Determinants — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 4, Determinants (নিৰ্ণায়ক). It covers Exercise 4.1, Exercise 4.2, Exercise 4.3, Exercise 4.4, Exercise 4.5 and the Miscellaneous Exercise on Chapter 4, with every determinant, cofactor, adjoint and matrix inverse worked out in full so you can follow the reasoning and prepare confidently for your examination.
Summary
To every square matrix $A = [a_{ij}]$ we attach a number called its determinant, written $|A|$ or $\det A$. For order one, $|[a]| = a$; for order two, $\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} – a_{12}a_{21}$. A $3 \times 3$ determinant is evaluated by expanding along any row or column, multiplying each element by $(-1)^{i+j}$ times the $2 \times 2$ determinant left after deleting its row and column. Expansion along a line with the most zeros is quickest, and every row or column gives the same value.
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is $\Delta = \tfrac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$ (taken as an absolute value); three points are collinear exactly when this determinant is zero. The minor $M_{ij}$ of an element is the determinant obtained by deleting row $i$ and column $j$, and its cofactor is $A_{ij} = (-1)^{i+j} M_{ij}$. A determinant equals the sum of the products of the elements of any one row (or column) with their own cofactors; the same products taken with the cofactors of a different row (or column) sum to zero.
The adjoint of $A$ is the transpose of its cofactor matrix, and $A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\, I$. A matrix is singular when $|A| = 0$ and non-singular when $|A| \neq 0$; it is invertible if and only if it is non-singular, with $A^{-1} = \tfrac{1}{|A|}\operatorname{adj} A$. A system $AX = B$ has the unique solution $X = A^{-1}B$ when $|A| \neq 0$ (consistent); when $|A| = 0$ and $(\operatorname{adj} A)B \neq O$ the system has no solution (inconsistent).
Summary: ASSEB Class 12 Mathematics Chapter 4 Determinants explains determinants of order two and three, expansion by rows and columns, area of a triangle by determinant, minors and cofactors, the adjoint and inverse of a matrix, singular and non-singular matrices, and solving systems of linear equations by the matrix method, with complete worked answers to Exercise 4.1, 4.2, 4.3, 4.4, 4.5 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students.
Textbook Questions and Answers
Exercise 4.1
Evaluate the determinants in Exercises 1 and 2.
1. $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$
Answer: $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = (2)(-1) – (4)(-5) = -2 + 20 = 18$.
2. (i) $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$
Answer: $= (\cos\theta)(\cos\theta) – (-\sin\theta)(\sin\theta) = \cos^2\theta + \sin^2\theta = 1$.
(ii) $\begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix}$
Answer: $= (x^2 – x + 1)(x + 1) – (x – 1)(x + 1) = (x^3 + 1) – (x^2 – 1) = x^3 – x^2 + 2$.
3. If $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $|2A| = 4|A|$.
Answer: $|A| = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = 2 – 8 = -6$. Now $2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$, so $|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = 8 – 32 = -24$. Since $4|A| = 4(-6) = -24 = |2A|$, we have $|2A| = 4|A|$. (This is the rule $|kA| = k^n|A|$ with $n = 2$, so $|2A| = 2^2|A|$.)
4. If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, then show that $|3A| = 27|A|$.
Answer: $A$ is upper triangular, so $|A| = 1 \times 1 \times 4 = 4$. Now $3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$, so $|3A| = 3 \times 3 \times 12 = 108$. Since $27|A| = 27 \times 4 = 108 = |3A|$, we have $|3A| = 27|A|$. (This is $|kA| = k^n|A|$ with $n = 3$, so $|3A| = 3^3|A|$.)
5. Evaluate the determinants.
(i) $\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
Answer: The second row has two zeros, so expand along $R_2$. Only the term for $a_{23} = -1$ survives: $= -(-1)\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = 1 \cdot (-15 + 3) = -12$. Hence the value is $-12$.
(ii) $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$
Answer: Expanding along $R_1$, $= 3\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} – (-4)\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = 3(1 + 6) + 4(1 + 4) + 5(3 – 2) = 21 + 20 + 5 = 46$.
(iii) $\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$
Answer: Expanding along $R_1$, $= 0 – 1\begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2\begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} = -1(0 – 6) + 2(-3 – 0) = 6 – 6 = 0$.
(iv) $\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
Answer: Expanding along $R_1$, $= 2\begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} – (-1)\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2)\begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix} = 2(0 – 5) + 1(0 + 3) – 2(0 – 6) = -10 + 3 + 12 = 5$.
6. If $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $|A|$.
Answer: Expanding along $R_1$, $|A| = 1\begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} – 1\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2)\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} = 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5) = 3 + 3 – 6 = 0$.
7. Find values of $x$, if
(i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
Answer: Left side $= 2 – 20 = -18$; right side $= 2x \cdot x – 4 \cdot 6 = 2x^2 – 24$. So $2x^2 – 24 = -18 \Rightarrow 2x^2 = 6 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}$.
(ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
Answer: Left side $= 10 – 12 = -2$; right side $= 5x – 6x = -x$. So $-x = -2 \Rightarrow x = 2$.
8. If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to (A) $6$ (B) $\pm 6$ (C) $-6$ (D) $0$
Answer: The correct option is (B) $\pm 6$. Left side $= x^2 – 36$; right side $= 36 – 36 = 0$. So $x^2 – 36 = 0 \Rightarrow x^2 = 36 \Rightarrow x = \pm 6$.
Exercise 4.2
1. Find area of the triangle with vertices at the point given in each of the following:
The area of a triangle is $\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$, taken as an absolute value.
(i) $(1, 0), (6, 0), (4, 3)$
Answer: $\Delta = \frac{1}{2}\begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} = \frac{1}{2}\left[1(0 – 3) – 0 + 1(18 – 0)\right] = \frac{1}{2}(-3 + 18) = \frac{15}{2}$ square units.
(ii) $(2, 7), (1, 1), (10, 8)$
Answer: $\Delta = \frac{1}{2}\begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix} = \frac{1}{2}\left[2(1 – 8) – 7(1 – 10) + 1(8 – 10)\right] = \frac{1}{2}(-14 + 63 – 2) = \frac{47}{2}$ square units.
(iii) $(-2, -3), (3, 2), (-1, -8)$
Answer: $\Delta = \frac{1}{2}\begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix} = \frac{1}{2}\left[-2(2 + 8) – (-3)(3 + 1) + 1(-24 + 2)\right] = \frac{1}{2}(-20 + 12 – 22) = \frac{1}{2}(-30)$. Taking the absolute value, the area $= 15$ square units.
2. Show that points $A(a, b + c)$, $B(b, c + a)$, $C(c, a + b)$ are collinear.
Answer: The area of the triangle formed by these points is $\Delta = \frac{1}{2}\begin{vmatrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{vmatrix}$. Apply $C_2 \to C_2 + C_1$, so every entry of $C_2$ becomes $a + b + c$: $\Delta = \frac{1}{2}(a + b + c)\begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix}$. Since columns $C_2$ and $C_3$ are identical, the determinant is $0$, so $\Delta = 0$. As the area is zero, the three points are collinear.
3. Find values of $k$ if area of triangle is 4 sq. units and vertices are
(i) $(k, 0), (4, 0), (0, 2)$
Answer: $\Delta = \frac{1}{2}\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \frac{1}{2}\left[k(0 – 2) – 0 + 1(8 – 0)\right] = \frac{1}{2}(-2k + 8)$. Setting the area equal to $4$, $\frac{1}{2}(-2k + 8) = \pm 4$, i.e. $-2k + 8 = \pm 8$. The $+$ sign gives $k = 0$ and the $-$ sign gives $k = 8$. Hence $k = 0$ or $8$.
(ii) $(-2, 0), (0, 4), (0, k)$
Answer: $\Delta = \frac{1}{2}\begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \frac{1}{2}\left[-2(4 – k) – 0 + 1(0 – 0)\right] = -(4 – k) = k – 4$. Setting the area equal to $4$, $|k – 4| = 4$, i.e. $k – 4 = \pm 4$. Hence $k = 0$ or $8$.
4. (i) Find equation of line joining $(1, 2)$ and $(3, 6)$ using determinants.
Answer: Let $(x, y)$ be any point on the line. Then the three points are collinear, so the area of their triangle is zero: $\frac{1}{2}\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$, i.e. $x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0 \Rightarrow -4x + 2y = 0 \Rightarrow y = 2x$. The required equation is $y = 2x$ (or $2x – y = 0$).
(ii) Find equation of line joining $(3, 1)$ and $(9, 3)$ using determinants.
Answer: With $(x, y)$ on the line, $\frac{1}{2}\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0$, i.e. $x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0 \Rightarrow -2x + 6y = 0 \Rightarrow x = 3y$. The required equation is $x = 3y$ (or $x – 3y = 0$).
5. If area of triangle is 35 sq units with vertices $(2, -6), (5, 4)$ and $(k, 4)$. Then $k$ is (A) $12$ (B) $-2$ (C) $-12, -2$ (D) $12, -2$
Answer: The correct option is (D) $12, -2$. $\Delta = \frac{1}{2}\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \frac{1}{2}\left[2(4 – 4) – (-6)(5 – k) + 1(20 – 4k)\right] = \frac{1}{2}(30 – 6k + 20 – 4k) = \frac{1}{2}(50 – 10k)$. Setting the area equal to $35$, $\frac{1}{2}(50 – 10k) = \pm 35$, i.e. $50 – 10k = \pm 70$. The $+$ sign gives $k = -2$ and the $-$ sign gives $k = 12$. Hence $k = 12, -2$.
Exercise 4.3
Write Minors and Cofactors of the elements of following determinants (Exercises 1 and 2). [Minor $M_{ij}$; cofactor $A_{ij} = (-1)^{i+j} M_{ij}$.]
1. (i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
Answer: Minors: $M_{11} = 3$, $M_{12} = 0$, $M_{21} = -4$, $M_{22} = 2$. Cofactors: $A_{11} = 3$, $A_{12} = -0 = 0$, $A_{21} = -(-4) = 4$, $A_{22} = 2$.
(ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$
Answer: Minors: $M_{11} = d$, $M_{12} = b$, $M_{21} = c$, $M_{22} = a$. Cofactors: $A_{11} = d$, $A_{12} = -b$, $A_{21} = -c$, $A_{22} = a$.
2. (i) $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$
Answer: Minors: $M_{11} = 1$, $M_{12} = 0$, $M_{13} = 0$, $M_{21} = 0$, $M_{22} = 1$, $M_{23} = 0$, $M_{31} = 0$, $M_{32} = 0$, $M_{33} = 1$. Cofactors: $A_{11} = 1$, $A_{12} = 0$, $A_{13} = 0$, $A_{21} = 0$, $A_{22} = 1$, $A_{23} = 0$, $A_{31} = 0$, $A_{32} = 0$, $A_{33} = 1$.
(ii) $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$
Answer: Minors: $M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = 11$, $M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = 6$, $M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3$, $M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = -4$, $M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = 2$, $M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$, $M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = -20$, $M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = -13$, $M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = 5$.
Cofactors: $A_{11} = 11$, $A_{12} = -6$, $A_{13} = 3$, $A_{21} = 4$, $A_{22} = 2$, $A_{23} = -1$, $A_{31} = -20$, $A_{32} = 13$, $A_{33} = 5$.
3. Using Cofactors of elements of second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.
Answer: The second-row elements are $a_{21} = 2$, $a_{22} = 0$, $a_{23} = 1$. Their cofactors are $A_{21} = -\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -(9 – 16) = 7$; $A_{22} = \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 15 – 8 = 7$; $A_{23} = -\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -(10 – 3) = -7$. Hence $\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) + 0(7) + 1(-7) = 14 – 7 = 7$.
4. Using Cofactors of elements of third column, evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$.
Answer: The third-column elements are $a_{13} = yz$, $a_{23} = zx$, $a_{33} = xy$. Their cofactors are $A_{13} = \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = z – y$; $A_{23} = -\begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(z – x) = x – z$; $A_{33} = \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = y – x$. Hence $\Delta = yz(z – y) + zx(x – z) + xy(y – x) = (x – y)(y – z)(z – x)$.
5. If $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ and $A_{ij}$ is Cofactor of $a_{ij}$, then value of $\Delta$ is given by
(A) $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}$ (B) $a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31}$ (C) $a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13}$ (D) $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$
Answer: The correct option is (D) $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$. The value of a determinant is the sum of the products of the elements of one row or column with their own cofactors. Option (D) uses the first-column elements $a_{11}, a_{21}, a_{31}$ with their own cofactors $A_{11}, A_{21}, A_{31}$, which is the correct expansion along $C_1$. In (A) and (C) elements of one row are multiplied by the cofactors of a different row, and such a sum is zero; (B) mixes elements and cofactors and is invalid.
Exercise 4.4
Find adjoint of each of the matrices in Exercises 1 and 2.
1. $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
Answer: For a $2 \times 2$ matrix the adjoint is obtained by interchanging the diagonal entries and changing the signs of the other two. Hence $\operatorname{adj} A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.
2. $\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$
Answer: The cofactors are $A_{11} = 3$, $A_{12} = -12$, $A_{13} = 6$, $A_{21} = 1$, $A_{22} = 5$, $A_{23} = 2$, $A_{31} = -11$, $A_{32} = -1$, $A_{33} = 5$. The adjoint is the transpose of the cofactor matrix, so $\operatorname{adj} A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$.
Verify $A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\, I$ in Exercises 3 and 4.
3. $\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$
Answer: $|A| = (2)(-6) – (3)(-4) = -12 + 12 = 0$ and $\operatorname{adj} A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$. Now $A(\operatorname{adj} A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}\begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$. Likewise $(\operatorname{adj} A)A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$. Since $|A|\, I = 0 \cdot I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, the identity $A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\, I$ holds.
4. $\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
Answer: $|A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0) = 11$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$. Then $A(\operatorname{adj} A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11 I = |A|\, I$. Likewise $(\operatorname{adj} A)A = 11 I$, so the identity holds.
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. [$A^{-1} = \frac{1}{|A|}\operatorname{adj} A$.]
5. $\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$
Answer: $|A| = (2)(3) – (-2)(4) = 6 + 8 = 14 \neq 0$, so $A^{-1}$ exists. $\operatorname{adj} A = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$, hence $A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.
6. $\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$
Answer: $|A| = (-1)(2) – (5)(-3) = -2 + 15 = 13 \neq 0$. $\operatorname{adj} A = \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$, hence $A^{-1} = \frac{1}{13}\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$.
7. $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$
Answer: $A$ is upper triangular, so $|A| = 1 \times 2 \times 5 = 10 \neq 0$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$, hence $A^{-1} = \frac{1}{10}\begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$.
8. $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$
Answer: Expanding along $R_1$, $|A| = 1\begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} = -3 \neq 0$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$, hence $A^{-1} = \frac{1}{-3}\begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{bmatrix}$.
9. $\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$
Answer: $|A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -2 – 4 + 3 = -3 \neq 0$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}$, hence $A^{-1} = -\frac{1}{3}\begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}$.
10. $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$
Answer: $|A| = 1(8 – 6) + 1(0 + 9) + 2(0 – 6) = 2 + 9 – 12 = -1 \neq 0$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$, hence $A^{-1} = \frac{1}{-1}\begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$.
11. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
Answer: $|A| = 1(-\cos^2\alpha – \sin^2\alpha) = -1 \neq 0$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$, hence $A^{-1} = \frac{1}{-1}\operatorname{adj} A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix} = A$. So this matrix is its own inverse.
12. Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.
Answer: $AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$, and $|AB| = 67 \times 61 – 87 \times 47 = 4087 – 4089 = -2$. So $(AB)^{-1} = -\frac{1}{2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$.
Now $|A| = 15 – 14 = 1$, so $A^{-1} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$; and $|B| = 54 – 56 = -2$, so $B^{-1} = -\frac{1}{2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$. Hence $B^{-1} A^{-1} = -\frac{1}{2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = -\frac{1}{2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$. Since this equals $(AB)^{-1}$, we have $(AB)^{-1} = B^{-1} A^{-1}$.
13. If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = O$. Hence find $A^{-1}$.
Answer: $A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$. Then $A^2 – 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Multiplying $A^2 – 5A + 7I = O$ by $A^{-1}$ (since $|A| = 7 \neq 0$): $A – 5I + 7A^{-1} = O \Rightarrow 7A^{-1} = 5I – A$, so $A^{-1} = \frac{1}{7}(5I – A) = \frac{1}{7}\left(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} – \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\right) = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$.
14. For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find the numbers $a$ and $b$ such that $A^2 + aA + bI = O$.
Answer: $A^2 = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}$. Then $A^2 + aA + bI = \begin{bmatrix} 11 + 3a + b & 8 + 2a \\ 4 + a & 3 + a + b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$. From the $(1,2)$ entry, $8 + 2a = 0 \Rightarrow a = -4$; from the $(1,1)$ entry, $11 + 3(-4) + b = 0 \Rightarrow b = 1$. (The $(2,1)$ and $(2,2)$ entries confirm these.) Hence $a = -4$, $b = 1$.
15. For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$, show that $A^3 – 6A^2 + 5A + 11I = O$. Hence, find $A^{-1}$.
Answer: $A^2 = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}$ and $A^3 = A^2 \cdot A = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$. Substituting gives $A^3 – 6A^2 + 5A + 11I = O$ (every entry is zero).
Here $|A| = -11 \neq 0$, so $A^{-1}$ exists. Multiplying the equation by $A^{-1}$: $A^2 – 6A + 5I + 11A^{-1} = O \Rightarrow A^{-1} = \frac{1}{11}(-A^2 + 6A – 5I) = \frac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$.
16. If $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$, verify that $A^3 – 6A^2 + 9A – 4I = O$ and hence find $A^{-1}$.
Answer: $A^2 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}$ and $A^3 = A^2 \cdot A = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}$. Substituting gives $A^3 – 6A^2 + 9A – 4I = O$.
Here $|A| = 4 \neq 0$. Multiplying the equation by $A^{-1}$: $A^2 – 6A + 9I – 4A^{-1} = O \Rightarrow 4A^{-1} = A^2 – 6A + 9I$, so $A^{-1} = \frac{1}{4}(A^2 – 6A + 9I) = \frac{1}{4}\begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$.
17. Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|\operatorname{adj} A|$ is equal to (A) $|A|$ (B) $|A|^2$ (C) $|A|^3$ (D) $3|A|$
Answer: The correct option is (B) $|A|^2$. For a matrix of order $n$, $|\operatorname{adj} A| = |A|^{n-1}$. Here $n = 3$, so $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
18. If $A$ is an invertible matrix of order 2, then $\det(A^{-1})$ is equal to (A) $\det(A)$ (B) $\frac{1}{\det(A)}$ (C) $1$ (D) $0$
Answer: The correct option is (B) $\frac{1}{\det(A)}$. Since $A A^{-1} = I$, taking determinants of both sides gives $\det(A)\cdot\det(A^{-1}) = \det(I) = 1$, so $\det(A^{-1}) = \frac{1}{\det(A)}$.
Exercise 4.5
Examine the consistency of the system of equations in Exercises 1 to 6. [For $AX = B$: if $|A| \neq 0$ the system is consistent with a unique solution; if $|A| = 0$ and $(\operatorname{adj} A)B \neq O$ it is inconsistent.]
1. $x + 2y = 2$; $2x + 3y = 3$
Answer: $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$, $|A| = 3 – 4 = -1 \neq 0$. So $A$ is non-singular and the system has a unique solution — it is consistent.
2. $2x – y = 5$; $x + y = 4$
Answer: $A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$, $|A| = 2 + 1 = 3 \neq 0$. So the system is consistent (unique solution).
3. $x + 3y = 5$; $2x + 6y = 8$
Answer: $A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$, $|A| = 6 – 6 = 0$, so $A$ is singular. Now $\operatorname{adj} A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$, so $(\operatorname{adj} A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O$. Hence the system is inconsistent.
4. $x + y + z = 1$; $2x + 3y + 2z = 2$; $ax + ay + 2az = 4$
Answer: $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{bmatrix}$, $|A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) = 4a – 2a – a = a$. If $a \neq 0$ then $|A| = a \neq 0$, so $A$ is non-singular and the system is consistent (unique solution). (If $a = 0$ the third equation becomes $0 = 4$, which is impossible, and the system is inconsistent.)
5. $3x – y – 2z = 2$; $2y – z = -1$; $3x – 5y = 3$
Answer: $A = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$, $|A| = 3(0 – 5) + 1(0 + 3) – 2(0 – 6) = -15 + 3 + 12 = 0$, so $A$ is singular. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$. Then $(\operatorname{adj} A)B = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq O$. Hence the system is inconsistent.
6. $5x – y + 4z = 5$; $2x + 3y + 5z = 2$; $5x – 2y + 6z = -1$
Answer: $A = \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix}$, $|A| = 5(18 + 10) + 1(12 – 25) + 4(-4 – 15) = 140 – 13 – 76 = 51 \neq 0$. So $A$ is non-singular and the system is consistent (unique solution).
Solve system of linear equations, using matrix method, in Exercises 7 to 14. [$X = A^{-1}B$.]
7. $5x + 2y = 4$; $7x + 3y = 5$
Answer: $A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$, $|A| = 15 – 14 = 1$, $A^{-1} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$. $X = A^{-1}B = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$. Hence $x = 2$, $y = -3$.
8. $2x – y = -2$; $3x + 4y = 3$
Answer: $A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$, $|A| = 8 + 3 = 11$, $A^{-1} = \frac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$. $X = \frac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2 \\ 3 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5 \\ 12 \end{bmatrix}$. Hence $x = -\frac{5}{11}$, $y = \frac{12}{11}$.
9. $4x – 3y = 3$; $3x – 5y = 7$
Answer: $A = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}$, $|A| = -20 + 9 = -11$, $A^{-1} = \frac{1}{-11}\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}$. $X = \frac{1}{-11}\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix} = \frac{1}{-11}\begin{bmatrix} 6 \\ 19 \end{bmatrix}$. Hence $x = -\frac{6}{11}$, $y = -\frac{19}{11}$.
10. $5x + 2y = 3$; $3x + 2y = 5$
Answer: $A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}$, $|A| = 10 – 6 = 4$, $A^{-1} = \frac{1}{4}\begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}$. $X = \frac{1}{4}\begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}\begin{bmatrix} 3 \\ 5 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4 \\ 16 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}$. Hence $x = -1$, $y = 4$.
11. $2x + y + z = 1$; $x – 2y – z = \frac{3}{2}$; $3y – 5z = 9$
Answer: $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}$, $|A| = 2(10 + 3) – 1(-5 – 0) + 1(3 – 0) = 26 + 5 + 3 = 34$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}$, so $A^{-1} = \frac{1}{34}\operatorname{adj} A$. $X = \frac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}\begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$. Hence $x = 1$, $y = \frac{1}{2}$, $z = -\frac{3}{2}$.
12. $x – y + z = 4$; $2x + y – 3z = 0$; $x + y + z = 2$
Answer: $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$, $|A| = 1(1 + 3) + 1(2 + 3) + 1(2 – 1) = 4 + 5 + 1 = 10$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$, so $A^{-1} = \frac{1}{10}\operatorname{adj} A$. $X = \frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$. Hence $x = 2$, $y = -1$, $z = 1$.
13. $2x + 3y + 3z = 5$; $x – 2y + z = -4$; $3x – y – 2z = 3$
Answer: $A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}$, $|A| = 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6) = 10 + 15 + 15 = 40$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$, so $A^{-1} = \frac{1}{40}\operatorname{adj} A$. $X = \frac{1}{40}\begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}\begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$. Hence $x = 1$, $y = 2$, $z = -1$.
14. $x – y + 2z = 7$; $3x + 4y – 5z = -5$; $2x – y + 3z = 12$
Answer: $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}$, $|A| = 1(12 – 5) + 1(9 + 10) + 2(-3 – 8) = 7 + 19 – 22 = 4$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$, so $A^{-1} = \frac{1}{4}\operatorname{adj} A$. $X = \frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$. Hence $x = 2$, $y = 1$, $z = 3$.
15. If $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, find $A^{-1}$. Using $A^{-1}$ solve the system of equations $2x – 3y + 5z = 11$; $3x + 2y – 4z = -5$; $x + y – 2z = -3$.
Answer: $|A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 – 2) = 0 – 6 + 5 = -1 \neq 0$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$, so $A^{-1} = \frac{1}{-1}\operatorname{adj} A = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$.
Writing the system as $AX = B$ with $B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$, $X = A^{-1}B = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$. Hence $x = 1$, $y = 2$, $z = 3$.
16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.
Answer: Let the cost per kg of onion, wheat and rice be $x$, $y$, $z$ respectively. Then $4x + 3y + 2z = 60$; $2x + 4y + 6z = 90$; $6x + 2y + 3z = 70$. $A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}$, $|A| = 4(12 – 12) – 3(6 – 36) + 2(4 – 24) = 0 + 90 – 40 = 50$. From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$, so $A^{-1} = \frac{1}{50}\operatorname{adj} A$.
$X = \frac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix} = \frac{1}{50}\begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$. Hence onion costs Rs 5 per kg, wheat Rs 8 per kg and rice Rs 8 per kg.
Miscellaneous Exercise on Chapter 4
1. Prove that the determinant $\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.
Answer: Expanding along $R_1$,
$= x(-x \cdot x – 1 \cdot 1) – \sin\theta(-\sin\theta \cdot x – 1 \cdot \cos\theta) + \cos\theta(-\sin\theta \cdot 1 – (-x)\cos\theta)$
$= x(-x^2 – 1) + \sin\theta(x\sin\theta + \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$
$= -x^3 – x + x\sin^2\theta + \sin\theta\cos\theta – \sin\theta\cos\theta + x\cos^2\theta = -x^3 – x + x(\sin^2\theta + \cos^2\theta) = -x^3 – x + x = -x^3$.
Since the value is $-x^3$, which contains no $\theta$, the determinant is independent of $\theta$. Hence proved.
2. Evaluate $\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$.
Answer: Expanding along $R_1$,
$= \cos\alpha\cos\beta(\cos\alpha\cos\beta – 0) – \cos\alpha\sin\beta(-\sin\beta\cos\alpha – 0) + (-\sin\alpha)(-\sin\beta \cdot \sin\alpha\sin\beta – \cos\beta \cdot \sin\alpha\cos\beta)$
$= \cos^2\alpha\cos^2\beta + \cos^2\alpha\sin^2\beta + \sin^2\alpha(\sin^2\beta + \cos^2\beta) = \cos^2\alpha(\cos^2\beta + \sin^2\beta) + \sin^2\alpha = \cos^2\alpha + \sin^2\alpha = 1$.
3. If $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$, find $(AB)^{-1}$.
Answer: $(AB)^{-1} = B^{-1} A^{-1}$. First find $B^{-1}$: $|B| = 1(3 – 0) – 2(-1 – 0) + (-2)(2 – 0) = 3 + 2 – 4 = 1$. From the cofactors, $\operatorname{adj} B = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$, so $B^{-1} = \operatorname{adj} B$.
Hence $(AB)^{-1} = B^{-1} A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}$.
4. Let $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that (i) $[\operatorname{adj} A]^{-1} = \operatorname{adj}(A^{-1})$, (ii) $(A^{-1})^{-1} = A$.
Answer: $|A| = 1(15 – 1) – 2(10 – 1) + 1(2 – 3) = 14 – 18 – 1 = -5 \neq 0$, so $A^{-1}$ exists.
(i) We know $\operatorname{adj} A = |A|\, A^{-1}$. So $[\operatorname{adj} A]^{-1} = (|A|\, A^{-1})^{-1} = \frac{1}{|A|} A$. Also $\operatorname{adj}(A^{-1}) = |A^{-1}|\, (A^{-1})^{-1} = \frac{1}{|A|} A$ (since $|A^{-1}| = \frac{1}{|A|}$ and $(A^{-1})^{-1} = A$). Both equal $-\frac{1}{5}\begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$, so $[\operatorname{adj} A]^{-1} = \operatorname{adj}(A^{-1})$ is verified.
(ii) Since $A^{-1} A = A A^{-1} = I$, the matrix $A$ is the inverse of $A^{-1}$, so $(A^{-1})^{-1} = A$ is verified.
5. Evaluate $\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$.
Answer: Apply $C_1 \to C_1 + C_2 + C_3$; each entry of $C_1$ becomes $2(x + y)$, so $= 2(x + y)\begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix}$. Now apply $R_2 \to R_2 – R_1$ and $R_3 \to R_3 – R_1$:
$= 2(x + y)\begin{vmatrix} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x \end{vmatrix} = 2(x + y)\left[x(-x) – (-y)(x – y)\right] = 2(x + y)(-x^2 + xy – y^2)$.
$= -2(x + y)(x^2 – xy + y^2) = -2(x^3 + y^3)$.
6. Evaluate $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$.
Answer: Apply $R_2 \to R_2 – R_1$ and $R_3 \to R_3 – R_1$: $= \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}$. Expanding along $C_1$, $= 1(y \cdot x – 0) = xy$.
7. Solve the system of equations $\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$; $\frac{4}{x} – \frac{6}{y} + \frac{5}{z} = 1$; $\frac{6}{x} + \frac{9}{y} – \frac{20}{z} = 2$.
Answer: Put $u = \frac{1}{x}$, $v = \frac{1}{y}$, $w = \frac{1}{z}$. The system becomes $2u + 3v + 10w = 4$; $4u – 6v + 5w = 1$; $6u + 9v – 20w = 2$. $A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}$, $|A| = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36) = 150 + 330 + 720 = 1200$.
From the cofactors, $\operatorname{adj} A = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$, so $\begin{bmatrix} u \\ v \\ w \end{bmatrix} = \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} = \frac{1}{1200}\begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{bmatrix}$.
So $\frac{1}{x} = \frac{1}{2}$, $\frac{1}{y} = \frac{1}{3}$, $\frac{1}{z} = \frac{1}{5}$, that is $x = 2$, $y = 3$, $z = 5$.
Choose the correct answer in Exercises 8 and 9.
8. If $x, y, z$ are nonzero real numbers, then the inverse of matrix $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ is
(A) $\begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$ (B) $xyz\begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$ (C) $\frac{1}{xyz}\begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ (D) $\frac{1}{xyz}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Answer: The correct option is (A). The inverse of a diagonal matrix is the diagonal matrix of the reciprocals of the entries, i.e. $A^{-1} = \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$. (Check: $A A^{-1} = I$.)
9. Let $A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$, where $0 \leq \theta \leq 2\pi$. Then
(A) $\det(A) = 0$ (B) $\det(A) \in (2, \infty)$ (C) $\det(A) \in (2, 4)$ (D) $\det(A) \in [2, 4]$
Answer: The correct option is (D) $\det(A) \in [2, 4]$. Expanding along $R_1$, $\det(A) = 1(1 + \sin^2\theta) – \sin\theta(-\sin\theta + \sin\theta) + 1(\sin^2\theta + 1) = (1 + \sin^2\theta) + (1 + \sin^2\theta) = 2 + 2\sin^2\theta$. Since $0 \leq \sin^2\theta \leq 1$, $\det(A)$ ranges from $2 + 0 = 2$ to $2 + 2 = 4$, so $\det(A) \in [2, 4]$.
Additional Questions and Answers
These extra questions on Determinants are designed for practice and revision by ASSEB Class 12 Mathematics students. They cover determinants, minors and cofactors, adjoint, inverse and the matrix method.
Multiple Choice Questions
1. If $A$ is a square matrix of order 3 and $|A| = 5$, then $|2A|$ is (A) $10$ (B) $20$ (C) $40$ (D) $80$
Answer: (C) $40$. For order $n$, $|kA| = k^n|A|$, so $|2A| = 2^3 \times 5 = 40$.
2. If $A$ is a square matrix of order 3 with $|A| = 4$, then $|\operatorname{adj} A|$ is (A) $4$ (B) $8$ (C) $16$ (D) $64$
Answer: (C) $16$. $|\operatorname{adj} A| = |A|^{n-1} = 4^2 = 16$.
3. The value of $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$ is (A) $-2$ (B) $2$ (C) $22$ (D) $-22$
Answer: (A) $-2$. $2 \times 5 – 3 \times 4 = 10 – 12 = -2$.
4. For a non-singular matrix $A$ of order $n$, $A(\operatorname{adj} A)$ equals (A) $I$ (B) $|A|\, I$ (C) $\operatorname{adj} A$ (D) $O$
Answer: (B) $|A|\, I$. By the fundamental property, $A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\, I$.
5. The area of the triangle with vertices $(0, 0), (3, 0), (0, 4)$ is (A) $5$ (B) $6$ (C) $7$ (D) $12$ (sq. units)
Answer: (B) $6$. $\Delta = \frac{1}{2}\begin{vmatrix} 0 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & 4 & 1 \end{vmatrix} = \frac{1}{2}|{-12}| = 6$ square units.
6. If two rows of a determinant are identical, then the value of the determinant is (A) $1$ (B) $0$ (C) $2$ (D) undefined
Answer: (B) $0$. A determinant with two identical rows (or columns) is always zero.
7. If $\begin{vmatrix} x & 1 \\ 3 & 2 \end{vmatrix} = 5$, then $x$ is (A) $2$ (B) $3$ (C) $4$ (D) $-4$
Answer: (C) $4$. $2x – 3 = 5 \Rightarrow 2x = 8 \Rightarrow x = 4$.
8. A square matrix $A$ is invertible if and only if $A$ is (A) singular (B) symmetric (C) non-singular (D) diagonal
Answer: (C) non-singular. $A$ has an inverse if and only if $|A| \neq 0$, i.e. $A$ is non-singular.
Fill in the Blanks
1. The determinant of the identity matrix of order 3 is __________.
Answer: $1$.
2. For a square matrix $A$, $A(\operatorname{adj} A) = $ __________.
Answer: $|A|\, I$.
3. The minor of an element $a_{ij}$ is the determinant obtained by deleting its __________ and __________.
Answer: $i$th row and $j$th column.
4. The cofactor of $a_{ij}$ is $A_{ij} = $ __________ $M_{ij}$.
Answer: $(-1)^{i+j}$.
5. The system $AX = B$ has a unique solution if $|A|$ __________ $0$.
Answer: $\neq$ (is not equal to).
True or False
1. Only square matrices have determinants.
Answer: True. A determinant is defined only for a square matrix.
2. For a matrix $A$, the symbol $|A|$ means the modulus (absolute value) of $A$.
Answer: False. $|A|$ is read as the determinant of $A$, not its modulus.
3. If $|A| = 0$, then $A$ is a non-singular matrix.
Answer: False. When $|A| = 0$ the matrix is singular; it is non-singular only when $|A| \neq 0$.
4. The area of the triangle formed by three collinear points is zero.
Answer: True. Collinear points make the area determinant zero.
5. For square matrices $A$ and $B$ of the same order, $|AB| = |A|\,|B|$.
Answer: True. The determinant of a product equals the product of the determinants.
Short Answer Questions
1. Evaluate $\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix}$.
Answer: $= \cos 15^\circ \cos 75^\circ – \sin 15^\circ \sin 75^\circ = \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0$.
2. Find the adjoint of $A = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}$.
Answer: Interchange the diagonal entries and change the signs of the other two: $\operatorname{adj} A = \begin{bmatrix} 5 & -4 \\ -3 & 2 \end{bmatrix}$.
3. Find $x$ if $\begin{vmatrix} x & 2 \\ 3 & x \end{vmatrix} = 10$.
Answer: $x^2 – 6 = 10 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$.
4. Find the area of the triangle with vertices $(2, 0), (-1, 0)$ and $(0, 3)$.
Answer: $\Delta = \frac{1}{2}\begin{vmatrix} 2 & 0 & 1 \\ -1 & 0 & 1 \\ 0 & 3 & 1 \end{vmatrix} = \frac{1}{2}\left[2(0 – 3) – 0 + 1(-3 – 0)\right] = \frac{1}{2}(-9)$. Taking the absolute value, the area $= \frac{9}{2}$ square units.
Key Terms
| Term | Meaning |
|---|---|
| Determinant (নিৰ্ণায়ক) | A number associated with a square matrix, written $|A|$ or $\det A$. |
| Expansion | Evaluating a determinant along a row or column using cofactors. |
| Minor ($M_{ij}$) | The determinant left after deleting the row and column of an element. |
| Cofactor ($A_{ij}$) | The signed minor $A_{ij} = (-1)^{i+j} M_{ij}$. |
| Adjoint ($\operatorname{adj} A$) | The transpose of the matrix of cofactors of $A$. |
| Inverse ($A^{-1}$) | The matrix $\frac{1}{|A|}\operatorname{adj} A$, existing when $|A| \neq 0$. |
| Singular matrix | A square matrix with $|A| = 0$; it has no inverse. |
| Non-singular matrix | A square matrix with $|A| \neq 0$; it is invertible. |
| Consistent system | A system of equations that has at least one solution. |
| Inconsistent system | A system of equations that has no solution. |
| Matrix method | Solving $AX = B$ as $X = A^{-1}B$ when $|A| \neq 0$. |