This exam looks at the determinatns and inverses of 2x2 matrices
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Calculate the determinants of the following matrices.
", "advice": "To find the determinant of a $2\\times2$ matrix, we use the following formula
\n$$
\\begin{vmatrix}
a & b \\\\
c & d \\\\
\\end{vmatrix} = ad-bc
$$
Hence, we have:
\na)
\n$$
\\det \\boldsymbol{A} = \\var{maA[0][0]} \\times \\var{maA[1][1]} - \\var{maA[0][1]} \\times \\var{maa[1][0]} = \\var{det(maA)}
$$
b)
\n$$
\\det \\boldsymbol{B} = \\var{maB[0][0]} \\times \\var{maB[1][1]} - \\var{maB[0][1]} \\times \\var{maB[1][0]} = \\var{det(maB)}
$$
Calculate the determinant of the matrix:
\n$$
\\boldsymbol{A} = \\simplify{{maA}}
$$
$\\det \\boldsymbol{A} = $ [[0]]
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\n$$
\\boldsymbol{B} = \\simplify{{maB}}
$$
$\\det \\boldsymbol{B} = $ [[0]]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Find the inverse of the following matrix:
", "advice": "To find the inverse of a matrix we first need to find the determinant:
\n$$
\\det \\boldsymbol{A} = \\var{a[0][0]} \\times \\var{a[1][1]} - \\var{a[0][1]} \\times \\var{a[1][0]} = \\var{det(a)}
$$
Since $\\det \\boldsymbol{A}\\neq 0$, we know that $\\boldsymbol{A}$ is invertible with:
\n$$
\\begin{aligned}
\\boldsymbol{A}^{-1} &= \\frac{1}{\\det \\boldsymbol{A}} \\begin{pmatrix} d & -b\\\\ -c& a \\end{pmatrix} \\\\
&= \\frac{1}{\\var{det(a)}} \\begin{pmatrix} \\var{a[1][1]} & \\var{-a[0][1]}\\\\ \\var{-a[1][0]}& \\var{a[0][0]} \\end{pmatrix} \\\\
\\end{aligned}
$$
Which gives us our inverse:
\n$$
\\simplify[fractionnumbers]{matrix:A^(-1)={inverse(a)}}
$$
Let:
\n$$
\\boldsymbol{A} = \\var{a}
$$
Calculate $\\boldsymbol{A}^{-1}$
\nInput all the elements of the matrix as fractions or integers and not as decimals.
\n$\\boldsymbol{A}^{-1} = $ [[0]]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Solve the set of simultaneous equations using matrices:
", "advice": "a)
\nBy using the coefficients of the variables in the equations, we can write the equations as:
\n$$
\\begin{pmatrix} \\var{a} & \\var{b}\\\\ \\var{a1}&\\var{b1} \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} \\var{c} \\\\ \\var{c1} \\end{pmatrix}
$$
Notice that carrying out the matrix multiplication will give you your system of equations.
\nb)
\nWe first find the determinant of $\\boldsymbol{A}$:
\n$$
\\det \\boldsymbol{A} = \\var{a} \\times \\var{b1} - \\var{b} \\times \\var{a1} = \\var{det}
$$
Since $\\det \\boldsymbol{A} \\neq 0$, $A$ is invertible with:
\n$$\\begin{aligned}
A^{-1} &= \\frac{1}{\\var{det}}\\begin{pmatrix}
{\\var{b1}}&{\\var{-b}} \\\\
{\\var{-a1}}&{\\var{a}}
\\end{pmatrix} \\\\
&= \\begin{pmatrix}
\\simplify[std]{{b1}/{det}}&\\simplify[std]{{-b}/{det}} \\\\
\\simplify[std]{{-a1}/{det}}&\\simplify[std]{{a}/{det}}
\\end{pmatrix}
\\end{aligned}
$$
c)
\nWe have the equation:
\n$$
A^{-1}b = \\begin{pmatrix}
\\simplify[std]{{b1}/{det}}&\\simplify[std]{{-b}/{det}} \\\\
\\simplify[std]{{-a1}/{det}}&\\simplify[std]{{a}/{det}}
\\end{pmatrix} \\begin{pmatrix}
\\var{c} \\\\
\\var{c1}
\\end{pmatrix}
$$
Using standard matrix multiplication, we can compute:
\n$$
A^{-1}b = \\begin{pmatrix}
\\simplify[std]{{b1}/{det}} \\times \\var{c} + \\simplify[std]{{-b}/{det}} \\times \\var{c1}\\\\
\\simplify[std]{{-a1}/{det}} \\times \\var{c} + \\simplify[std]{{a}/{det}} \\times \\var{c1}
\\end{pmatrix} = \\begin{pmatrix} \\simplify[std]{{c*b1-c1*b}/{det}}\\\\\\simplify[std]{{c1*a-c*a1}/{det}}\\end{pmatrix}
$$
d)
\nNotice that
\n$$
\\boldsymbol{A}v = b \\Rightarrow v = \\boldsymbol{A}^{-1}b
$$
Hence, we can use our solution to part $c)$ to find:
\n$$
\\begin{pmatrix}
x \\\\
y
\\end{pmatrix} = \\boldsymbol{A}^{-1}b = \\begin{pmatrix}
\\simplify[std]{{c*b1-c1*b}/{det}} \\\\
\\simplify[std]{{c1*a-c*a1}/{det}}
\\end{pmatrix}
$$
And therefore, we have:
\n$$
\\begin{eqnarray*} x&=& \\simplify[std]{{c*b1-c1*b}/{det}}\\\\ y&=& \\simplify[std]{{c1*a-c*a1}/{det}} \\end{eqnarray*}
$$
Write the following set of equations as a matrix equation:
$$
\\begin{eqnarray*} \\simplify[std]{{a}x+{b}y}&=&\\var{c}\\\\ \\simplify[std]{{a1}x+{b1}y}&=&\\var{c1} \\end{eqnarray*}
$$
That is, in the form:
\n$$
\\boldsymbol{A}v=b
$$
for some matrix $\\boldsymbol{A}$ and column vectors $v$ and $b$.
\n
[[0]]$
\\begin{pmatrix}
x \\\\
y \\\\
\\end{pmatrix} = \\ $[[1]]
Find the inverse of $\\boldsymbol{A}$, the $2 \\times 2$ matrix you defined in the previous question
\nInput all numbers as fractions or integers and not as decimals. Simplify your fractions as much as possible!
\n$\\boldsymbol{A}^{-1} = $ [[0]]
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\nGive your answer as fractions or integers and not as decimals.
\n$\\boldsymbol{A}^{-1}b = $ [[0]]
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\nGive your answer as fractions or integers and not as decimals.
\n$x = $ [[0]]
\n$y = $ [[1]]
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