Adjoint, Inverse and Systems of Equations — Worked Solutions

Ques­tion 1: Adjoint and inverse of a 2 × 2 matrix#

The prob­lem#

Find $\operatorname{adj}A$ and $A^{-1}$ for $A = \begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix}$.

Under­stand­ing the prob­lem#

You need two things: the adjoint, and the inverse, which exists only if $\lvert A \rvert \ne 0$.

The idea#

For a $2 \times 2$ matrix, the adjoint is found by swap­ping the diag­o­nal entries and chang­ing the signs of the other two. Then $\displaystyle A^{-1} = \frac{1}{\lvert A \rvert}\operatorname{adj}A$.

Step-by-step solu­tion#

Step 1. Deter­mi­nant.

$$\lvert A \rvert = 4(2) - (-1)(3) = 8 + 3 = 11 \ne 0$$

so $A$ is invert­ible.

Step 2. Adjoint: swap $4$ and $2$; change the signs of $-1$ and $3$.

$$\operatorname{adj}A = \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}$$

Step 3. Inverse.

$$\displaystyle A^{-1} = \frac{1}{11}\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}$$

Check­ing the answer#

$$A\,(\operatorname{adj}A) = \begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 8 + 3 & 4 - 4 \\ 6 - 6 & 3 + 8 \end{bmatrix} = \begin{bmatrix} 11 & 0 \\ 0 & 11 \end{bmatrix} = 11I \;✓$$

Answer#

$\operatorname{adj}A = \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}$, $\displaystyle A^{-1} = \frac{1}{11}\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}$.

Ques­tion 2: Inverse of a 3 × 3 matrix#

The prob­lem#

Find $A^{-1}$ for $A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 3 \end{bmatrix}$ and ver­ify $AA^{-1} = I$.

Under­stand­ing the prob­lem#

For a $3 \times 3$ matrix you need all nine cofac­tors. The cofac­tor $A_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor: the deter­mi­nant left after delet­ing row $i$ and col­umn $j$.

The idea#

Find the cofac­tors, use them to find $\lvert A \rvert$ (expand­ing along row 1), trans­pose the cofac­tor matrix to get $\operatorname{adj}A$, and divide by $\lvert A \rvert$.

Step-by-step solu­tion#

Step 1. Cofac­tors of row 1.

$$A_{11} = +\begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = 3, \quad A_{12} = -\begin{vmatrix} 0 & 1 \\ 1 & 3 \end{vmatrix} = -(0 - 1) = 1, \quad A_{13} = +\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -1$$

Step 2. Cofac­tors of row 2.

$$A_{21} = -\begin{vmatrix} 1 & 0 \\ 0 & 3 \end{vmatrix} = -3, \quad A_{22} = +\begin{vmatrix} 2 & 0 \\ 1 & 3 \end{vmatrix} = 6, \quad A_{23} = -\begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = -(0 - 1) = 1$$

Step 3. Cofac­tors of row 3.

$$A_{31} = +\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1, \quad A_{32} = -\begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = -2, \quad A_{33} = +\begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} = 2$$

Step 4. Deter­mi­nant, expand­ing along row 1.

$$\lvert A \rvert = 2(3) + 1(1) + 0(-1) = 7$$

Step 5. Adjoint = trans­pose of the cofac­tor matrix (rows become columns).

$$\displaystyle \operatorname{adj}A = \begin{bmatrix} 3 & -3 & 1 \\ 1 & 6 & -2 \\ -1 & 1 & 2 \end{bmatrix}, \qquad A^{-1} = \frac17\begin{bmatrix} 3 & -3 & 1 \\ 1 & 6 & -2 \\ -1 & 1 & 2 \end{bmatrix}$$

Step 6. Ver­ify $AA^{-1} = I$.

$$\displaystyle \frac17\begin{bmatrix} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 3 & -3 & 1 \\ 1 & 6 & -2 \\ -1 & 1 & 2 \end{bmatrix} = \frac17\begin{bmatrix} 6 + 1 & -6 + 6 & 2 - 2 \\ 1 - 1 & 6 + 1 & -2 + 2 \\ 3 - 3 & -3 + 3 & 1 + 6 \end{bmatrix} = \frac17\begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{bmatrix} = I$$

Check­ing the answer#

Step 6 is itself the check: the prod­uct is the iden­tity. ✓

Answer#

$\displaystyle A^{-1} = \frac17\begin{bmatrix} 3 & -3 & 1 \\ 1 & 6 & -2 \\ -1 & 1 & 2 \end{bmatrix}$, and $AA^{-1} = I$.

Com­mon mis­take to avoid#

For­get­ting to trans­pose: the cofac­tors of row 1 ($3, 1, -1$) become col­umn 1 of the adjoint.

Ques­tion 3: Ver­i­fy­ing A(adj A) = |A| I#

The prob­lem#

Ver­ify $A(\operatorname{adj}A) = \lvert A \rvert I$ for $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & 1 \\ 3 & 0 & -1 \end{bmatrix}$.

Under­stand­ing the prob­lem#

"Ver­ify" means cal­cu­late both sides sep­a­rately and show they are equal.

The idea#

Find all cofac­tors, form $\operatorname{adj}A$, find $\lvert A \rvert$, then mul­ti­ply $A$ by $\operatorname{adj}A$.

Step-by-step solu­tion#

Step 1. Cofac­tors.

$$\begin{aligned} A_{11} &= +(2 \cdot (-1) - 1 \cdot 0) = -2, & A_{12} &= -(0 \cdot (-1) - 1 \cdot 3) = 3, & A_{13} &= +(0 - 6) = -6 \\ A_{21} &= -((-1)(-1) - 2 \cdot 0) = -1, & A_{22} &= +(-1 - 6) = -7, & A_{23} &= -(0 - (-3)) = -3 \\ A_{31} &= +((-1)(1) - 2 \cdot 2) = -5, & A_{32} &= -(1 - 0) = -1, & A_{33} &= +(2 - 0) = 2 \end{aligned}$$

Step 2. Deter­mi­nant along row 1.

$$\lvert A \rvert = 1(-2) + (-1)(3) + 2(-6) = -2 - 3 - 12 = -17$$

Step 3. Adjoint (trans­pose of cofac­tors).

$$\operatorname{adj}A = \begin{bmatrix} -2 & -1 & -5 \\ 3 & -7 & -1 \\ -6 & -3 & 2 \end{bmatrix}$$

Step 4. Mul­ti­ply.

$$A(\operatorname{adj}A) = \begin{bmatrix} -2 - 3 - 12 & -1 + 7 - 6 & -5 + 1 + 4 \\ 0 + 6 - 6 & 0 - 14 - 3 & 0 - 2 + 2 \\ -6 + 0 + 6 & -3 + 0 + 3 & -15 + 0 - 2 \end{bmatrix} = \begin{bmatrix} -17 & 0 & 0 \\ 0 & -17 & 0 \\ 0 & 0 & -17 \end{bmatrix}$$

Step 5. This is $-17I = \lvert A \rvert I$, as required.

Check­ing the answer#

The off-diag­o­nal zeros are expected: a row times the cofac­tors of a dif­fer­ent row always gives $0$. ✓

Answer#

$\lvert A \rvert = -17$ and $A(\operatorname{adj}A) = -17I = \lvert A \rvert I$.

Ques­tion 4: Which matri­ces are sin­gu­lar?#

The prob­lem#

Which are sin­gu­lar? (a) $\begin{bmatrix} 3 & 6 \\ 2 & 4 \end{bmatrix}$; (b) $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$; (c) $\begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix}$.

Under­stand­ing the prob­lem#

A square matrix is sin­gu­lar when its deter­mi­nant is $0$ (it has no inverse).

The idea#

Eval­u­ate each deter­mi­nant.

Step-by-step solu­tion#

Part (a)

$$3(4) - 6(2) = 12 - 12 = 0 \quad\Rightarrow\quad \text{singular}$$

Part (b)

Step 1. Expand along row 1.

$$1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9 = 0$$

So it is sin­gu­lar. (Also, row 2 $-$ row 1 $= (3, 3, 3) =$ row 3 $-$ row 2, so the rows are depen­dent.)

Part (c)

Step 1. Expand along row 1.

$$2(1 - 0) - 0(1 - 0) + 1(2 - 0) = 2 + 2 = 4 \ne 0 \quad\Rightarrow\quad \text{non-singular}$$

Check­ing the answer#

In (a) row 1 is $\displaystyle \tfrac32 \times$ row 2, so a zero deter­mi­nant is expected. ✓

Answer#

(a) and (b) are sin­gu­lar; (c) is non-sin­gu­lar ($\lvert A \rvert = 4$).

Ques­tion 5: Con­sis­tent or incon­sis­tent?#

The prob­lem#

Con­sis­tent or incon­sis­tent? (a) $x + 3y = 4$, $2x + 6y = 9$; (b) $3x - y = 2$, $6x - 2y = 4$.

Under­stand­ing the prob­lem#

A sys­tem is con­sis­tent if it has at least one solu­tion, incon­sis­tent if it has none.

The idea#

Write each as $AX = B$. If $\lvert A \rvert = 0$, com­pute $(\operatorname{adj}A)B$: if it is not $O$, the sys­tem is incon­sis­tent; if it is $O$, check fur­ther.

Step-by-step solu­tion#

Part (a)

Step 1. $A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$, $B = \begin{bmatrix} 4 \\ 9 \end{bmatrix}$; $\lvert A \rvert = 6 - 6 = 0$.

Step 2. $\operatorname{adj}A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$.

$$(\operatorname{adj}A)B = \begin{bmatrix} 24 - 27 \\ -8 + 9 \end{bmatrix} = \begin{bmatrix} -3 \\ 1 \end{bmatrix} \ne O$$

So the sys­tem is incon­sis­tent.

Part (b)

Step 1. $A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$, $B = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$; $\lvert A \rvert = -6 + 6 = 0$.

Step 2. $\operatorname{adj}A = \begin{bmatrix} -2 & 1 \\ -6 & 3 \end{bmatrix}$.

$$(\operatorname{adj}A)B = \begin{bmatrix} -4 + 4 \\ -12 + 12 \end{bmatrix} = O$$

Step 3. Check fur­ther: the sec­ond equa­tion is exactly twice the first, so both describe the same line $y = 3x - 2$. Every point on it is a solu­tion.

Check­ing the answer#

(a) Dou­bling the first equa­tion gives $2x + 6y = 8$, which con­tra­dicts $2x + 6y = 9$: par­al­lel lines. ✓

Answer#

(a) Incon­sis­tent (no solu­tion); (b) con­sis­tent with infi­nitely many solu­tions (the same line).

Ques­tion 6: Solv­ing a 2 × 2 sys­tem by matri­ces#

The prob­lem#

Solve by matri­ces: $3x - 2y = 7$, $5x + 4y = 19$.

Under­stand­ing the prob­lem#

Write the sys­tem as $AX = B$ and solve with $X = A^{-1}B$.

The idea#

$A = \begin{bmatrix} 3 & -2 \\ 5 & 4 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 7 \\ 19 \end{bmatrix}$.

Step-by-step solu­tion#

Step 1. $\lvert A \rvert = 12 + 10 = 22 \ne 0$, so there is a unique solu­tion.

Step 2. $\operatorname{adj}A = \begin{bmatrix} 4 & 2 \\ -5 & 3 \end{bmatrix}$.

Step 3. Mul­ti­ply.

$$\displaystyle X = \frac{1}{22}\begin{bmatrix} 4 & 2 \\ -5 & 3 \end{bmatrix}\begin{bmatrix} 7 \\ 19 \end{bmatrix} = \frac{1}{22}\begin{bmatrix} 28 + 38 \\ -35 + 57 \end{bmatrix} = \frac{1}{22}\begin{bmatrix} 66 \\ 22 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$$

Check­ing the answer#

$3(3) - 2(1) = 7$ ✓ and $5(3) + 4(1) = 19$ ✓.

Answer#

$x = 3$, $y = 1$.

Ques­tion 7: Solv­ing a 3 × 3 sys­tem by matri­ces#

The prob­lem#

Solve by matri­ces: $2x + y + z = 7$, $x - y + 2z = 5$, $3x + 2y - z = 6$.

Under­stand­ing the prob­lem#

Three equa­tions, three unknowns. Write $AX = B$ with

$$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 5 \\ 6 \end{bmatrix}.$$

The idea#

If $\lvert A \rvert \ne 0$, then $\displaystyle X = A^{-1}B = \frac{1}{\lvert A \rvert}(\operatorname{adj}A)B$.

Step-by-step solu­tion#

Step 1. Cofac­tors.

$$\begin{aligned} A_{11} &= (1 - 4) = -3, & A_{12} &= -(-1 - 6) = 7, & A_{13} &= (2 + 3) = 5 \\ A_{21} &= -(-1 - 2) = 3, & A_{22} &= (-2 - 3) = -5, & A_{23} &= -(4 - 3) = -1 \\ A_{31} &= (2 + 1) = 3, & A_{32} &= -(4 - 1) = -3, & A_{33} &= (-2 - 1) = -3 \end{aligned}$$

Step 2. Deter­mi­nant along row 1.

$$\lvert A \rvert = 2(-3) + 1(7) + 1(5) = 6 \ne 0$$

Step 3. Adjoint.

$$\operatorname{adj}A = \begin{bmatrix} -3 & 3 & 3 \\ 7 & -5 & -3 \\ 5 & -1 & -3 \end{bmatrix}$$

Step 4. Mul­ti­ply.

$$\displaystyle X = \frac16\begin{bmatrix} -21 + 15 + 18 \\ 49 - 25 - 18 \\ 35 - 5 - 18 \end{bmatrix} = \frac16\begin{bmatrix} 12 \\ 6 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}$$

Check­ing the answer#

$2(2) + 1 + 2 = 7$ ✓; $2 - 1 + 4 = 5$ ✓; $6 + 2 - 2 = 6$ ✓.

Answer#

$x = 2$, $y = 1$, $z = 2$.

Ques­tion 8: The rever­sal law for inverses#

The prob­lem#

For $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$, ver­ify $(AB)^{-1} = B^{-1}A^{-1}$.

Under­stand­ing the prob­lem#

Com­pute the left side (mul­ti­ply first, then invert) and the right side (invert each, then mul­ti­ply in the reverse order), and show they are equal.

The idea#

Use the $2 \times 2$ inverse rule each time.

Step-by-step solu­tion#

Step 1. Prod­uct $AB$.

$$AB = \begin{bmatrix} 2 - 3 & -4 + 9 \\ 1 + 4 & -2 - 12 \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}$$

Step 2. Its inverse: $\lvert AB \rvert = 14 - 25 = -11$.

$$\displaystyle (AB)^{-1} = \frac{1}{-11}\begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$$

Step 3. Inverses of $A$ and $B$: $\lvert A \rvert = -8 - 3 = -11$, $\lvert B \rvert = 3 - 2 = 1$.

$$\displaystyle A^{-1} = \frac{1}{-11}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}, \qquad B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$$

Step 4. Mul­ti­ply $B^{-1}A^{-1}$ in that order.

$$\displaystyle B^{-1}A^{-1} = \frac{1}{-11}\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{-11}\begin{bmatrix} -12 - 2 & -9 + 4 \\ -4 - 1 & -3 + 2 \end{bmatrix} = \frac{1}{-11}\begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$$

Step 5. Steps 2 and 4 give the same matrix, so $(AB)^{-1} = B^{-1}A^{-1}$.

Check­ing the answer#

$\lvert AB \rvert = \lvert A \rvert\lvert B \rvert = (-11)(1) = -11$, match­ing Step 2. ✓

Answer#

Both sides equal $\displaystyle \frac{1}{-11}\begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}$.

Com­mon mis­take to avoid#

The order reverses: $(AB)^{-1} = B^{-1}A^{-1}$, not $A^{-1}B^{-1}$.

Ques­tion 9: A matrix equa­tion and the inverse from it#

The prob­lem#

If $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find $a, b$ with $A^2 + aA + bI = O$ and use it to find $A^{-1}$.

Under­stand­ing the prob­lem#

You must find num­bers $a$ and $b$ that make the matrix $A^2 + aA + bI$ the zero matrix, then use that equa­tion (not the adjoint) to get $A^{-1}$.

The idea#

Com­pute $A^2$, write $A^2 + aA + bI$ entry by entry, and set each entry to $0$. Then mul­ti­ply the equa­tion by $A^{-1}$.

Step-by-step solu­tion#

Step 1. Square $A$.

$$A^2 = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 + 2 & 6 + 2 \\ 3 + 1 & 2 + 1 \end{bmatrix} = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}$$

Step 2. Form the sum.

$$A^2 + aA + bI = \begin{bmatrix} 11 + 3a + b & 8 + 2a \\ 4 + a & 3 + a + b \end{bmatrix}$$

Step 3. Set each entry to $0$. From $4 + a = 0$: $a = -4$ (and $8 + 2a = 0$ agrees). Then $11 - 12 + b = 0$ gives $b = 1$ (and $3 - 4 + 1 = 0$ agrees).

Step 4. So $A^2 - 4A + I = O$. Since $\lvert A \rvert = 3 - 2 = 1 \ne 0$, $A^{-1}$ exists. Mul­ti­ply by $A^{-1}$.

$$A - 4I + A^{-1} = O \;\Rightarrow\; A^{-1} = 4I - A$$

Step 5. Com­pute.

$$A^{-1} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$$

Check­ing the answer#

$$AA^{-1} = \begin{bmatrix} 3 - 2 & -6 + 6 \\ 1 - 1 & -2 + 3 \end{bmatrix} = I \;✓$$

Answer#

$a = -4$, $b = 1$; $A^{-1} = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$.

Ques­tion 10: Prices of fruit from three pur­chases#

The prob­lem#

Three friends buy fruit: $2$ apples, $3$ oranges, $1$ mango cost ₹115; $1$ apple, $2$ oranges, $2$ man­goes cost ₹95; $3$ apples, $1$ orange, $2$ man­goes cost ₹130. Find the price of each fruit.

Under­stand­ing the prob­lem#

Let the prices (in rupees) be $a$ for an apple, $o$ for an orange and $m$ for a mango. Each pur­chase gives one equa­tion:

$$2a + 3o + m = 115, \qquad a + 2o + 2m = 95, \qquad 3a + o + 2m = 130.$$

The idea#

Write $AX = B$ with $A = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 2 & 2 \\ 3 & 1 & 2 \end{bmatrix}$, $X = \begin{bmatrix} a \\ o \\ m \end{bmatrix}$, $B = \begin{bmatrix} 115 \\ 95 \\ 130 \end{bmatrix}$, and use $X = A^{-1}B$.

Step-by-step solu­tion#

Step 1. Cofac­tors.

$$\begin{aligned} A_{11} &= (4 - 2) = 2, & A_{12} &= -(2 - 6) = 4, & A_{13} &= (1 - 6) = -5 \\ A_{21} &= -(6 - 1) = -5, & A_{22} &= (4 - 3) = 1, & A_{23} &= -(2 - 9) = 7 \\ A_{31} &= (6 - 2) = 4, & A_{32} &= -(4 - 1) = -3, & A_{33} &= (4 - 3) = 1 \end{aligned}$$

Step 2. Deter­mi­nant along row 1.

$$\lvert A \rvert = 2(2) + 3(4) + 1(-5) = 11$$

Step 3. Adjoint.

$$\operatorname{adj}A = \begin{bmatrix} 2 & -5 & 4 \\ 4 & 1 & -3 \\ -5 & 7 & 1 \end{bmatrix}$$

Step 4. Mul­ti­ply.

$$\displaystyle X = \frac{1}{11}\begin{bmatrix} 230 - 475 + 520 \\ 460 + 95 - 390 \\ -575 + 665 + 130 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 275 \\ 165 \\ 220 \end{bmatrix} = \begin{bmatrix} 25 \\ 15 \\ 20 \end{bmatrix}$$

Check­ing the answer#

$2(25) + 3(15) + 20 = 115$ ✓; $25 + 30 + 40 = 95$ ✓; $75 + 15 + 40 = 130$ ✓.

Answer#

Apple ₹25, orange ₹15, mango ₹20.