Identity and Inverse Matrices}\)
\(\,\,\,\,\,\,\,\,A^{-1}=\displaystyle\frac{1}{ad-bc}\left[{\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right]\)

\(\bullet\text{ Transpose Matrix}\)
\(\,\,\,\,\,\,\,\,\left[{\begin{array}{ccc} 1 \\ 2 \\ 5 \\ \end{array} } \right]\Rightarrow\left[{\begin{array}{c} 1 & 2 & 5 \end{array} } \right]\)

\(\bullet\text{ Rotation Matrix}\)
\(\,\,\,\,\,\,\,\,\)\(R(\theta)=\left[{\begin{array}{cc}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \\\end{array} } \right]\)

\(\bullet\text{ Eigenvectors and Eigenvalues}\)
\(\,\,\,\,\,\,\,\,(A-\lambda I)\vec{v}=\vec{0}\)