Determinants, Area and Cofactors — Worked Solutions

Ques­tion 1: Two $2 \times 2$ deter­mi­nants#

The prob­lem#

Eval­u­ate $\begin{vmatrix} 7 & 3 \\ -2 & 5 \end{vmatrix}$ and $\begin{vmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{vmatrix}$.

Under­stand­ing the prob­lem#

Each is a $2 \times 2$ deter­mi­nant; you need a sin­gle num­ber (or expres­sion) for each.

The idea#

Use $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$: the prod­uct of the lead­ing diag­o­nal minus the prod­uct of the other diag­o­nal.

Step-by-step solu­tion#

Part (a)

Step 1. Mul­ti­ply the diag­o­nals.

$$ad = 7 \times 5 = 35, \qquad bc = 3 \times (-2) = -6.$$

Step 2. Sub­tract.

$$35 - (-6) = 35 + 6 = 41.$$

Part (b)

Step 1. Mul­ti­ply the diag­o­nals.

$$ad = \cos\theta \cdot \cos\theta = \cos^2\theta, \qquad bc = \sin\theta \cdot (-\sin\theta) = -\sin^2\theta.$$

Step 2. Sub­tract and use $\sin^2\theta + \cos^2\theta = 1$.

$$\cos^2\theta - (-\sin^2\theta) = \cos^2\theta + \sin^2\theta = 1.$$

Check­ing the answer#

In (b), try $\theta = 0$: $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$ ✓.

Answer#

$41$; $1$.

Com­mon mis­take to avoid#

$3 \times (-2) = -6$, and sub­tract­ing $-6$ means adding $6$. Writ­ing $35 - 6 = 29$ is the most com­mon slip here.

Ques­tion 2: Expand­ing along the sec­ond row#

The prob­lem#

Eval­u­ate $\begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & 1 \\ -2 & 4 & 5 \end{vmatrix}$ along the sec­ond row.

Under­stand­ing the prob­lem#

You are told which row to use: the sec­ond row, $(3, 0, 1)$. It con­tains a $0$, which saves one cal­cu­la­tion.

The idea#

Expand along row $2$. The signs for row $2$ are $-\ +\ -$. Each entry is mul­ti­plied by its sign and by the $2 \times 2$ minor left after delet­ing its row and col­umn.

Step-by-step solu­tion#

Step 1. Write the expan­sion with the row-2 signs.

$$\Delta = -3\,M_{21} + 0\cdot M_{22} - 1\cdot M_{23}.$$

Step 2. Minor $M_{21}$: delete row 2 and col­umn 1.

$$M_{21} = \begin{vmatrix} -1 & 2 \\ 4 & 5 \end{vmatrix} = (-1)(5) - (2)(4) = -5 - 8 = -13.$$

Step 3. The mid­dle term is $0 \times M_{22} = 0$, so you need not com­pute $M_{22}$.

Step 4. Minor $M_{23}$: delete row 2 and col­umn 3.

$$M_{23} = \begin{vmatrix} 1 & -1 \\ -2 & 4 \end{vmatrix} = (1)(4) - (-1)(-2) = 4 - 2 = 2.$$

Step 5. Com­bine.

$$\Delta = -3(-13) + 0 - 1(2) = 39 - 2 = 37.$$

Check­ing the answer#

Expand­ing along the first row instead: $1(0 - 4) - (-1)(15 + 2) + 2(12 - 0) = -4 + 17 + 24 = 37$ ✓.

Answer#

$37$

Com­mon mis­take to avoid#

The first entry of the sec­ond row car­ries a minus sign, not a plus. Start­ing with $+$ in every row is a fre­quent error.

Ques­tion 3: Solv­ing an equa­tion of deter­mi­nants#

The prob­lem#

Find $x$ if $\begin{vmatrix} x & 4 \\ 3 & x \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ -1 & 5 \end{vmatrix}$.

Under­stand­ing the prob­lem#

Each side is a num­ber once eval­u­ated. Set­ting them equal gives an equa­tion in $x$.

The idea#

Expand both $2 \times 2$ deter­mi­nants, then solve the result­ing qua­dratic.

Step-by-step solu­tion#

Step 1. Left side.

$$x \cdot x - 4 \cdot 3 = x^2 - 12.$$

Step 2. Right side.

$$2 \cdot 5 - 3 \cdot (-1) = 10 + 3 = 13.$$

Step 3. Equate and solve.

$$x^2 - 12 = 13 \quad\Longrightarrow\quad x^2 = 25 \quad\Longrightarrow\quad x = \pm 5.$$

Check­ing the answer#

$x = 5$: $25 - 12 = 13$ ✓. $x = -5$: $25 - 12 = 13$ ✓.

Answer#

$x = 5$ or $x = -5$

Ques­tion 4: Deter­mi­nant of $2A$ and of $A^T$#

The prob­lem#

If $A$ is $3 \times 3$ with $\lvert A \rvert = 4$, find $\lvert 2A \rvert$ and $\lvert A^T \rvert$.

Under­stand­ing the prob­lem#

You do not know the entries of $A$, only its deter­mi­nant. You must use prop­er­ties of deter­mi­nants.

The idea#

For a matrix of order $n$, $\lvert kA \rvert = k^n\lvert A \rvert$ (each of the $n$ rows is mul­ti­plied by $k$). Also $\lvert A^T \rvert = \lvert A \rvert$.

Step-by-step solu­tion#

Step 1. Here $n = 3$ and $k = 2$.

$$\lvert 2A \rvert = 2^3\lvert A \rvert = 8 \times 4 = 32.$$

Step 2. Trans­pos­ing does not change a deter­mi­nant.

$$\lvert A^T \rvert = \lvert A \rvert = 4.$$

Check­ing the answer#

Try $A = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$, which has $\lvert A \rvert = 4$. Then $2A$ has diag­o­nal $8, 2, 2$ and deter­mi­nant $32$ ✓.

Answer#

$\lvert 2A \rvert = 32$; $\lvert A^T \rvert = 4$.

Com­mon mis­take to avoid#

$\lvert 2A \rvert$ is not $2\lvert A \rvert = 8$. The fac­tor $2$ comes out once from each of the three rows.

Ques­tion 5: Area of a tri­an­gle#

The prob­lem#

Find the area of the tri­an­gle $(-2, -1)$, $(3, 4)$, $(5, -2)$.

Under­stand­ing the prob­lem#

You have the three ver­tices. You need the area in square units.

The idea#

Use $\displaystyle \Delta = \tfrac{1}{2}\lvert D \rvert$, where $D = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$. Take the absolute value at the end, since an area can­not be neg­a­tive.

Step-by-step solu­tion#

Step 1. Set up the deter­mi­nant.

$$D = \begin{vmatrix} -2 & -1 & 1 \\ 3 & 4 & 1 \\ 5 & -2 & 1 \end{vmatrix}.$$

Step 2. Expand along row $1$ (signs $+\ -\ +$).

$$D = -2\begin{vmatrix} 4 & 1 \\ -2 & 1 \end{vmatrix} - (-1)\begin{vmatrix} 3 & 1 \\ 5 & 1 \end{vmatrix} + 1\begin{vmatrix} 3 & 4 \\ 5 & -2 \end{vmatrix}.$$

Step 3. Eval­u­ate the three minors.

$$\begin{vmatrix} 4 & 1 \\ -2 & 1 \end{vmatrix} = 4 + 2 = 6, \quad \begin{vmatrix} 3 & 1 \\ 5 & 1 \end{vmatrix} = 3 - 5 = -2, \quad \begin{vmatrix} 3 & 4 \\ 5 & -2 \end{vmatrix} = -6 - 20 = -26.$$

Step 4. Com­bine.

$$D = -2(6) + 1(-2) + 1(-26) = -12 - 2 - 26 = -40.$$

Step 5. Area.

$$\displaystyle \Delta = \frac{1}{2}\lvert -40 \rvert = 20.$$

Check­ing the answer#

The neg­a­tive sign of $D$ only means the ver­tices were listed clock­wise; the area is still pos­i­tive. A rough sketch shows a tri­an­gle span­ning about $7$ units across and $6$ up, so an area of $20$ is sen­si­ble.

Answer#

Area $= 20$ square units.

Ques­tion 6: Prov­ing three points are collinear#

The prob­lem#

Show that $(1, -1)$, $(3, 5)$, $(-1, -7)$ are collinear.

Under­stand­ing the prob­lem#

"Collinear" means the three points lie on one straight line. You must prove it.

The idea#

Three points are collinear exactly when the "tri­an­gle" they form has zero area, i.e. $D = 0$.

Step-by-step solu­tion#

Step 1. Set up the deter­mi­nant.

$$D = \begin{vmatrix} 1 & -1 & 1 \\ 3 & 5 & 1 \\ -1 & -7 & 1 \end{vmatrix}.$$

Step 2. Expand along row $1$.

$$D = 1\begin{vmatrix} 5 & 1 \\ -7 & 1 \end{vmatrix} - (-1)\begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} + 1\begin{vmatrix} 3 & 5 \\ -1 & -7 \end{vmatrix}.$$

Step 3. Eval­u­ate the minors.

$$5 + 7 = 12, \qquad 3 + 1 = 4, \qquad -21 + 5 = -16.$$

Step 4. Com­bine.

$$D = 12 + 4 - 16 = 0.$$

Step 5. Since $D = 0$, the area is $0$, so the points are collinear.

Check­ing the answer#

Slopes: from $(1, -1)$ to $(3, 5)$ is $\displaystyle \tfrac{6}{2} = 3$; from $(1, -1)$ to $(-1, -7)$ is $\displaystyle \tfrac{-6}{-2} = 3$. Equal slopes through a com­mon point ✓.

Answer#

$D = 0$, so the area is zero and the three points are collinear.

Ques­tion 7: Equa­tion of a line using a deter­mi­nant#

The prob­lem#

Use deter­mi­nants to find the line through $(-1, 4)$ and $(3, -2)$.

Under­stand­ing the prob­lem#

Any point $(x, y)$ on the line is collinear with the two given points. That con­di­tion, writ­ten as a deter­mi­nant, is the equa­tion of the line.

The idea#

Set $\begin{vmatrix} x & y & 1 \\ -1 & 4 & 1 \\ 3 & -2 & 1 \end{vmatrix} = 0$ and expand.

Step-by-step solu­tion#

Step 1. Expand along row $1$.

$$x\begin{vmatrix} 4 & 1 \\ -2 & 1 \end{vmatrix} - y\begin{vmatrix} -1 & 1 \\ 3 & 1 \end{vmatrix} + 1\begin{vmatrix} -1 & 4 \\ 3 & -2 \end{vmatrix} = 0.$$

Step 2. Eval­u­ate the minors.

$$4 + 2 = 6, \qquad -1 - 3 = -4, \qquad 2 - 12 = -10.$$

Step 3. Sub­sti­tute.

$$6x - y(-4) - 10 = 0 \quad\Longrightarrow\quad 6x + 4y - 10 = 0.$$

Step 4. Divide by $2$.

$$3x + 2y - 5 = 0.$$

Check­ing the answer#

$(-1, 4)$: $-3 + 8 - 5 = 0$ ✓. $(3, -2)$: $9 - 4 - 5 = 0$ ✓.

Answer#

$3x + 2y - 5 = 0$

Ques­tion 8: Minors and cofac­tors of a row#

The prob­lem#

Find the minors and cofac­tors of the sec­ond row of $\begin{vmatrix} 2 & -1 & 0 \\ 1 & 3 & 5 \\ 4 & 0 & -2 \end{vmatrix}$ and eval­u­ate the deter­mi­nant with them.

Under­stand­ing the prob­lem#

The sec­ond row is $(1, 3, 5)$. For each of these three entries you must find the minor $M_{2j}$ and the cofac­tor $A_{2j} = (-1)^{2+j}M_{2j}$. Then use "entries times their own cofac­tors, added" to get the deter­mi­nant.

The idea#

Delete row $2$ and the rel­e­vant col­umn to get each minor. The signs for row $2$ are $-\ +\ -$.

Step-by-step solu­tion#

Step 1. $M_{21}$: delete row 2, col­umn 1.

$$M_{21} = \begin{vmatrix} -1 & 0 \\ 0 & -2 \end{vmatrix} = 2 - 0 = 2, \qquad A_{21} = (-1)^3(2) = -2.$$

Step 2. $M_{22}$: delete row 2, col­umn 2.

$$M_{22} = \begin{vmatrix} 2 & 0 \\ 4 & -2 \end{vmatrix} = -4 - 0 = -4, \qquad A_{22} = (-1)^4(-4) = -4.$$

Step 3. $M_{23}$: delete row 2, col­umn 3.

$$M_{23} = \begin{vmatrix} 2 & -1 \\ 4 & 0 \end{vmatrix} = 0 - (-4) = 4, \qquad A_{23} = (-1)^5(4) = -4.$$

Step 4. Deter­mi­nant $= a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$.

$$\Delta = 1(-2) + 3(-4) + 5(-4) = -2 - 12 - 20 = -34.$$

Check­ing the answer#

Expand­ing along row $1$: $2(-6 - 0) - (-1)(-2 - 20) + 0 = -12 - 22 = -34$ ✓.

Answer#

Minors $M_{21} = 2$, $M_{22} = -4$, $M_{23} = 4$; cofac­tors $A_{21} = -2$, $A_{22} = -4$, $A_{23} = -4$; deter­mi­nant $= -34$.

Ques­tion 9: A deter­mi­nant that van­ishes with­out expand­ing#

The prob­lem#

Show with­out expand­ing that $\begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{vmatrix} = 0$ (add col­umn $2$ to col­umn $3$).

Under­stand­ing the prob­lem#

You must prove the deter­mi­nant is $0$ using prop­er­ties, not by mul­ti­ply­ing every­thing out. The hint tells you the first move.

The idea#

Adding one col­umn to another does not change the value of a deter­mi­nant. After the oper­a­tion, col­umn $3$ becomes a mul­ti­ple of col­umn $1$, and a deter­mi­nant with pro­por­tional (or equal) columns is $0$.

Step-by-step solu­tion#

Step 1. Replace col­umn 3 by col­umn 3 + col­umn 2 ($C_3 \to C_3 + C_2$). The value is unchanged.

$$\begin{vmatrix} 1 & a & a + b + c \\ 1 & b & a + b + c \\ 1 & c & a + b + c \end{vmatrix}.$$

Step 2. Every entry of col­umn 3 is now $a + b + c$. Take this com­mon fac­tor out of col­umn 3.

$$(a + b + c)\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix}.$$

Step 3. Now col­umn 1 and col­umn 3 are iden­ti­cal. A deter­mi­nant with two equal columns is $0$ (just as for two equal rows, since $\lvert A^T \rvert = \lvert A \rvert$).

$$(a + b + c) \times 0 = 0.$$

Check­ing the answer#

Try $a = 1$, $b = 2$, $c = 3$: rows $(1, 1, 5)$, $(1, 2, 4)$, $(1, 3, 3)$. Expand­ing: $1(6 - 12) - 1(3 - 4) + 5(3 - 2) = -6 + 1 + 5 = 0$ ✓.

Answer#

After $C_3 \to C_3 + C_2$, col­umn 3 is $(a + b + c)$ times col­umn 1, so the deter­mi­nant is $0$.

Ques­tion 10: Find­ing $k$ from a given area#

The prob­lem#

Find $k$ if the area of the tri­an­gle $(1, k)$, $(4, 0)$, $(0, 2)$ is $5$.

Under­stand­ing the prob­lem#

The area is known but one coor­di­nate is unknown. Because the area for­mula uses an absolute value, expect two pos­si­ble answers.

The idea#

Com­pute $D$ in terms of $k$, then solve $\displaystyle \tfrac{1}{2}\lvert D \rvert = 5$, i.e. $D = 10$ or $D = -10$.

Step-by-step solu­tion#

Step 1. Set up and expand along row 1.

$$D = \begin{vmatrix} 1 & k & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = 1(0 - 2) - k(4 - 0) + 1(8 - 0) = -2 - 4k + 8 = 6 - 4k.$$

Step 2. Area con­di­tion.

$$\displaystyle \frac{1}{2}\lvert 6 - 4k \rvert = 5 \quad\Longrightarrow\quad \lvert 6 - 4k \rvert = 10.$$

Step 3. Solve both cases.

$$\begin{aligned} 6 - 4k = 10 &\quad\Longrightarrow\quad -4k = 4 \quad\Longrightarrow\quad k = -1,\\ 6 - 4k = -10 &\quad\Longrightarrow\quad -4k = -16 \quad\Longrightarrow\quad k = 4. \end{aligned}$$

Check­ing the answer#

$k = -1$: $D = 6 + 4 = 10$, area $5$ ✓. $k = 4$: $D = 6 - 16 = -10$, area $5$ ✓.

Answer#

$k = -1$ or $k = 4$

Com­mon mis­take to avoid#

Solv­ing only $6 - 4k = 10$ loses the sec­ond answer. The absolute value always gives two cases.