Exercises — Properties of determinants

Look at columns 1 and 3 — they are identical $(3,1,6)$. $P1$ says column properties equal row properties, and $P3$ says two equal rows (hence columns) force the determinant to $0$. $\boxed{=0}$ No arithmetic needed — that is the whole point of Recognition.

• $\det (A^{\top })=\det A=7$ by $P1$.
• $2A$ scales all three rows by $2$, so by the $P4$ corollary $\det (2A)=2^3\det A=8\cdot 7=56$. $\boxed{\det (A^{\top })=7,\det (2A)=56}$

Column 3 is all zeros. By $P7$ (via $P1$), a zero row/column gives $\det =0$. $\boxed{=0}$

$P8$: for a triangular matrix the determinant is just the product of the diagonal. $\det A=2\cdot (-3)\cdot 5=\boxed{-30}$

Step 1 (why): Use $P6$ to clear column 1 — these operations don't change the value. $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$: $\left|\begin{array}{ccc}1& 2& -1\\ 0& -1& 3\\ 0& -3& 5\end{array}\right|.$ Step 2 (why): Clear column 2 below the diagonal. $R_3\to R_3-3R_2$: $\left|\begin{array}{ccc}1& 2& -1\\ 0& -1& 3\\ 0& 0& -4\end{array}\right|.$ Step 3: Now triangular. By $P8$, $\det =1\cdot (-1)\cdot (-4)=\boxed{4}$.

Step 1 (why): Row 1 $=3\cdot (1,2,3)$. $P4$ lets us factor $3$ out of that one row. $=3\left|\begin{array}{ccc}1& 2& 3\\ 1& 0& 2\\ 2& 2& 0\end{array}\right|.$ Step 2 (why): $P6$ to make zeros. $R_2\to R_2-R_1$, $R_3\to R_3-2R_1$: $=3\left|\begin{array}{ccc}1& 2& 3\\ 0& -2& -1\\ 0& -2& -6\end{array}\right|.$ Step 3: $R_3\to R_3-R_2$: $\left|\begin{array}{ccc}1& 2& 3\\ 0& -2& -1\\ 0& 0& -5\end{array}\right|$. By $P8$: $=3\cdot (1\cdot (-2)\cdot (-5))=3\cdot 10=\boxed{30}.$

• By $P9$ corollary, $\det (A^{-1})=\frac{1}{\det A}=\frac{1}{4}$.
• $2A^{-1}$ scales all $3$ rows by $2$, so $P4$ corollary gives $2^3\cdot \frac{1}{4}=8\cdot \frac{1}{4}=\boxed{2}.$

Track each operation's factor:

• Multiply $R_1$ by $5$: factor $\times 5$ ($P4$). Running $=5\cdot 6=30$.
• Swap $R_2,R_3$: factor $\times (-1)$ ($P2$). Running $=-30$.
• $R_1\to R_1+4R_3$: factor $\times 1$ ($P6$, no change). Running $=-30$. $\boxed{\det B=-30}$

Idea: make row 2 collapse using $P6$. $R_2\to R_2-R_1$ (subtract row 1; $P6$, value unchanged): $\left|\begin{array}{ccc}a& b& c\\ 2& 2& 2\\ 1& 1& 1\end{array}\right|.$ Now row 2 $=2\cdot (1,1,1)=2\cdot$ row 3. Pull out the $2$ ($P4$): $=2\left|\begin{array}{ccc}a& b& c\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|.$ Rows 2 and 3 are equal $\Rightarrow \det =0$ by $P3$. Hence the whole thing is $2\cdot 0=\boxed{0}$.

Take $A=B=I_2$ (the identity). Then $\det A=\det B=1$, so $\det A+\det B=2$. But $A+B=2I_2=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$, and by $P8$ (diagonal) $\det (2I_2)=2\cdot 2=4$. Since $4\ne 2$, the "rule" fails. $P5$ is linearity in one row at a time; adding whole matrices changes every row at once, so it does not factor. $\boxed{\det (A+B)=4\ne 2=\det A+\det B}$

Step 1 (why): kill the first column below the top using column operations ($P1$ lets column ops mirror row ops). $C_3\to C_3-C_2$ then $C_2\to C_2-C_1$: $\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-b\\ a^2& b^2-a^2& c^2-b^2\end{array}\right|.$ Step 2: factor each of the new columns ($P4$): $b^2-a^2=(b-a)(b+a)$, $c^2-b^2=(c-b)(c+b)$: $=(b-a)(c-b)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^2& b+a& c+b\end{array}\right|.$ Step 3: expand along row 1 (only the top-left $1$ survives), leaving the $2\times 2$: $=(b-a)(c-b)[(c+b)-(b+a)]=(b-a)(c-b)(c-a).$ $\boxed{=(b-a)(c-a)(c-b)}$

Chain three properties:

• $\det (M^{\top })=\det M$ by $P1$.
• $M^{\top }=-M$ means we multiplied all $3$ rows by $-1$, so $\det (M^{\top })=(-1{)}^3\det M=-\det M$ ($P4$ corollary). Combine: $\det M=-\det M\Rightarrow 2\det M=0\Rightarrow \boxed{\det M=0}$. (Key: $(-1{)}^3=-1$ because $3$ is odd — this fails for even-sized skew matrices, where $\det$ can be nonzero.)

Break the product with $P9$, $P1$, $P4$ corollary:

• $\det (A^2)=(\det A{)}^2=9$ (apply $P9$ twice).
• $\det (B^{-1})=1/\det B=-\frac{1}{2}$.
• $\det (2A^{\top })=2^3\det (A^{\top })=8\cdot \det A=24$ ($P4$ corollary + $P1$). Multiply (order doesn't matter, $\det$ of a product is the product of $\det$s): $9\cdot \left(-\frac{1}{2}\right)\cdot 24=9\cdot (-12)=\boxed{-108}.$

The determinant is the area-scaling factor of the map (see Linear Transformations & Scaling of Area). $\det T=2\cdot 3-1\cdot 1=5.$ New area $=|\det T|\times \text{old area}=5\times 6=\boxed{30}$. Since $\det T=5>0$, orientation is preserved (no flip). Look at the red parallelogram in the figure: the unit square of area $1$ becomes a region of area $5$ — every shape scales by the same factor.

If $A^{-1}$ exists, then $A{A}^{-1}=I$. Apply $P9$: $\det A\cdot \det (A^{-1})=\det I=1.$ A product equalling $1$ cannot have a zero factor, so $\det A\ne 0$ (and $\det (A^{-1})=1/\det A$, the $P9$ corollary). Conversely, if $\det A=0$ the map squashes area to $0$ (rows are linearly dependent, $P3$-style), so it cannot be undone — no inverse exists. This is exactly the bridge to Inverse of a Matrix via Adjoint. $■$