// Numbas version: finer_feedback_settings {"name": "3x3 Determinant Problems", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "3x3 Determinant Problems", "tags": [], "metadata": {"description": "

This question tests learner's knowledge of the inverse matrix method for a 3x3 matrix.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

$3 \\times 3$ Determinant Problems

\n

Calculate the two values for $x$ that satisfy the equation $|A|=\\var{k}$

\n

where the matrix $A$ is given by:

\n

$$
A=\\begin{pmatrix} x&\\var{a}&\\var{b}\\\\ \\var{c}&x&\\var{d}\\\\\\var{e1}&\\var{f}&\\var{g} \\end{pmatrix}
$$

", "advice": "

We must calculate the determinant as usual, using the expansion of the first row:

\n

$$
\\begin{aligned}
\\det{A} &= x\\begin{vmatrix}
x &\\var{d} \\\\
\\var{f} & \\var{g} \\\\
\\end{vmatrix} -\\var{a} \\begin{vmatrix}
\\var{c} & \\var{d} \\\\
\\var{e1} & \\var{g} \\\\
\\end{vmatrix} + \\var{b} \\begin{vmatrix}
\\var{c} & x \\\\
\\var{e1} & \\var{f}
\\end{vmatrix} \\\\ 
&= x(x-\\simplify{{d}{f}})-\\var{a}(\\simplify{{c}*{g}}-\\simplify{{e1}*{d}})+\\var{b}(\\simplify{{c}*{f}}-\\var{e1}x) \\\\
&= x^2-\\simplify{{d}*{f}}x-\\simplify{{a}*{c}*{g}}+\\simplify{{a}*{e1}*{d}}+\\simplify{{b}*{c}*{f}}-\\simplify{{b}*{e1}}x \\\\
&= x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{a}*{e1}*{d}+{b}*{c}*{f}-{a}*{c}*{g}}
\\end{aligned}
$$

\n

Since we know that $\\det A = \\var{k}$:

\n

$$
\\begin{aligned}
x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{a}*{e1}*{d}+{b}*{c}*{f}-{a}*{c}*{g}} &= \\var{k} \\\\
x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{b}*{e1}*{f}*{d}} &= 0
\\end{aligned}
$$

\n

This can be solved by formula or by finding factors:

\n

$$
(x-\\simplify{{f}*{d}})(x-\\simplify{{b}*{e1}})=0
$$

\n

So we have:

\n

$$
x=\\simplify{{f}*{d}}\\,\\,\\,\\,\\,or\\,\\,\\,\\,\\,x=\\simplify{{b}*{e1}}
$$

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Enter the smaller of the two values

\n

\\(x=\\) [[0]]

\n

Enter the larger of the two values

\n

\\(x=\\) [[1]]

\n

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