Questions and Answers (Board Exam)

Board Exam Presentation Strategy

When solving Determinant and Matrix problems in Board Exams, the method of presentation is very important because step marks are awarded even if the final answer is incorrect.

Define matrices clearly

If a question is about solving a system of equations by matrix method, first write the system in the form $$AX = B,$$ and clearly state what $A$, $X$, and $B$ represent.

State the formula before using it

For example, if the matrix is non-singular, explicitly write $$X = A^{-1}B$$ and if you are finding inverse, write $$A^{-1} = \frac{1}{|A|}\operatorname{adj} A.$$

Check the determinant first

Before finding the inverse, always calculate $|A|$ and state whether it is zero or non-zero. This establishes whether the matrix is singular or non-singular.

Show cofactor work in $3 \times 3$ problems

Do not jump directly to the adjoint matrix. Show at least a few cofactor calculations such as $A_{11}, A_{12}, A_{13}$, because that is where many step marks are awarded.

Write proper conclusions

After calculations, write a final statement such as:

• 'Hence, the matrix is singular.'
• 'Hence, the system is inconsistent.'
• 'Therefore, $x=2, y=1, z=3$.'

This style helps secure full marks in descriptive board-exam answers.

Question 1

Evaluate the determinant: $$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix}$$

Solution: Using the $2 \times 2$ determinant formula, $$\Delta = (\cos 15^\circ)(\cos 75^\circ) - (\sin 15^\circ)(\sin 75^\circ).$$ Now use the identity $$\cos A \cos B - \sin A \sin B = \cos(A+B).$$ So, $$\Delta = \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0.$$

Answer: $0$

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Question 2

Find the value of $x$ if $$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}.$$

Solution: Evaluate the determinant on the left-hand side: $$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = (2)(1) - (4)(5) = 2 - 20 = -18.$$ Now evaluate the determinant on the right-hand side: $$\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} = (2x)(x) - (4)(6) = 2x^2 - 24.$$ Equating both values, $$-18 = 2x^2 - 24.$$ So, $$2x^2 = 6 \Rightarrow x^2 = 3.$$ Hence, $$x = \pm \sqrt{3}.$$

Answer: $x = \pm \sqrt{3}$

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Question 3

For what value of $k$ is the matrix $$A = \begin{bmatrix} k & 8 \\ 4 & 2k \end{bmatrix}$$ a singular matrix?

Solution: A matrix is singular if its determinant is zero. Therefore, $$|A| = 0.$$ Now, $$\begin{vmatrix} k & 8 \\ 4 & 2k \end{vmatrix} = k(2k) - 8(4) = 2k^2 - 32.$$ Set this equal to zero: $$2k^2 - 32 = 0.$$ So, $$2k^2 = 32 \Rightarrow k^2 = 16.$$ Hence, $$k = \pm 4.$$

Answer: $k = 4, -4$

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Question 4

If $A$ is a square matrix of order 3 and $|A| = 5$, find $|3A|$.

Solution: For a matrix of order $n$, $$|kA| = k^n|A|.$$ Here, $k=3$ and $n=3$. Therefore, $$|3A| = 3^3|A| = 27 \times 5 = 135.$$

Answer: $135$

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Question 5

Write the value of the determinant $$\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix}.$$

Solution: Apply the row operation $$R_1 \to R_1 + R_2 + R_3.$$ Then each element of the first row becomes $$(a-b)+(b-c)+(c-a)=0,$$ $$(b-c)+(c-a)+(a-b)=0,$$ $$(c-a)+(a-b)+(b-c)=0.$$ So the determinant becomes $$\begin{vmatrix} 0 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix}.$$ A determinant with an entire row of zeros is zero.

Answer: $0$

Question 6

Find the maximum value of $$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{vmatrix}.$$

Solution: Apply the column operations $$C_2 \to C_2 - C_1, \qquad C_3 \to C_3 - C_1.$$ Then the determinant becomes $$\begin{vmatrix} 1 & 0 & 0 \\ 1 & \sin\theta & 0 \\ 1 & 0 & \cos\theta \end{vmatrix}.$$ Now expand along the first row: $$\Delta = 1 \cdot \begin{vmatrix} \sin\theta & 0 \\ 0 & \cos\theta \end{vmatrix} = \sin\theta \cos\theta.$$ Use the identity $$2\sin\theta\cos\theta = \sin 2\theta.$$ So, $$\Delta = \frac{1}{2}\sin 2\theta.$$ The maximum value of $\sin 2\theta$ is 1. Hence the maximum value of the determinant is $$\frac{1}{2}.$$

Answer: $\frac{1}{2}$

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Question 7

Find the area of the triangle whose vertices are $(3, 8)$, $(-4, 2)$, and $(5, -1)$.

Solution: The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is $$\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|.$$ Substituting the given points, $$\text{Area} = \frac{1}{2}\left|\begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & -1 & 1 \end{vmatrix}\right|.$$ Expand along the first row: $$= \frac{1}{2}\left| 3\begin{vmatrix} 2 & 1 \\ -1 & 1 \end{vmatrix} - 8\begin{vmatrix} -4 & 1 \\ 5 & 1 \end{vmatrix} + 1\begin{vmatrix} -4 & 2 \\ 5 & -1 \end{vmatrix} \right|.$$ Now, $$\begin{vmatrix} 2 & 1 \\ -1 & 1 \end{vmatrix} = 2(1)-1(-1)=3,$$ $$\begin{vmatrix} -4 & 1 \\ 5 & 1 \end{vmatrix} = -4(1)-1(5)=-9,$$ $$\begin{vmatrix} -4 & 2 \\ 5 & -1 \end{vmatrix} = (-4)(-1)-2(5)=4-10=-6.$$ Therefore, $$\text{Area} = \frac{1}{2}|3(3)-8(-9)-6| = \frac{1}{2}|9+72-6| = \frac{1}{2}(75) = 37.5.$$

Answer: $37.5$ sq. units

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Question 8

Find the value of $k$ if the area of the triangle is 35 sq. units and vertices are $(2, -6)$, $(5, 4)$, and $(k, 4)$.

Solution: Using the area formula, $$\frac{1}{2}\left|\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix}\right| = 35.$$ So, $$\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 70.$$ Expand along the first row: $$2\begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} - (-6)\begin{vmatrix} 5 & 1 \\ k & 1 \end{vmatrix} + 1\begin{vmatrix} 5 & 4 \\ k & 4 \end{vmatrix} = \pm 70.$$ Now, $$\begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} = 0,$$ $$\begin{vmatrix} 5 & 1 \\ k & 1 \end{vmatrix} = 5-k,$$ $$\begin{vmatrix} 5 & 4 \\ k & 4 \end{vmatrix} = 20-4k.$$ Thus, $$0 + 6(5-k) + (20-4k) = \pm 70.$$ So, $$30-6k+20-4k = \pm 70,$$ $$50-10k = \pm 70.$$ Case 1: $$50-10k=70 \Rightarrow -10k=20 \Rightarrow k=-2.$$ Case 2: $$50-10k=-70 \Rightarrow -10k=-120 \Rightarrow k=12.$$

Answer: $k = 12, -2$

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Question 9

Using determinants, show that the points $A(a, 0)$, $B(0, b)$, and $C(x, y)$ are collinear if $$\frac{x}{a} + \frac{y}{b} = 1.$$

Solution: For collinearity of the three points, the determinant $$\begin{vmatrix} a & 0 & 1 \\ 0 & b & 1 \\ x & y & 1 \end{vmatrix}$$ must be zero. Expand along the first row: $$a\begin{vmatrix} b & 1 \\ y & 1 \end{vmatrix} - 0 + 1\begin{vmatrix} 0 & b \\ x & y \end{vmatrix} = 0.$$ So, $$a(b-y) + (0-byx? )$$ Carefully evaluating the last minor, $$\begin{vmatrix} 0 & b \\ x & y \end{vmatrix} = 0\cdot y - bx = -bx.$$ Hence, $$a(b-y) - bx = 0,$$ $$ab - ay - bx = 0,$$ $$bx + ay = ab.$$ Dividing by $ab$, $$\frac{x}{a} + \frac{y}{b} = 1.$$ Thus the given condition implies the determinant is zero, so the points are collinear.

Answer: Proved.

Question 10

Find the equation of the line joining $A(1, 2)$ and $B(3, 6)$ using determinants.

Solution: Let $P(x,y)$ be any point on the required line. Then the three points are collinear, so $$\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0.$$ Expand along the first row: $$x\begin{vmatrix} 2 & 1 \\ 6 & 1 \end{vmatrix} - y\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} + 1\begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = 0.$$ Now, $$\begin{vmatrix} 2 & 1 \\ 6 & 1 \end{vmatrix} = 2-6=-4,$$ $$\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = 1-3=-2,$$ $$\begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = 6-6=0.$$ Hence, $$-4x - y(-2) = 0 \Rightarrow -4x + 2y = 0.$$ Therefore, $$y = 2x.$$

Answer: $y = 2x$

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Question 11

Write the cofactor of $a_{21}$ in the matrix $$A = \begin{bmatrix} 1 & 2 & 3 \\ -4 & 3 & 6 \\ 2 & -7 & 9 \end{bmatrix}.$$

Solution: The element $a_{21}$ is $-4$. Delete row 2 and column 1 to get its minor: $$M_{21} = \begin{vmatrix} 2 & 3 \\ -7 & 9 \end{vmatrix} = 2(9) - 3(-7) = 18 + 21 = 39.$$ Now, $$A_{21} = (-1)^{2+1}M_{21} = (-1)^3(39) = -39.$$

Answer: $-39$

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Question 12

If $A$ is a square matrix of order 3 and $|A| = 4$, write the value of $|\operatorname{adj} A|$.

Solution: For a non-singular matrix of order $n$, $$|\operatorname{adj} A| = |A|^{n-1}.$$ Here $n=3$ and $|A|=4$. Therefore, $$|\operatorname{adj} A| = 4^{3-1} = 4^2 = 16.$$

Answer: $16$

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Question 13

If $A$ is an invertible matrix of order $2 \times 2$ such that $|A| = 8$, find the value of $|A^{-1}|$.

Solution: We use the property $$|A^{-1}| = \frac{1}{|A|}.$$ Since $|A|=8$, $$|A^{-1}| = \frac{1}{8}.$$

Answer: $\frac{1}{8}$

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Question 14

Find the adjoint of matrix $$A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}.$$

Solution: For a $2 \times 2$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ its adjoint is $$\operatorname{adj} A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$ Here, $$a=2, \quad b=-1, \quad c=3, \quad d=4.$$ So, $$\operatorname{adj} A = \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}.$$

Answer: $$\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$$

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Question 15

Given $$A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix},$$ find $A^{-1}$.

Solution: First find the determinant: $$|A| = 2(7) - (-3)(-4) = 14 - 12 = 2.$$ Since $|A| \neq 0$, the inverse exists. Now, $$\operatorname{adj} A = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}.$$ Therefore, $$A^{-1} = \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 7/2 & 3/2 \\ 2 & 1 \end{bmatrix}.$$

Answer: $$\begin{bmatrix} 7/2 & 3/2 \\ 2 & 1 \end{bmatrix}$$

Question 16

If $$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix},$$ show that $A^2 - 5A + 7I = O$. Hence find $A^{-1}$.

Solution: First compute $$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}.$$ Now, $$A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$ Hence, $$A^2 - 5A + 7I = O.$$ Now post-multiply by $A^{-1}$: $$A - 5I + 7A^{-1} = O.$$ So, $$7A^{-1} = 5I - A.$$ Thus, $$A^{-1} = \frac{1}{7}\left(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\right) = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}.$$

Answer: $$\frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$$

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Question 17

Solve the following system of linear equations using matrix method: $$x - y + 2z = 7$$ $$3x + 4y - 5z = -5$$ $$2x - y + 3z = 12$$

Solution: Write the system as $$AX = B,$$ where $$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \qquad B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}.$$ Now compute $$|A| = 1\begin{vmatrix} 4 & -5 \\ -1 & 3 \end{vmatrix} - (-1)\begin{vmatrix} 3 & -5 \\ 2 & 3 \end{vmatrix} + 2\begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix}.$$ Thus, $$|A| = 1(12-5) + 1(9+10) + 2(-3-8) = 7 + 19 - 22 = 4.$$ Since $|A| \neq 0$, a unique solution exists.

Cofactors are: $$A_{11}=7, \quad A_{12}=-19, \quad A_{13}=-11,$$ $$A_{21}=1, \quad A_{22}=-1, \quad A_{23}=-1,$$ $$A_{31}=-3, \quad A_{32}=11, \quad A_{33}=7.$$ Therefore, $$\operatorname{adj} A = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}.$$ Now, $$X = A^{-1}B = \frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}.$$ Multiply: $$= \frac{1}{4}\begin{bmatrix} 49 - 5 - 36 \\ -133 + 5 + 132 \\ -77 + 5 + 84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}.$$ Hence, $$x=2, \quad y=1, \quad z=3.$$

Answer: $x=2, y=1, z=3$

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Question 18

If $$A = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 2 & 2 \\ -3 & 1 & -1 \end{bmatrix},$$ find $A^{-1}$. Using $A^{-1}$, solve the system $$2x + y - 3z = 13,$$ $$3x + 2y + z = 4,$$ $$x + 2y - z = 8.$$

Solution: Notice that the coefficient matrix of the system is $$A^T = \begin{bmatrix} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{bmatrix}.$$ So the system is $$A^T X = B,$$ where $$B = \begin{bmatrix} 13 \\ 4 \\ 8 \end{bmatrix}.$$ Therefore, $$X = (A^T)^{-1}B = (A^{-1})^T B.$$ First compute $A^{-1}$.

Determinant of $A$: $$|A| = 2\begin{vmatrix} 2 & 2 \\ 1 & -1 \end{vmatrix} - 3\begin{vmatrix} 1 & 2 \\ -3 & -1 \end{vmatrix} + 1\begin{vmatrix} 1 & 2 \\ -3 & 1 \end{vmatrix}.$$ So, $$|A| = 2(-2-2) - 3(-1+6) + (1+6) = -8 - 15 + 7 = -16.$$ Cofactors are: $$A_{11}=-4, \quad A_{12}=-5, \quad A_{13}=7,$$ $$A_{21}=4, \quad A_{22}=1, \quad A_{23}=-11,$$ $$A_{31}=4, \quad A_{32}=-3, \quad A_{33}=1.$$ Thus, $$\operatorname{adj} A = \begin{bmatrix} -4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1 \end{bmatrix}.$$ Hence, $$A^{-1} = \frac{1}{-16}\begin{bmatrix} -4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1 \end{bmatrix}.$$ So, $$(A^{-1})^T = \frac{1}{-16}\begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix}.$$ Now, $$X = (A^{-1})^T B = \frac{1}{-16}\begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix}\begin{bmatrix} 13 \\ 4 \\ 8 \end{bmatrix}.$$ Multiply: $$= \frac{1}{-16}\begin{bmatrix} -52 - 20 + 56 \\ 52 + 4 - 88 \\ 52 - 12 + 8 \end{bmatrix} = \frac{1}{-16}\begin{bmatrix} -16 \\ -32 \\ 48 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix}.$$ Thus, $$x=1, \quad y=2, \quad z=-3.$$

Answer: $x=1, y=2, z=-3$

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Question 19

Find the product $$\begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}$$ and use it to solve the system of equations: $$x - y + z = 4,$$ $$x - 2y - 2z = 9,$$ $$2x + y + 3z = 1.$$

Solution: Let $$C = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix}, \qquad D = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}.$$ Now multiply $CD$: $$CD = \begin{bmatrix} -4+7+5 & 4-1-3 & 4-3-1 \\ -4+14-10 & 4-2+6 & 4-6+2 \\ -8-7+15 & 8+1-9 & 8+3-3 \end{bmatrix} = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} = 8I.$$ Thus, $$C^{-1} = \frac{1}{8}D.$$ The system is $$CX = B,$$ where $$B = \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix}.$$ Hence, $$X = C^{-1}B = \frac{1}{8}DB.$$ Now, $$DB = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}\begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} = \begin{bmatrix} -16 + 36 + 4 \\ -28 + 9 + 3 \\ 20 - 27 - 1 \end{bmatrix} = \begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix}.$$ So, $$X = \frac{1}{8}\begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix}.$$ Therefore, $$x=3, \quad y=-2, \quad z=-1.$$

Answer: $x=3, y=-2, z=-1$

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Question 20

Solve the following system of equations: $$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4,$$ $$\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1,$$ $$\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2.$$

Solution: Let $$u = \frac{1}{x}, \quad v = \frac{1}{y}, \quad w = \frac{1}{z}.$$ Then the system becomes $$2u + 3v + 10w = 4,$$ $$4u - 6v + 5w = 1,$$ $$6u + 9v - 20w = 2.$$ So, $$A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}, \qquad X = \begin{bmatrix} u \\ v \\ w \end{bmatrix}, \qquad B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}.$$ Now, $$|A| = 2(120-45) - 3(-80-30) + 10(36+36) = 150 + 330 + 720 = 1200.$$ Hence the inverse exists. The adjoint is $$\operatorname{adj} A = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}.$$ Therefore, $$X = \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}.$$ Multiply: $$= \frac{1}{1200}\begin{bmatrix} 300+150+150 \\ 440-100+60 \\ 288+0-48 \end{bmatrix} = \frac{1}{1200}\begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix}.$$ So, $$u=\frac{1}{2}, \quad v=\frac{1}{3}, \quad w=\frac{1}{5}.$$ Hence, $$x=2, \quad y=3, \quad z=5.$$

Answer: $x=2, y=3, z=5$

Question 21

Given $$A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix},$$ find $AB$. Use this to solve the system: $$x - y = 3,$$ $$2x + 3y + 4z = 17,$$ $$y + 2z = 7.$$

Solution: First compute $AB$: $$AB = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}.$$ Carrying out multiplication, $$AB = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I.$$ Hence, $$A^{-1} = \frac{1}{6}B.$$ Now write the system as $$AX = C,$$ where $$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \qquad C = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}.$$ So, $$X = A^{-1}C = \frac{1}{6}BC.$$ Now, $$BC = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} = \begin{bmatrix} 6+34-28 \\ -12+34-28 \\ 6-17+35 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix}.$$ Thus, $$X = \frac{1}{6}\begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}.$$ So, $$x=2, \quad y=-1, \quad z=4.$$

Answer: $x=2, y=-1, z=4$

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Question 22

Check the consistency of the system: $$3x - y - 2z = 2,$$ $$2y - z = -1,$$ $$3x - 5y = 3.$$

Solution: The coefficient matrix is $$A = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}.$$ Now, $$|A| = 3\begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} - (-1)\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2)\begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}.$$ So, $$|A| = 3(0-5) + 1(0+3) + (-2)(0-6) = -15 + 3 + 12 = 0.$$ Since $|A|=0$, compute $(\operatorname{adj} A)B$. The adjoint is $$\operatorname{adj} A = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}.$$ Also, $$B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}.$$ Thus, $$(\operatorname{adj} A)B = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = \begin{bmatrix} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq O.$$ Hence the system is inconsistent.

Answer: Inconsistent

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Question 23

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find the cost of each item per kg by matrix method.

Solution: Let the cost per kg of onion, wheat, and rice be $x, y, z$ respectively. Then, $$4x + 3y + 2z = 60,$$ $$2x + 4y + 6z = 90,$$ $$6x + 2y + 3z = 70.$$ To simplify, divide the second equation by 2: $$x + 2y + 3z = 45.$$ So the equivalent system is $$4x + 3y + 2z = 60,$$ $$x + 2y + 3z = 45,$$ $$6x + 2y + 3z = 70.$$ Thus, $$A = \begin{bmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix}, \qquad B = \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix}.$$ Now, $$|A| = 4(6-6) - 3(3-18) + 2(2-12) = 0 + 45 - 20 = 25.$$ So a unique solution exists. The adjoint is $$\operatorname{adj} A = \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix}.$$ Hence, $$X = A^{-1}B = \frac{1}{25}\begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix}\begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix}.$$ Multiply: $$= \frac{1}{25}\begin{bmatrix} 0 - 225 + 350 \\ 900 + 0 - 700 \\ -600 + 450 + 350 \end{bmatrix} = \frac{1}{25}\begin{bmatrix} 125 \\ 200 \\ 200 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}.$$ Therefore, $$x=5, \quad y=8, \quad z=8.$$

Answer: Onion = ₹5/kg, Wheat = ₹8/kg, Rice = ₹8/kg

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Question 24

Determine the inverse of the matrix $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta \end{bmatrix}.$$

Solution: First find the determinant. Since the matrix is block-triangular with first row and first column simple, $$|A| = 1\cdot \begin{vmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{vmatrix} = 1\cdot(-\cos^2\theta - \sin^2\theta) = -1.$$ So the inverse exists. Now compute cofactors: $$A_{11} = -1, \quad A_{12}=0, \quad A_{13}=0,$$ $$A_{21}=0, \quad A_{22}=-\cos\theta, \quad A_{23}=-\sin\theta,$$ $$A_{31}=0, \quad A_{32}=-\sin\theta, \quad A_{33}=\cos\theta.$$ Thus, $$\operatorname{adj} A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\theta & -\sin\theta \\ 0 & -\sin\theta & \cos\theta \end{bmatrix}.$$ Therefore, $$A^{-1} = \frac{1}{-1}\operatorname{adj} A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta \\ 0 & \sin\theta & -\cos\theta \end{bmatrix}.$$ So the inverse is the matrix itself.

Answer: $A^{-1} = A$

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Question 25

Prove that $$|\operatorname{adj}(\operatorname{adj} A)| = |A|^{(n-1)^2}$$ for a non-singular matrix $A$ of order $n$.

Solution: Let $$B = \operatorname{adj} A.$$ For any non-singular matrix $X$ of order $n$, $$|\operatorname{adj} X| = |X|^{n-1}.$$ Applying this to $X=B$, $$|\operatorname{adj} B| = |B|^{n-1}.$$ Since $B = \operatorname{adj} A$, $$|\operatorname{adj}(\operatorname{adj} A)| = |\operatorname{adj} A|^{n-1}.$$ Again using $$|\operatorname{adj} A| = |A|^{n-1},$$ we get $$|\operatorname{adj}(\operatorname{adj} A)| = (|A|^{n-1})^{n-1} = |A|^{(n-1)^2}.$$ Hence proved.

Answer: Proved.