Exercise 4.3 · Q5 (MCQ): Which expression equals $\Delta$⁠?

Basics you need

Solution

1. The cofactors we will need

   Delete the row and column of each position and attach $(-1)^{i+j}$⁠:

   $$\begin{aligned} A_{11} &= +(a_{22}a_{33} - a_{23}a_{32}),\\ A_{12} &= -(a_{21}a_{33} - a_{23}a_{31}),\\ A_{13} &= +(a_{21}a_{32} - a_{22}a_{31}),\\ A_{21} &= -(a_{12}a_{33} - a_{13}a_{32}),\\ A_{31} &= +(a_{12}a_{23} - a_{13}a_{22}),\\ A_{32} &= -(a_{11}a_{23} - a_{13}a_{21}),\\ A_{33} &= +(a_{11}a_{22} - a_{12}a_{21}). \end{aligned}$$
   For example, $A_{21}$⁠: delete $R_2$ and $C_1$⁠, leaving $a_{12}, a_{13}$ over $a_{32}, a_{33}$⁠, whose determinant is $a_{12}a_{33} - a_{13}a_{32}$⁠; position (2, 1) has sign $-$⁠. The others are found the same way (Example 10 of the book does $A_{11}$ and $A_{21}$⁠).
   Section 4.4, Definitions 1 and 2; Example 10. Lesson: minors and cofactors

2. (D) equals $\Delta$ for every determinant

   Substitute the cofactors of column 1:

   $$\begin{aligned} &a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}\\ &= a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32}\\ &\quad - a_{21}a_{12}a_{33} + a_{21}a_{13}a_{32}\\ &\quad + a_{31}a_{12}a_{23} - a_{31}a_{13}a_{22}. \end{aligned}$$

   Compare with Basic 2, reordering the factors in each product:

   Term of (D)	Same as term of $\Delta$
   $a_{11}a_{22}a_{33}$	$a_{11}a_{22}a_{33}$
   $-a_{11}a_{23}a_{32}$	$-a_{11}a_{23}a_{32}$
   $-a_{21}a_{12}a_{33}$	$-a_{12}a_{21}a_{33}$
   $+a_{21}a_{13}a_{32}$	$+a_{13}a_{21}a_{32}$
   $+a_{31}a_{12}a_{23}$	$+a_{12}a_{23}a_{31}$
   $-a_{31}a_{13}a_{22}$	$-a_{13}a_{22}a_{31}$

   The six terms match exactly, so (D) $= \Delta$ for every $3\times3$ determinant.

   (D) pairs each element of column 1 with its own cofactor, so it is the expansion along $C_1$⁠. The book states that every row and column expansion gives $\Delta$ (Section 4.2.3, equation (3), and the Remark after Example 10); the table shows it directly for this case.
   Section 4.2.3, expansion along $C_1$⁠. Lesson: expanding a 3 × 3 determinant

3. (A) is always 0

   (A) uses the elements of row 1 with the cofactors of row 3:

   $$\begin{aligned} &a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}\\ &= a_{11}a_{12}a_{23} - a_{11}a_{13}a_{22}\\ &\quad - a_{12}a_{11}a_{23} + a_{12}a_{13}a_{21}\\ &\quad + a_{13}a_{11}a_{22} - a_{13}a_{12}a_{21}\\ &= 0. \end{aligned}$$

   The terms cancel in pairs: first with third, second with fifth, fourth with sixth.

   This is the Note in Section 4.4: elements of one row times the cofactors of another row always give 0. (It is the expansion of a determinant whose rows 1 and 3 are equal.) So (A) equals $\Delta$ only for determinants that happen to be 0, not in general.
   Section 4.4, Note; Example 11 checks the same sum for a numerical determinant.

4. (C) is always 0

   (C) uses the elements of row 2 with the cofactors of row 1:

   $$\begin{aligned} &a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13}\\ &= a_{21}a_{22}a_{33} - a_{21}a_{23}a_{32}\\ &\quad - a_{22}a_{21}a_{33} + a_{22}a_{23}a_{31}\\ &\quad + a_{23}a_{21}a_{32} - a_{23}a_{22}a_{31}\\ &= 0. \end{aligned}$$

   Again the terms cancel in pairs (first with third, second with fifth, fourth with sixth).

   Same reason as (A): the elements and the cofactors come from different rows. This is exactly the example written out in the Note of Section 4.4.

5. (B) mixes a row with a column

   (B) multiplies the elements of row 1 ($a_{11}, a_{12}, a_{13}$⁠) by the cofactors of column 1 ($A_{11}, A_{21}, A_{31}$⁠). Only the first pair has matching subscripts. Substituting:

   $$\begin{aligned} &a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31}\\ &= a_{11}(a_{22}a_{33} - a_{23}a_{32})\\ &\quad - a_{12}(a_{12}a_{33} - a_{13}a_{32})\\ &\quad + a_{13}(a_{12}a_{23} - a_{13}a_{22}). \end{aligned}$$

   This contains $-a_{12}^2a_{33}$ and $-a_{13}^2a_{22}$⁠, squares of elements, which never occur in $\Delta$ (each term of $\Delta$ uses one element from each row). So (B) is not $\Delta$ in general.

   The second and third terms of (B) pair $a_{12}$ with $A_{21}$ and $a_{13}$ with $A_{31}$⁠: the subscripts are swapped, so the element is multiplied by a cofactor that belongs to a different position. Step 6 gives a numerical determinant where (B) fails.

6. One determinant that rules out (A), (B), (C)

   Take

   $$\Delta = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1\cdot(1\cdot1 - 1\cdot0) = 1,$$

   expanding along $R_3$⁠, whose only nonzero entry is $a_{33} = 1$ (sign $+$⁠, minor: rows 1 and 2, columns 1 and 2). The cofactors needed:

   $$\begin{aligned} A_{11} &= +\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1,\\ A_{12} &= -\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0,\\ A_{13} &= +\begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0,\\ A_{21} &= -\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = -1,\\ A_{31} &= +\begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} = 0,\\ A_{32} &= -\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = 0,\\ A_{33} &= +\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1. \end{aligned}$$

   Now evaluate each option ($a_{11} = a_{12} = 1$⁠, $a_{13} = 0$⁠; $a_{21} = 0$⁠, $a_{22} = 1$⁠, $a_{23} = 0$⁠; $a_{31} = a_{32} = 0$⁠, $a_{33} = 1$⁠):

   Option	Value	Equals $\Delta = 1$⁠?
   (A)	$1(0) + 1(0) + 0(1) = 0$	No
   (B)	$1(1) + 1(-1) + 0(0) = 0$	No
   (C)	$0(1) + 1(0) + 0(0) = 0$	No
   (D)	$1(1) + 0(-1) + 0(0) = 1$	Yes

   One explicit determinant on which an option gives the wrong value proves that the option is not "the value of $\Delta$" (Basic 4). This one has a single 1 off the diagonal, which is just enough to separate (B) from (D): their only difference is $a_{12}A_{21} + a_{13}A_{31}$ against $a_{21}A_{21} + a_{31}A_{31}$⁠.
   One counterexample disproves a "for all" claim.

Every fact used, and where it comes from