// Numbas version: finer_feedback_settings {"name": "3x3 Matrices", "metadata": {"description": "
This exam test students understanding of determinants and inverses of 3x3 matrices.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Calculate the determinants of the following matrices.
", "advice": "For simplicity, we will use the expansion of the first row to calculate our determinant
\na)
\nWe have:
\n$$
\\begin{aligned}
\\det{A} &= a_1A_1 + b_1B_1+ c_1C_1 \\\\
&= \\var{maA[0][0]}
\\begin{vmatrix}
\\var{maA[1][1]} & \\var{maA[1][2]} \\\\
\\var{maA[2][1]} & \\var{maA[2][2]} \\\\
\\end{vmatrix} - \\var{maA[0][1]} \\begin{vmatrix}
\\var{maA[1][0]} & \\var{maA[1][2]} \\\\
\\var{maA[2][0]} & \\var{maA[2][2]} \\\\
\\end{vmatrix} + \\var{maA[0][2]} \\begin{vmatrix}
\\var{maA[1][0]} & \\var{maA[1][1]} \\\\
\\var{maA[2][0]} & \\var{maA[2][1]} \\\\
\\end{vmatrix} \\\\
&= \\var{maA[0][0]} \\times \\var{a1a} - \\left(\\var{maA[0][1]} \\times \\var{b1a}\\right) + \\var{maA[0][2]} \\times \\var{c1a} \\\\
&= \\var{deta}
\\end{aligned}
$$
b)
\nWe have:
\n$$
\\begin{aligned}
\\det{B} &= a_1A_1 + b_1B_1+ c_1C_1 \\\\
&= \\var{maB[0][0]}
\\begin{vmatrix}
\\var{maB[1][1]} & \\var{maB[1][2]} \\\\
\\var{maB[2][1]} & \\var{maB[2][2]} \\\\
\\end{vmatrix} - \\var{maB[0][1]} \\begin{vmatrix}
\\var{maB[1][0]} & \\var{maB[1][2]} \\\\
\\var{maB[2][0]} & \\var{maB[2][2]} \\\\
\\end{vmatrix} + \\var{maB[0][2]} \\begin{vmatrix}
\\var{maB[1][0]} & \\var{maB[1][1]} \\\\
\\var{maB[2][0]} & \\var{maB[2][1]} \\\\
\\end{vmatrix} \\\\
&= \\var{maB[0][0]} \\times \\var{a1b} - \\left(\\var{maB[0][1]} \\times \\var{b1b}\\right) + \\var{maB[0][2]} \\times \\var{c1b} \\\\
&= \\var{detb}
\\end{aligned}
$$
Calculate the determinant of $A = \\simplify{{maA}}$
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", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "Calculate the two values for $x$ that satisfy the equation $|A|=\\var{k}$
\nwhere the matrix $A$ is given by:
\n$$
A=\\begin{pmatrix} x&\\var{a}&\\var{b}\\\\ \\var{c}&x&\\var{d}\\\\\\var{e1}&\\var{f}&\\var{g} \\end{pmatrix}
$$
We must calculate the determinant as usual, using the expansion of the first row:
\n$$
\\begin{aligned}
\\det{A} &= x\\begin{vmatrix}
x &\\var{d} \\\\
\\var{f} & \\var{g} \\\\
\\end{vmatrix} -\\var{a} \\begin{vmatrix}
\\var{c} & \\var{d} \\\\
\\var{e1} & \\var{g} \\\\
\\end{vmatrix} + \\var{b} \\begin{vmatrix}
\\var{c} & x \\\\
\\var{e1} & \\var{f}
\\end{vmatrix} \\\\
&= x(x-\\simplify{{d}{f}})-\\var{a}(\\simplify{{c}*{g}}-\\simplify{{e1}*{d}})+\\var{b}(\\simplify{{c}*{f}}-\\var{e1}x) \\\\
&= x^2-\\simplify{{d}*{f}}x-\\simplify{{a}*{c}*{g}}+\\simplify{{a}*{e1}*{d}}+\\simplify{{b}*{c}*{f}}-\\simplify{{b}*{e1}}x \\\\
&= x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{a}*{e1}*{d}+{b}*{c}*{f}-{a}*{c}*{g}}
\\end{aligned}
$$
Since we know that $\\det A = \\var{k}$:
\n$$
\\begin{aligned}
x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{a}*{e1}*{d}+{b}*{c}*{f}-{a}*{c}*{g}} &= \\var{k} \\\\
x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{b}*{e1}*{f}*{d}} &= 0
\\end{aligned}
$$
This can be solved by formula or by finding factors:
\n$$
(x-\\simplify{{f}*{d}})(x-\\simplify{{b}*{e1}})=0
$$
So we have:
\n$$
x=\\simplify{{f}*{d}}\\,\\,\\,\\,\\,or\\,\\,\\,\\,\\,x=\\simplify{{b}*{e1}}
$$
Enter the smaller of the two values
\n\\(x=\\) [[0]]
\nEnter the larger of the two values
\n\\(x=\\) [[1]]
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\nYou will have to define the dimension of the transposed matrix before you enter your answer.
", "advice": "The transposition process results in rows becoming columns and columns becoming rows.
\nIt may help to imagine the matrix being \"flipped\" about its diagonal.
\n\nUsing this technique gives us:
\na)
\n$$
A=\\var{A} \\;\\;\\; A^{T}=\\var{TA}
$$
b)
\n$$
B=\\var{B} \\;\\;\\; B^{T}=\\var{TB}
$$
c)
\n$$
C=\\var{C} \\;\\;\\; C^{T}=\\var{TC}
$$
d)
\n$$
D=\\var{D} \\;\\;\\; D^{T}=\\var{TD}
$$
e)
\n$$
E=\\var{EE} \\;\\;\\; E^{T}=\\var{TE}
$$
Let:
\n$$
A=\\var{A}
$$
Find $A^{T}$
\n$A^{T}=$ [[0]]
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\n$$
B=\\var{B}
$$
Find $B^{T}$
\n$B^{T}=$ [[0]]
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\n$$
C=\\var{C}
$$
Find $C^{T}$
\n$C^{T}=$ [[0]]
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\n$$
D=\\var{D}
$$
Find $D^{T}$
\n$D^{T}=$ [[0]]
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\n$$
E=\\var{EE}
$$
Find $E^{T}$
\n$E^{T}=$ [[0]]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Follow the steps in the questions to find the inverse of a $3 \\times 3$ matrix
", "advice": "a)
\nFor simplicity, we will use the expansion of the first row to find the determinant
\n$$
\\begin{aligned}
\\det{A} &= a_1A_1 + b_1B_1+ c_1C_1 \\\\
&= \\var{matrixA[0][0]}
\\begin{vmatrix}
\\var{matrixA[1][1]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][1]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} - \\var{matrixA[0][1]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} + \\var{matrixA[0][2]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][1]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][1]} \\\\
\\end{vmatrix} \\\\
&= \\var{matrixA[0][0]} \\times \\var{cof11} - \\left(\\var{matrixA[0][1]} \\times \\var{cof12}\\right) + \\var{matrixA[0][2]} \\times \\var{cof13} \\\\
&= \\var{deta}
\\end{aligned}
$$
b)
\nGiven arbitrary matrix
\n$$
A = \\begin{pmatrix}
a & b & c \\\\
d & e & f \\\\
g & h & j \\\\
\\end{pmatrix}
$$
It's cofactors are given by
\n$$
\\begin{aligned}
A_{11} &= +\\begin{vmatrix}
e & f \\\\
h & j \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a22} & \\var{a23} \\\\
\\var{a32} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof11}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{12} &= -\\begin{vmatrix}
d & f \\\\
g & j \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a21} & \\var{a23} \\\\
\\var{a31} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof12}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{13} &= +\\begin{vmatrix}
d & e \\\\
g & h \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a21} & \\var{a22} \\\\
\\var{a31} & \\var{a32} \\\\
\\end{vmatrix} \\\\
&= \\var{cof13}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{21} &= -\\begin{vmatrix}
b & c \\\\
h & j \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a12} & \\var{a13} \\\\
\\var{a32} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof21}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{22} &= +\\begin{vmatrix}
a & c \\\\
g & j \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a11} & \\var{a13} \\\\
\\var{a31} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof22}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{23} &= -\\begin{vmatrix}
a & b \\\\
g & h \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a11} & \\var{a12} \\\\
\\var{a31} & \\var{a32} \\\\
\\end{vmatrix} \\\\
&= \\var{cof23}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{31} &= +\\begin{vmatrix}
b & c \\\\
e & f \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a12} & \\var{a13} \\\\
\\var{a22} & \\var{a23} \\\\
\\end{vmatrix} \\\\
&= \\var{cof31}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{32} &= -\\begin{vmatrix}
a & c \\\\
d & f \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a11} & \\var{a13} \\\\
\\var{a21} & \\var{a23} \\\\
\\end{vmatrix} \\\\
&= \\var{cof32}
\\end{aligned}
$$
$$
\\begin{aligned}
A_{33} &= +\\begin{vmatrix}
a & b \\\\
d & e \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a11} & \\var{a12} \\\\
\\var{a21} & \\var{a22} \\\\
\\end{vmatrix} \\\\
&= \\var{cof33}
\\end{aligned}
$$
c)
\nUsing our answer from the previous question, we simply write the cofactors in the form
\n$$
\\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\\\
A_{21} & A_{22} & A_{23} \\\\
A_{31} & A_{32} & A_{33} \\\\
\\end{pmatrix}
$$
Giving us our matrix of cofactors
\n$$
\\begin{pmatrix}
\\var{cof11} & \\var{cof12} & \\var{cof13} \\\\
\\var{cof21} & \\var{cof22} & \\var{cof23} \\\\
\\var{cof31} & \\var{cof32} & \\var{cof33} \\\\
\\end{pmatrix}
$$
d)
\nThe transposition process turns rows into columns and columns into rows
\nCarrying out this process on our matrix of cofactors gives us the adjugate
\n$$
\\begin{pmatrix}
\\var{cof11} & \\var{cof21} & \\var{cof31} \\\\
\\var{cof12} & \\var{cof22} & \\var{cof32} \\\\
\\var{cof13} & \\var{cof23} & \\var{cof33} \\\\
\\end{pmatrix}
$$
e)
\nWe can find the inverse of $A$ using our determinant and adjugate, using the formula
\n$$
A^{-1} = \\frac{1}{\\det A}(adj \\; A)
$$
Therefore, we can calculate $A^{-1}$ by
\n$$
\\begin{aligned}
A^{-1} &= \\frac{1}{\\var{deta}} \\begin{pmatrix}
\\var{cof11} & \\var{cof21} & \\var{cof31} \\\\
\\var{cof12} & \\var{cof22} & \\var{cof32} \\\\
\\var{cof13} & \\var{cof23} & \\var{cof33} \\\\
\\end{pmatrix} \\\\
&= \\var{inverseA}
\\end{aligned}
$$
cof23
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\n$$
A=\\var{matrixA}
$$
Find the determinant of $A$
\n$\\det A =$ [[0]]
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\nThe cofactor $A_{ij}$ denotes the cofactor in row $i$ and column $j$
\n$A _{11}=$ [[0]]
\n$A_{12}=$ [[1]]
\n$A_{13}=$ [[2]]
\n$A_{21}=$ [[3]]
\n$A_{22}=$ [[4]]
\n$A_{23}=$ [[5]]
\n$A_{31}=$ [[6]]
\n$A_{32}=$ [[7]]
\n$A_{33}=$ [[8]]
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\n[[0]]
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\n[[0]]
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\nElements will be accepted as fractions or correct to 2 decimal places
\n$A^{-1}=$ [[0]]
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