Exercise 4.5 · Q11: Solve $2x+y+z=1$⁠, $x-2y-z=\frac{3}{2}$⁠, $3y-5z=9$ by the matrix method

Basics you need

Solution

1. Write the system as $AX = B$

   The third equation has no $x$-term, so its coefficient of $x$ is 0. With columns in the order $x, y, z$⁠:

   $$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix},$$ $$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix}.$$

   $A$ is $3\times3$ and $X$ is $3\times1$⁠, so $AX$ is defined and $3\times1$⁠, like $B$⁠. Row by row,

   $$AX = \begin{bmatrix} 2x + y + z \\ x - 2y - z \\ 0x + 3y - 5z \end{bmatrix},$$

   which reproduces the three left-hand sides.

   Leaving out the 0 would make the third row $3, -5$⁠, only two entries, and $A$ would not even be a matrix. Each unknown keeps its own column, so a missing unknown leaves a 0 in its column.
   Section 4.6.1 (matrix form); Section 3.4.5 (row times column). Lesson: solving systems AX = B

2. Compute $|A|$ by expanding along row 3

   Row 3 is $(0, 3, -5)$ and contains a 0, so expand along it. The signs for row 3 are $+\,-\,+$⁠:

   $$\begin{aligned} |A| &= 0\begin{vmatrix} 1 & 1 \\ -2 & -1 \end{vmatrix} - 3\begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}\\ &\qquad + (-5)\begin{vmatrix} 2 & 1 \\ 1 & -2 \end{vmatrix}\\ &= 0(1) - 3(-3) + (-5)(-5)\\ &= 0 + 9 + 25 = 34. \end{aligned}$$

   The two minors that matter are

   $$\begin{aligned} \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} &= (2)(-1) - (1)(1) = -3,\\ \begin{vmatrix} 2 & 1 \\ 1 & -2 \end{vmatrix} &= (2)(-2) - (1)(1) = -5. \end{aligned}$$

   So $|A| = 34 \ne 0$⁠: $A$ is nonsingular and the system has a unique solution.

   Each small determinant is the minor of the row-3 entry in front of it: delete row 3 and that entry's column. The first term is 0 whatever its minor is, which is why a row with a zero is the cheapest choice. The $-$ before the 3 comes from the sign $(-1)^{3+2} = -1$⁠, not from the entry itself.
   Section 4.2.3: a $3\times3$ determinant can be expanded along any row or column, all six ways giving the same value. Section 4.6.1, Case I. Lesson: expanding a 3 × 3 determinant

3. Find all nine minors and cofactors

   For each entry, delete its row and column, evaluate the $2\times2$ determinant left ($ad - bc$⁠), then attach the sign $(-1)^{i+j}$⁠.

   Row 1 (entries $2, 1, 1$⁠):

   $$\begin{aligned} M_{11} &= \begin{vmatrix} -2 & -1 \\ 3 & -5 \end{vmatrix}\\ &= (-2)(-5) - (-1)(3) = 13,\\ A_{11} &= (-1)^{1+1}(13) = 13,\\ M_{12} &= \begin{vmatrix} 1 & -1 \\ 0 & -5 \end{vmatrix}\\ &= (1)(-5) - (-1)(0) = -5,\\ A_{12} &= (-1)^{1+2}(-5) = 5,\\ M_{13} &= \begin{vmatrix} 1 & -2 \\ 0 & 3 \end{vmatrix}\\ &= (1)(3) - (-2)(0) = 3,\\ A_{13} &= (-1)^{1+3}(3) = 3. \end{aligned}$$

   Row 2 (entries $1, -2, -1$⁠):

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 1 & 1 \\ 3 & -5 \end{vmatrix}\\ &= (1)(-5) - (1)(3) = -8,\\ A_{21} &= (-1)^{2+1}(-8) = 8,\\ M_{22} &= \begin{vmatrix} 2 & 1 \\ 0 & -5 \end{vmatrix}\\ &= (2)(-5) - (1)(0) = -10,\\ A_{22} &= (-1)^{2+2}(-10) = -10,\\ M_{23} &= \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix}\\ &= (2)(3) - (1)(0) = 6,\\ A_{23} &= (-1)^{2+3}(6) = -6. \end{aligned}$$

   Row 3 (entries $0, 3, -5$⁠):

   $$\begin{aligned} M_{31} &= \begin{vmatrix} 1 & 1 \\ -2 & -1 \end{vmatrix}\\ &= (1)(-1) - (1)(-2) = 1,\\ A_{31} &= (-1)^{3+1}(1) = 1,\\ M_{32} &= \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}\\ &= (2)(-1) - (1)(1) = -3,\\ A_{32} &= (-1)^{3+2}(-3) = 3,\\ M_{33} &= \begin{vmatrix} 2 & 1 \\ 1 & -2 \end{vmatrix}\\ &= (2)(-2) - (1)(1) = -5,\\ A_{33} &= (-1)^{3+3}(-5) = -5. \end{aligned}$$
   Every cofactor is needed, including those of row 3 already used for $|A|$⁠, because $\operatorname{adj} A$ contains all nine. Note that $a_{31} = 0$ does not make its cofactor 0: $A_{31}$ depends only on the other entries, and it is 1. The sign $(-1)^{i+j}$ is $+$ at positions where $i + j$ is even (the corners and the centre) and $-$ at the four others.
   Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

4. Build $\operatorname{adj} A$ and $A^{-1}$

   Matrix of cofactors (each $A_{ij}$ in position $(i, j)$⁠):

   $$\begin{bmatrix} 13 & 5 & 3 \\ 8 & -10 & -6 \\ 1 & 3 & -5 \end{bmatrix}$$

   Its transpose (rows become columns) is the adjoint:

   $$\operatorname{adj} A = \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix},$$

   and, since $A^{-1} = \frac{1}{|A|}\operatorname{adj} A$⁠,

   $$A^{-1} = \frac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}.$$
   The cofactors of row 1 of $A$ become column 1 of $\operatorname{adj} A$⁠. This transpose is what makes row $i$ of $A$ meet its own cofactors in the product $A(\operatorname{adj} A)$⁠, giving $|A|$ on the diagonal.
   Definition 3, Section 4.5.1; Theorem 4, Section 4.5. Lesson: the adjoint

5. Check: $A(\operatorname{adj} A) = 34\,I$

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

   $$\begin{aligned} (1,1)&:\ 2(13) + 1(5) + 1(3)\\ &= 26 + 5 + 3 = 34,\\ (1,2)&:\ 2(8) + 1(-10) + 1(-6)\\ &= 16 - 10 - 6 = 0,\\ (1,3)&:\ 2(1) + 1(3) + 1(-5)\\ &= 2 + 3 - 5 = 0,\\ (2,1)&:\ 1(13) + (-2)(5) + (-1)(3)\\ &= 13 - 10 - 3 = 0,\\ (2,2)&:\ 1(8) + (-2)(-10) + (-1)(-6)\\ &= 8 + 20 + 6 = 34,\\ (2,3)&:\ 1(1) + (-2)(3) + (-1)(-5)\\ &= 1 - 6 + 5 = 0,\\ (3,1)&:\ 0(13) + 3(5) + (-5)(3)\\ &= 0 + 15 - 15 = 0,\\ (3,2)&:\ 0(8) + 3(-10) + (-5)(-6)\\ &= 0 - 30 + 30 = 0,\\ (3,3)&:\ 0(1) + 3(3) + (-5)(-5)\\ &= 0 + 9 + 25 = 34. \end{aligned}$$ $$A(\operatorname{adj} A) = \begin{bmatrix} 34 & 0 & 0 \\ 0 & 34 & 0 \\ 0 & 0 & 34 \end{bmatrix} = 34\,I,$$

   so $AA^{-1} = \frac{1}{34}(34\,I) = I$⁠.

   The diagonal entries are "a row times its own cofactors", which is $|A|$ (the expansion along that row). The off-diagonal entries are "a row times another row's cofactors", which is always 0. Seeing exactly this pattern confirms all nine cofactors at once.
   Theorem 1, Section 4.5, and the Note in Section 4.4 (a row times the cofactors of a different row gives 0). Basics: why the inverse divides by |A|

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

   $A^{-1}$ is $3\times3$⁠, $B$ is $3\times1$⁠, so $A^{-1}B$ is $3\times1$⁠. First $(\operatorname{adj} A)B$⁠, row by row:

   $$\begin{aligned} \text{row 1}&:\ 13(1) + 8\left(\tfrac{3}{2}\right) + 1(9)\\ &= 13 + 12 + 9 = 34,\\ \text{row 2}&:\ 5(1) + (-10)\left(\tfrac{3}{2}\right) + 3(9)\\ &= 5 - 15 + 27 = 17,\\ \text{row 3}&:\ 3(1) + (-6)\left(\tfrac{3}{2}\right) + (-5)(9)\\ &= 3 - 9 - 45 = -51. \end{aligned}$$ $$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}.$$

   So $x = 1$⁠, $y = \frac{1}{2}$⁠, $z = -\frac{3}{2}$.

   Multiplying $AX = B$ on the left by $A^{-1}$ gives $(A^{-1}A)X = A^{-1}B$⁠, that is $X = A^{-1}B$⁠; so this is the only possible solution, and Step 7 confirms it works. The fraction in $B$ is harmless: it is multiplied by whole numbers, and the factor $\frac{1}{34}$ comes last.
   Section 4.6.1, Case I; associativity of matrix multiplication, Section 3.4.6. Lesson: inverse of a matrix

7. Check in all three original equations

   $$\begin{aligned} 2x + y + z &= 2(1) + \tfrac{1}{2} + \left(-\tfrac{3}{2}\right) = 1,\\ x - 2y - z &= 1 - 2\left(\tfrac{1}{2}\right) - \left(-\tfrac{3}{2}\right) = \tfrac{3}{2},\\ 3y - 5z &= 3\left(\tfrac{1}{2}\right) - 5\left(-\tfrac{3}{2}\right) = 9. \end{aligned}$$

   In detail: $2 + \frac{1}{2} - \frac{3}{2} = 1$⁠; $1 - 1 + \frac{3}{2} = \frac{3}{2}$⁠; $\frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9$⁠. All three hold.

   A solution of the system must satisfy every equation. Checking all three in the original equations also guards against a slip in writing down $A$ or $B$ in Step 1.

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