// Numbas version: finer_feedback_settings {"name": "2x2 Matrix Problems", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "2x2 Matrix Problems", "tags": [], "metadata": {"description": "
Putting a pair of linear equations into matrix notation and then solving by finding the inverse of the coefficient matrix.
", "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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