// Numbas version: finer_feedback_settings
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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": "Rewrite the following system of equations as a matrix equation
\n\n\\begin{align}
\\simplify[std]{ {ma[0][0]}x + {ma[0][1]}y} &= \\var{mb[0][0]} \\\\
\\simplify[std]{ {ma[1][0]}x + {ma[1][1]}y} &= \\var{mb[1][0]}
\\end{align}
\nInput all numbers as fractions or integers and not as decimals.
", "advice": "a)
\nThe equations can be written in the matrix form
\n\\[ \\var{ma}\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\var{mb} \\]
\nb)
\n$\\mathrm{det}(\\mathbf{A}) = \\simplify[]{ {ma[0][0]}*{ma[1][1]} - {ma[0][1]}*{ma[1][0]}} = \\var{det(ma)} \\neq 0$, so $\\mathbf{A}$ is invertible.
\n\\[ \\mathbf{A}^{-1} = \\simplify[fractionnumbers]{{ma_inverse}} \\]
\nc)
\nWe have
\n\\begin{align}
\\mathbf{A}^{-1}\\mathbf{b} &= \\simplify[fractionnumbers]{{ma_inverse}*{mb}} \\\\
&= \\simplify[fractionnumbers]{{ma_inverse*mb}}
\\end{align}
\nd)
\nRearrange the equation $\\mathbf{Av}=\\mathbf{b}$ to make $\\mathbf{v}$ the subject:
\n\\begin{align}
\\mathbf{A}^{-1}\\mathbf{A}\\mathbf{v} &= \\mathbf{A}^{-1}\\mathbf{b} \\\\
\\mathbf{v} &= \\mathbf{A}^{-1}\\mathbf{b} \\\\ \\\\
\\end{align}
\nHence,
\n\\[ \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\simplify[fractionnumbers]{{ma_inverse*mb}} \\]
\nThat is,
\n\\begin{align}
x &= \\simplify[fractionnumbers]{{x}}, \\\\ \\\\
y &= \\simplify[fractionnumbers]{{y}}
\\end{align}
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\n$A^{-1} = $ [[0]] use fractions or integers in your answer
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\n
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\n$A^{-1}b = $ [[0]]
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\n$x = $ [[0]]
\n$y = $ [[1]]
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