Exercise 4.5 · Q6: Is $5x-y+4z=5$⁠, $2x+3y+5z=2$⁠, $5x-2y+6z=-1$ consistent?

Basics you need

Solution

1. Write the system as $AX = B$

   $$\begin{gathered} A=\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}\\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}5\\2\\-1\end{bmatrix} \end{gathered}$$

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

   $$AX=\begin{bmatrix}5x-y+4z\\2x+3y+5z\\5x-2y+6z\end{bmatrix},$$

   and $AX = B$ is the given system, entry by entry.

   The $-y$ in the first equation is the entry $-1$⁠, and the $-2y$ in the third is $-2$⁠. Each row of $A$ is one equation; each column belongs to one unknown.
   Basics: matrix facts the chapter uses

2. Expand $|A|$ along row 1

   No entry is zero, so every row costs three minors; row 1 is as good as any. Signs $+\,-\,+$⁠:

   $$\begin{aligned} |A|&=5\begin{vmatrix}3&5\\-2&6\end{vmatrix}-(-1)\begin{vmatrix}2&5\\5&6\end{vmatrix}\\ &\quad+4\begin{vmatrix}2&3\\5&-2\end{vmatrix} \end{aligned}$$ $$\begin{aligned} \begin{vmatrix}3&5\\-2&6\end{vmatrix}&=18-(-10)=28\\ \begin{vmatrix}2&5\\5&6\end{vmatrix}&=12-25=-13\\ \begin{vmatrix}2&3\\5&-2\end{vmatrix}&=-4-15=-19 \end{aligned}$$ $$\begin{aligned} |A|&=5(28)+1(-13)+4(-19)\\ &=140-13-76=51 \end{aligned}$$

   So $|A| = 51 \ne 0$⁠.

   The middle term is $-(-1)\times(-13)$⁠: pattern sign $-$⁠, entry $-1$⁠, minor $-13$⁠, giving $-13$ overall. In the first minor, $3\cdot6 = 18$ and $(-2)\cdot5 = -10$⁠, and subtracting $-10$ adds 10.
   Expansion along $R_1$⁠, Section 4.2.3. Lesson: expanding a 3 × 3 determinant

3. Conclude: the system is consistent

   $|A| = 51 \ne 0$⁠, so $A$ is non-singular and invertible (Theorem 4). Then $X = A^{-1}B$ satisfies $AX = A(A^{-1}B) = B$⁠, and any solution must equal it. So the system is consistent, with exactly one solution.

   This is Case I of Section 4.6.1, and it fully answers the question. The remaining steps find the solution, which the matrix method is designed to do, and give a numerical confirmation.
   Lesson: solving systems AX = B

4. All nine cofactors

   $$\begin{aligned} A_{11}&=(-1)^{1+1}\begin{vmatrix}3&5\\-2&6\end{vmatrix}\\ &=18+10=28\\ A_{12}&=(-1)^{1+2}\begin{vmatrix}2&5\\5&6\end{vmatrix}\\ &=-(12-25)=13\\ A_{13}&=(-1)^{1+3}\begin{vmatrix}2&3\\5&-2\end{vmatrix}\\ &=-4-15=-19\\ A_{21}&=(-1)^{2+1}\begin{vmatrix}-1&4\\-2&6\end{vmatrix}\\ &=-(-6+8)=-2\\ A_{22}&=(-1)^{2+2}\begin{vmatrix}5&4\\5&6\end{vmatrix}\\ &=30-20=10\\ A_{23}&=(-1)^{2+3}\begin{vmatrix}5&-1\\5&-2\end{vmatrix}\\ &=-(-10+5)=5\\ A_{31}&=(-1)^{3+1}\begin{vmatrix}-1&4\\3&5\end{vmatrix}\\ &=-5-12=-17\\ A_{32}&=(-1)^{3+2}\begin{vmatrix}5&4\\2&5\end{vmatrix}\\ &=-(25-8)=-17\\ A_{33}&=(-1)^{3+3}\begin{vmatrix}5&-1\\2&3\end{vmatrix}\\ &=15+2=17 \end{aligned}$$
   Each minor deletes the row and column of its entry. For example, $A_{21}$ deletes row 2 and column 1, leaving $(-1, 4)$ from row 1 and $(-2, 6)$ from row 3; its value is $(-1)(6) - (-2)(4)$⁠, that is $-6 + 8 = 2$⁠, and the sign $(-1)^3$ makes it $-2$⁠. The first row of cofactors, $28, 13, -19$⁠, reproduces $|A|$⁠: $5(28) + (-1)(13)$ $+\, 4(-19) = 51$⁠.
   Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

5. $\operatorname{adj}A$⁠, checked by $A(\operatorname{adj}A) = 51I$

   Transpose the matrix of cofactors: row 1 of cofactors becomes column 1, and so on.

   $$\operatorname{adj}A=\begin{bmatrix}28&-2&-17\\13&10&-17\\-19&5&17\end{bmatrix}$$

   Row of $A$ times column of $\operatorname{adj}A$⁠:

   $$\begin{aligned} (1,1)&:\ 140-13-76=51\\ (1,2)&:\ -10-10+20=0\\ (1,3)&:\ -85+17+68=0\\ (2,1)&:\ 56+39-95=0\\ (2,2)&:\ -4+30+25=51\\ (2,3)&:\ -34-51+85=0\\ (3,1)&:\ 140-26-114=0\\ (3,2)&:\ -10-20+30=0\\ (3,3)&:\ -85+34+102=51 \end{aligned}$$

   So $A(\operatorname{adj}A) = 51I = |A|I$ ✓, and therefore

   $$A^{-1}=\frac{1}{51}\begin{bmatrix}28&-2&-17\\13&10&-17\\-19&5&17\end{bmatrix}.$$
   For instance, entry $(2,1)$ is row 2 of $A$⁠, $(2, 3, 5)$⁠, times column 1 of $\operatorname{adj}A$⁠, $(28, 13, -19)$⁠: $56 + 39 - 95 = 0$⁠. Getting $|A|$ on the diagonal and 0 elsewhere (Theorem 1) confirms every cofactor at once.
   Definition 3, Theorem 1 (Section 4.5.1) and Theorem 4. Lesson: inverse of a matrix

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

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

   $$\begin{aligned} &\begin{bmatrix}28(5)+(-2)(2)+(-17)(-1)\\13(5)+10(2)+(-17)(-1)\\-19(5)+5(2)+17(-1)\end{bmatrix}\\ &=\begin{bmatrix}140-4+17\\65+20+17\\-95+10-17\end{bmatrix}=\begin{bmatrix}153\\102\\-102\end{bmatrix} \end{aligned}$$

   Then divide by 51:

   $$X=\frac{1}{51}\begin{bmatrix}153\\102\\-102\end{bmatrix}=\begin{bmatrix}3\\2\\-2\end{bmatrix}$$

   So $x = 3$⁠, $y = 2$⁠, $z = -2$⁠.

   $X = \frac{1}{51}(\operatorname{adj}A)B$⁠. Doing the whole-number product first and dividing at the end avoids fractions; $153 = 3\cdot51$ and $102 = 2\cdot51$⁠, so the division comes out exact.
   Basics: matrix facts the chapter uses

7. Check in all three equations

   $$\begin{aligned} 5x-y+4z&=15-2-8=5\ \checkmark\\ 2x+3y+5z&=6+6-10=2\ \checkmark\\ 5x-2y+6z&=15-4-12=-1\ \checkmark \end{aligned}$$
   Substituting into the equations exactly as printed confirms the solution, and with it the consistency: a solution has been exhibited.

Every fact used, and where it comes from