Exercise 4.5 · Q2: Is $2x-y=5$⁠, $x+y=4$ consistent?

Basics you need

Solution

1. Write the system as $AX = B$

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

   $A$ is $2\times2$ and $X$ is $2\times1$⁠, so $AX$ is defined and is $2\times1$⁠, like $B$⁠:

   $$AX=\begin{bmatrix}2\cdot x+(-1)\cdot y\\1\cdot x+1\cdot y\end{bmatrix}=\begin{bmatrix}2x-y\\x+y\end{bmatrix}.$$

   Setting this equal to $B$ entry by entry gives back $2x - y = 5$ and $x + y = 4$⁠.

   The first row of $A$ holds the coefficients of the first equation, in the order $x$⁠, then $y$⁠. The $-y$ supplies the entry $-1$⁠; the second equation $x + y$ supplies two 1s. Multiplying $AX$ out confirms nothing has been misplaced.
   Equal matrices have equal entries, so $AX = B$ is the same as both equations together. Basics: matrix facts the chapter uses

2. Compute $|A|$

   $$\begin{aligned} |A|&=\begin{vmatrix}2&-1\\1&1\end{vmatrix}=2\cdot1-1\cdot(-1)\\ &=2+1=3 \end{aligned}$$

   So $|A| = 3 \ne 0$⁠, and $A$ is non-singular.

   The other diagonal is $1$ and $-1$⁠, with product $-1$⁠. Subtracting $-1$ adds 1, which is why the answer is $3$⁠, not $1$⁠. Brackets around $-1$ keep that sign visible.
   $ad - cb$ with $a = 2$⁠, $b = -1$⁠, $c = 1$⁠, $d = 1$⁠. Lesson: determinant of a square matrix

3. Conclude: the system is consistent

   Since $|A| \ne 0$⁠, $A^{-1}$ exists (Theorem 4, Section 4.5). Pre-multiplying $AX = B$ by $A^{-1}$ gives $X = A^{-1}B$⁠, and this $X$ does satisfy the system because $A(A^{-1}B) = B$⁠. So a solution exists, and it is the only one: the system is consistent.

   This is Case I of Section 4.6.1. The determinant alone settles consistency here; the solution is found next as confirmation and because the matrix method is the point of the exercise.
   $|A| \ne 0$ $\Rightarrow$ exactly one solution. Lesson: solving systems AX = B

4. Cofactors and $\operatorname{adj}A$

   For each entry, delete its row and column; the single entry left is the minor $M_{ij}$⁠. Then $A_{ij} = (-1)^{i+j}M_{ij}$⁠:

   $$\begin{aligned} A_{11}&=(-1)^{1+1}(1)=1\\ A_{12}&=(-1)^{1+2}(1)=-1\\ A_{21}&=(-1)^{2+1}(-1)=1\\ A_{22}&=(-1)^{2+2}(2)=2 \end{aligned}$$

   The adjoint is the transpose of the matrix of cofactors:

   $$\operatorname{adj}A=\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}=\begin{bmatrix}1&1\\-1&2\end{bmatrix}$$
   $A_{21}$ is the cofactor of the entry in row 2, column 1. Deleting row 2 and column 1 leaves the entry $-1$⁠, and the sign $(-1)^{3} = -1$ turns it into $+1$⁠. In the adjoint, $A_{21}$ moves to row 1, column 2: that is the transpose. Shortcut check: swap 2 and 1 on the diagonal, change the signs of $-1$ and $1$⁠.
   Definition 3, Section 4.5.1. Lesson: the adjoint

5. $A^{-1}$⁠, checked by $A(\operatorname{adj}A) = |A|I$

   $$A^{-1}=\frac{1}{|A|}\operatorname{adj}A=\frac{1}{3}\begin{bmatrix}1&1\\-1&2\end{bmatrix}$$

   Check with Theorem 1: $A(\operatorname{adj}A)$ should be $3I$⁠. Entry by entry:

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

   So $A(\operatorname{adj}A) = 3I$⁠, and dividing by 3 gives $AA^{-1} = I$⁠. ✓

   Leaving the $\frac13$ outside the matrix keeps the arithmetic in whole numbers. The check is worth the half minute: a wrong sign in a cofactor would show up as a non-zero off-diagonal entry.
   $A(\operatorname{adj}A) = |A|I$ (Theorem 1, Section 4.5.1). Basics: why the inverse divides by |A|

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

   $$X=A^{-1}B=\frac{1}{3}\begin{bmatrix}1&1\\-1&2\end{bmatrix}\begin{bmatrix}5\\4\end{bmatrix}$$ $$\begin{aligned} x&=\tfrac{1}{3}(1\cdot5+1\cdot4)=\tfrac{9}{3}=3\\ y&=\tfrac{1}{3}((-1)\cdot5+2\cdot4)=\tfrac{3}{3}=1 \end{aligned}$$

   So $x = 3$⁠, $y = 1$⁠.

   The product of the $2\times2$ matrix and the $2\times1$ column $B$ is a $2\times1$ column, one entry per unknown. Each entry is a row times $B$⁠, then the factor $\frac13$ is applied.
   Row times column, then the scalar. Basics: matrix facts the chapter uses

7. Check in both original equations

   $$\begin{aligned} 2x-y&=2(3)-1=5\ \checkmark\\ x+y&=3+1=4\ \checkmark \end{aligned}$$
   Both equations hold, so $(3, 1)$ really is a solution, which is exactly what "consistent" asks for. Geometrically, rewriting the equations as $y = 2x - 5$ and $y = 4 - x$ shows the slopes 2 and $-1$ are different, so the lines cannot be parallel.
   Lines with different slopes cross at one point. Basics: two lines, cross, parallel or the same

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