Exercise 4.5 · Q4: Consistency of a $3\times3$ system containing a constant $a$

Basics you need

Solution

1. Write the system as $AX = B$

   $$\begin{gathered} A=\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}\\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}1\\2\\4\end{bmatrix} \end{gathered}$$

   $A$ is $3\times3$ and $X$ is $3\times1$⁠, so $AX$ is $3\times1$⁠, and row by row

   $$AX=\begin{bmatrix}x+y+z\\2x+3y+2z\\ax+ay+2az\end{bmatrix},$$

   which is the left side of the system. So $AX = B$ is the given system.

   Each row of $A$ lists the coefficients of one equation in the order $x, y, z$⁠. The letter $a$ is simply part of the coefficients of the third equation; it is a fixed number, not a fourth unknown.
   Basics: matrix facts the chapter uses

2. Expand $|A|$ along row 1

   No row or column contains a zero, so no expansion is shorter than another. Row 1 is the convenient choice because its entries are all 1, so each minor is multiplied by just 1:

   $$\begin{aligned} |A|&=1\begin{vmatrix}3&2\\a&2a\end{vmatrix}-1\begin{vmatrix}2&2\\a&2a\end{vmatrix}\\ &\quad+1\begin{vmatrix}2&3\\a&a\end{vmatrix} \end{aligned}$$

   The three $2\times2$ determinants:

   $$\begin{aligned} \begin{vmatrix}3&2\\a&2a\end{vmatrix}&=3(2a)-a(2)=4a\\ \begin{vmatrix}2&2\\a&2a\end{vmatrix}&=2(2a)-a(2)=2a\\ \begin{vmatrix}2&3\\a&a\end{vmatrix}&=2(a)-a(3)=-a \end{aligned}$$

   Signed sum, with the pattern $+\,-\,+$⁠:

   $$\begin{aligned} |A|&=1(4a)-1(2a)+1(-a)\\ &=4a-2a-a=a \end{aligned}$$
   Each minor comes from deleting row 1 and the column of the entry being used: column 1 for the first, column 2 for the second, column 3 for the third. The middle term carries a minus sign from the pattern. Everything collapses to the single letter $a$⁠, and that is the key to the whole question.
   Expansion along $R_1$⁠, Section 4.2.3. Lesson: expanding a 3 × 3 determinant

3. Two cases

   Since $|A| = a$⁠:

   • if $a \ne 0$⁠, then $|A| \ne 0$⁠: unique solution, the system is consistent (Part (i) below);
   • if $a = 0$⁠, then $|A| = 0$⁠: we must compute $(\operatorname{adj}A)B$ (Part (ii) below).
   The question gives no value for $a$⁠, so we cannot just say "consistent" or "inconsistent" without looking at every possible value. Only one value, $a = 0$⁠, makes the determinant vanish, so there are exactly two cases.
   Section 4.6.1, Cases I and II. Lesson: solving systems AX = B

Part (i): the case $a \ne 0$

1. Consistent

   For $a \ne 0$⁠, $|A| = a \ne 0$⁠, so $A$ is non-singular, $A^{-1}$ exists (Theorem 4), and $X = A^{-1}B$ is the unique solution (Section 4.6.1, Case I). So for every $a \ne 0$ the system is consistent.

   This already answers the question for $a \ne 0$⁠. The next steps find the solution itself, which confirms the conclusion and shows how it depends on $a$⁠.
   Lesson: solving systems AX = B

2. All nine cofactors

   Delete row $i$ and column $j$ to get the minor, then attach the sign $(-1)^{i+j}$⁠. These formulas hold for every value of $a$⁠:

   $$\begin{aligned} A_{11}&=(-1)^{1+1}\begin{vmatrix}3&2\\a&2a\end{vmatrix}\\ &=6a-2a=4a\\ A_{12}&=(-1)^{1+2}\begin{vmatrix}2&2\\a&2a\end{vmatrix}\\ &=-(4a-2a)=-2a\\ A_{13}&=(-1)^{1+3}\begin{vmatrix}2&3\\a&a\end{vmatrix}\\ &=2a-3a=-a\\ A_{21}&=(-1)^{2+1}\begin{vmatrix}1&1\\a&2a\end{vmatrix}\\ &=-(2a-a)=-a\\ A_{22}&=(-1)^{2+2}\begin{vmatrix}1&1\\a&2a\end{vmatrix}\\ &=2a-a=a\\ A_{23}&=(-1)^{2+3}\begin{vmatrix}1&1\\a&a\end{vmatrix}\\ &=-(a-a)=0\\ A_{31}&=(-1)^{3+1}\begin{vmatrix}1&1\\3&2\end{vmatrix}\\ &=2-3=-1\\ A_{32}&=(-1)^{3+2}\begin{vmatrix}1&1\\2&2\end{vmatrix}\\ &=-(2-2)=0\\ A_{33}&=(-1)^{3+3}\begin{vmatrix}1&1\\2&3\end{vmatrix}\\ &=3-2=1 \end{aligned}$$

   Check: row 1 times its own cofactors gives

   $$1(4a)+1(-2a)+1(-a)=a=|A|.\ \checkmark$$
   $A_{21}$ and $A_{22}$ have the same minor. Deleting row 2 leaves rows 1 and 3, and in those rows columns 1 and 2 of $A$ are identical (1 above $a$⁠). So deleting column 1 or deleting column 2 leaves the same block, with rows $(1, 1)$ and $(a, 2a)$⁠. Only the sign differs. The check against $|A|$ confirms the first row of cofactors.
   Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

3. $\operatorname{adj}A$⁠, and $A(\operatorname{adj}A) = |A|I$

   Transpose the matrix of cofactors (cofactors of row $i$ go down column $i$⁠):

   $$\operatorname{adj}A=\begin{bmatrix}4a&-a&-1\\-2a&a&0\\-a&0&1\end{bmatrix}$$

   Check with Theorem 1, row of $A$ times column of $\operatorname{adj}A$⁠:

   $$\begin{aligned} (1,1)&:\ 4a-2a-a=a\\ (1,2)&:\ -a+a+0=0\\ (1,3)&:\ -1+0+1=0\\ (2,1)&:\ 2(4a)+3(-2a)+2(-a)=0\\ (2,2)&:\ 2(-a)+3(a)+2(0)=a\\ (2,3)&:\ 2(-1)+3(0)+2(1)=0\\ (3,1)&:\ a(4a)+a(-2a)+2a(-a)=0\\ (3,2)&:\ a(-a)+a(a)+2a(0)=0\\ (3,3)&:\ a(-1)+a(0)+2a(1)=a \end{aligned}$$

   So $A(\operatorname{adj}A) = aI = |A|I$⁠. ✓ Hence $A^{-1} = \frac1a\operatorname{adj}A$⁠.

   The first column of $\operatorname{adj}A$ is the first row of cofactors, $4a, -2a, -a$⁠, and so on. The product check is the strongest test available: all nine entries must match $|A|I$⁠, so a wrong sign anywhere would show.
   Definition 3 and Theorem 1, Section 4.5.1. Lesson: the adjoint

4. The solution $X = A^{-1}B$

   First $(\operatorname{adj}A)B$⁠, a $3\times3$ times a $3\times1$⁠, so a $3\times1$ column:

   $$\begin{aligned} &\begin{bmatrix}4a(1)+(-a)(2)+(-1)(4)\\-2a(1)+a(2)+0(4)\\-a(1)+0(2)+1(4)\end{bmatrix}\\ &=\begin{bmatrix}2a-4\\0\\4-a\end{bmatrix} \end{aligned}$$

   Then divide by $|A| = a$ (allowed, as $a \ne 0$⁠):

   $$\begin{aligned} x&=\frac{2a-4}{a}=2-\frac{4}{a}\\ y&=\frac{0}{a}=0\\ z&=\frac{4-a}{a}=\frac{4}{a}-1 \end{aligned}$$
   $X = A^{-1}B = \frac{1}{a}(\operatorname{adj}A)B$⁠, so computing $(\operatorname{adj}A)B$ first keeps the fractions to the last line. The answer depends on $a$⁠, as it must: changing $a$ changes the third equation.
   Row times column, then the scalar $\frac1a$⁠. Basics: matrix facts the chapter uses

5. Check in all three equations

   $$\begin{aligned} x+y+z&=2-\tfrac{4}{a}+0+\tfrac{4}{a}-1\\ &=1\ \checkmark\\ 2x+3y+2z&=4-\tfrac{8}{a}+0+\tfrac{8}{a}-2\\ &=2\ \checkmark\\ ax+ay+2az&=2a-4+0+8-2a\\ &=4\ \checkmark \end{aligned}$$
   In each line the terms with $a$ cancel, leaving the right-hand side 1, 2, 4. In the third line, $a\cdot(2 - \frac4a) = 2a - 4$ and $2a(\frac4a - 1) = 8 - 2a$⁠. The check works for every $a \ne 0$ at once.

Part (ii): the case $a = 0$

1. $|A| = 0$⁠, so compute $(\operatorname{adj}A)B$

   With $a = 0$⁠, $|A| = a = 0$⁠, so $A$ is singular and Case II of Section 4.6.1 applies. The cofactor formulas of Part (i), Step 2 hold for every $a$⁠, so put $a = 0$ in them:

   $$\operatorname{adj}A=\begin{bmatrix}0&0&-1\\0&0&0\\0&0&1\end{bmatrix}$$ $$\begin{aligned} (\operatorname{adj}A)B&=\begin{bmatrix}0(1)+0(2)+(-1)(4)\\0(1)+0(2)+0(4)\\0(1)+0(2)+1(4)\end{bmatrix}\\ &=\begin{bmatrix}-4\\0\\4\end{bmatrix}\ne O \end{aligned}$$

   So for $a = 0$ the system is inconsistent.

   The reason for the rule: if some $X_0$ satisfied $AX_0 = B$⁠, then $(\operatorname{adj}A)B = (\operatorname{adj}A)AX_0$⁠, which equals $|A|X_0 = O$ by Theorem 1. Here $(\operatorname{adj}A)B$ has non-zero entries, so no solution can exist.
   Section 4.6.1, Case II, and Theorem 1. Lesson: solving systems AX = B

2. See it directly

   With $a = 0$ the third equation reads $0x + 0y + 0z = 4$⁠. The left side is 0 for every $x, y, z$⁠, and $0 \ne 4$⁠, so no triple satisfies it, and the whole system has no solution.

   This confirms the matrix test in the plainest possible way. It also shows why the one-word answer "Consistent" holds only with the condition $a \ne 0$⁠.

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