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Suppose that the matrix $A$ below has determinant $−3$. Find the determinants of $B$, $C$ and $D$. $$ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} $$ $$ B = 2 \begin{pmatrix} b & a & c \\ e & d & f \\ h & g & i \\ \end{pmatrix} $$ $$ C = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{pmatrix} $$ $$ D = \begin{pmatrix} a - 5d & b - 5e & c - 5f \\ d & e & f \\ 2g & 2h & 2i \\ \end{pmatrix} $$

My attempt:

Firstly I am unsure for B if it means $|2X|$ or $2|X|$:

For $B$ assuming $|2X|$: There is one column swap, $C1 \longleftrightarrow C2$ therefore the sign of the determinant is changed, then the whole matrix is multiplied by two, which means each row is multiplied by two this has the affect of the determinant being multiplied by $2^3$ .

$$|B| = 3 \times 2^3 = 24$$

For $B$ assuming $2|X|$:There is one column swap, $C1 \longleftrightarrow C2$ therefore the sign of the determinant is changed, then the whole matrix is multiplied by $2$. This has the effect of multiplying the determinant by $2$.

$$|B| = 3 \times 2 = 6$$

For $C$: This is simply the matrix multiplied by the transpose. Since transposition has no effect on the determinant of a matrix its simply $|A| \times |A|$

$$|C| = |A| \times |A^T| = |A| \times |A| = -3 \times -3 = 9$$

For $D$: $R1 - 5R2$ has no effect on the determinant, however $2 \times R3$ has the effect of multiplying the determinant of the matrix by $2$

$$|D| = -3 \times 2 = -6$$

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    $\begingroup$ Why would you assume $\;|B|=2|X|\;$ when it is given $\;B=2X\implies |B|=|2X|=8|X|\;$ ? All the rest it looks very well to me. $\endgroup$
    – user312943
    Apr 22, 2016 at 10:22
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    $\begingroup$ Multiplying a matrix by a scalar means multiplying each element of the matrix by the scalar. Otherwise your answers are fine. $\endgroup$
    – almagest
    Apr 22, 2016 at 10:28
  • $\begingroup$ Ok understood . $\endgroup$
    – Mark Donaghue
    Apr 22, 2016 at 10:41
  • $\begingroup$ Why did you remove your attempts at solving this multi-part problem? It is not an improvement, and will likely cause the Question to be placed on-hold. $\endgroup$
    – hardmath
    Apr 22, 2016 at 13:28
  • $\begingroup$ @hardmath fixed sorry, didnt want the question to be unanswered. $\endgroup$
    – Mark Donaghue
    Apr 23, 2016 at 3:05

1 Answer 1

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For $B$: There is one column swap, $C1 \longleftrightarrow C2$ therefore the sign of the determinant is changed, then the whole matrix is multiplied by two, which means each row is multiplied by $2$ this has the affect of the determinant being multiplied by $2^3$ .

$$|B| = 3 \times 2^3 = 24$$

For $C$: This is simply the matrix multiplied by the transpose. Since transposition has no effect on the determinant of a matrix its simply $|A| \times |A|$

$$|C| = |A| \times |A^T| = |A| \times |A| = -3 \times -3 = 9$$

For $D$: $R1 - 5R2$ has no effect on the determinant, however $2 \times R3$ has the effect of multiplying the determinant of the matrix by $2$

$$|D| = -3 \times 2 = -6$$

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