Notes

 

Problems & Video

\(\textbf{1)}\) Use Cramer’s Rule to solve the following system of equations.
\(3x+4y=18\)
\(2x-y=1\)

Show Answer

The answer is \(x=2,y=3\)

 

\(\textbf{2)}\) \(x-4y=12\)
\(\hspace{11pt}x+3y=-2\)

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The answer is \((4,-2)\)

Show Work

\(\,\,\,\,\,x=\frac{\left|{\begin{array}{cc}
12 & -4 \\
-2 & 3 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -4 \\
1 & 3 \\
\end{array} }\right|}=\frac{(12)(3)-(-4)(-2)}{(1)(3)-(-4)(1)}=\frac{36-8}{3+4}=\frac{28}{7}=4\)
\(\,\,\,\,\,y=\frac{\left|{\begin{array}{cc}
1 & 12 \\
1 & -2 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -4 \\
1 & 3 \\
\end{array} }\right|}=\frac{(1)(-2)-(12)(1)}{(1)(3)-(-4)(1)}=\frac{-2-12}{3+4}=\frac{-14}{7}=-2\)
\(\,\,\,\,\,\text{The answer is } (4,-2)\)

 

\(\textbf{3)}\) \(2x+3y=11\)
\(\hspace{11pt}x+y=3\)

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The answer is \((-2,5)\)

Show Work

\(\,\,\,\,\,x=\frac{\left|{\begin{array}{cc}
11 & 3 \\
3 & 1 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
2 & 3 \\
1 & 1 \\
\end{array} }\right|}=\frac{(11)(1)-(3)(3)}{(2)(1)-(3)(1)}=\frac{11-9}{2-3}=\frac{2}{-1}=-2\)
\(\,\,\,\,\,y=\frac{\left|{\begin{array}{cc}
2 & 11 \\
1 & 3 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
2 & 3 \\
1 & 1 \\
\end{array} }\right|}=\frac{(2)(3)-(1)(11)}{(2)(1)-(3)(1)}=\frac{6-11}{2-3}=\frac{-5}{-1}=5\)
\(\,\,\,\,\,\text{The answer is } (-2,5)\)

 

\(\textbf{4)}\) \(x-y=-6\)
\(\hspace{11pt}5x+2y=12\)

Show Answer

The answer is \((0,6)\)

Show Work

\(\,\,\,\,\,x=\frac{\left|{\begin{array}{cc}
-6 & -1 \\
12 & 2 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -1 \\
5 & 2 \\
\end{array} }\right|}=\frac{(-6)(2)-(-1)(12)}{(1)(2)-(-1)(5)}=\frac{-12+12}{2+5}=\frac{0}{7}=0\)
\(\,\,\,\,\,y=\frac{\left|{\begin{array}{cc}
1 & -6 \\
5 & 12 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -1 \\
5 & 2 \\
\end{array} }\right|}=\frac{(1)(12)-(-6)(5)}{(1)(2)-(-1)(5)}=\frac{12+30}{2+5}=\frac{42}{7}=6\)
\(\,\,\,\,\,\text{The answer is } (0,6)\)

 

\(\textbf{5)}\) \(7x+4y=31\)
\(\hspace{11pt}3x+2y=15\)

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The answer is \((1,6)\)

 

\(\textbf{6)}\) \(5x-4y=9\)
\(\hspace{11pt}7x+3y=4\)

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The answer is \((1,-1)\)

 

\(\textbf{7)}\) \(15x-3y=126\)
\(\hspace{11pt}2x+5y=33\)

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The answer is \((9,3)\)

 

\(\textbf{8)}\) \(2x-y=-1\)
\(\hspace{11pt}3x+4y=-7\)

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The answer is \((-1,-1)\)

 

\(\textbf{9)}\) \(2x-7y=-18\)
\(\hspace{11pt}3x+8y=47\)

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The answer is \((5,4)\)

 

\(\textbf{10)}\) \(3x+2y=12\)
\(\hspace{15pt}5x-4y=-2\)

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The answer is \((2,3)\)

 

\(\textbf{11)}\) \(3x+y=10\)
\(\hspace{11pt}x-y=2\)

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The answer is \((3,1)\)

 

 

See Related Pages\(\)

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

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

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

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

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

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

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

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

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

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