Question 1

If any two rows or any two columns are identical or proportional, then the value of determinant is

Accepted Answer

0 — [{"id": "3b1e6ed2-0874-6aaf-9cdf-62251cbd4785", "type": "html", "value": " If any two rows or any two columns are identical or proportional, then the value of determinant is zero. "}]

Question 2

If the area of triangle is 4 square unit and vertices are (k, 0), (4, 0), (0, 2) then the value of k is..

Accepted Answer

0 — [{"id": "33bb15e5-1cec-ec43-1c58-227c29978a2e", "type": "html", "value": " "}]

Question 3

If (6, 0), (4, 3), (2, 1) is the vertices of a triangle then the area of the triangle is

Accepted Answer

5 square unit — [{"id": "53285455-c45d-2f35-cf6a-6c40276f0d2e", "type": "html", "value": " "}]

Question 4

Evaluate the determinants in Exercises 1 and 2.

1. $\left| \begin{array}{cc} 2 & 4 \ -5 & -1 \end{array} \right|$
2. (i) $\left| \begin{array}{cc} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{array} \right|$
(ii) $\left| \begin{array}{cc} x^{2} - x + 1 & x - 1 \ x + 1 & x + 1 \end{array} \right|$
3. If $A = \begin{bmatrix} 1 & 2 \ 4 & 2 \end{bmatrix}$, then show that $|2A| = 4 |A|$
4. If $A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4 \end{bmatrix}$, then show that $|3A| = 27 |A|$
5. Evaluate the determinants
(i) $\left| \begin{array}{ccc} 3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0 \end{array} \right|$
(ii) $\left| \begin{array}{ccc} 3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1 \end{array} \right|$
(iii)  $\left| \begin{array}{lll}0 & 1 & 2\ -1 & 0 & -3\ -2 & 3 & 0 \end{array} \right|$
(iv)  $\left| \begin{array}{rrr}2 & -1 & -2\ 0 & 2 & -1\ 3 & -5 & 0 \end{array} \right|$
6. If  $\mathrm{A} = \left[ \begin{array}{rrr}1 & 1 & -2\ 2 & 1 & -3\ 5 & 4 & -9 \end{array} \right]$ , find  $|\mathrm{A}|$
7. Find values of  $x$ , if

(i)  $\left| \begin{array}{lll}2 & 4\ 5 & 1 \end{array} \right| = \left| \begin{array}{lll}2x & 4\ 6 & x \end{array} \right|$

(ii)  $\left| \begin{array}{ll}2 & 3\ 4 & 5 \end{array} \right| = \left| \begin{array}{ll}x & 3\ 2x & 5 \end{array} \right|$
8. If  $\left| \begin{array}{ll}x & 2\ 18 & x \end{array} \right| = \left| \begin{array}{ll}6 & 2\ 18 & 6 \end{array} \right|$ , then  $x$  is equal to

(A) 6

(B)  $\pm 6$

(C) -6

(D) 0

Accepted Answer

1. Evaluate $\left| \begin{array}{cc} 2 & 4 \ -5 & -1 \end{array} \right| = (2)(-1) - (4)(-5) = -2 + 20 = 18.\n
2.(i) \(\left| \begin{array}{cc} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{array} \right| = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1.\n
(ii) \(\left| \begin{array}{cc} x^{2} - x + 1 & x - 1 \ x + 1 & x + 1 \end{array} \right| = (x^{2} - x + 1)(x + 1) - (x - 1)(x + 1) = (x^{3} + x^{2} - x^{2} - x + x + 1) - (x^{2} - 1) = (x^{3} + 1) - (x^{2} - 1) = x^{3} + 1 - x^{2} + 1 = x^{3} - x^{2} + 2.\n
3. Given \(A = \begin{bmatrix} 1 & 2 \ 4 & 2 \end{bmatrix}$, $|A| = (1)(2) - (2)(4) = 2 - 8 = -6.$\n
Now, $2A = \begin{bmatrix} 2 & 4 \ 8 & 4 \end{bmatrix}$, so $|2A| = (2)(4) - (4)(8) = 8 - 32 = -24.$\n
Since $|2A| = 4 |A|$ means $-24 = 4 \times (-6)$, which is true. Hence proved.\n
4. $A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4 \end{bmatrix}$. $|A|$ is product of diagonal entries since it is upper triangular: $1 \times 1 \times 4 = 4.$\n
$3A = \begin{bmatrix} 3 & 0 & 3 \ 0 & 3 & 6 \ 0 & 0 & 12 \end{bmatrix}$, determinant is $3 \times 3 \times 12 = 108.$\n
Since $|3A| = 27 |A|$ means $108 = 27 \times 4$, which is true. Hence proved.\n
5. (i) $\left| \begin{array}{ccc} 3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0 \end{array} \right|$
= 3 \times \left| \begin{array}{cc} 0 & -1 \ -5 & 0 \end{array} \right| - (-1) \times \left| \begin{array}{cc} 0 & -1 \ 3 & 0 \end{array} \right| + (-2) \times \left| \begin{array}{cc} 0 & 0 \ 3 & -5 \end{array} \right|
= 3(0 - 5) + 1(0 - (-3)) - 2(0 - 0) = 3(-5) + 3 - 0 = -15 + 3 = -12.

(ii) $\left| \begin{array}{ccc} 3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1 \end{array} \right|$
= 3 \times \left| \begin{array}{cc} 1 & -2 \ 3 & 1 \end{array} \right| - (-4) \times \left| \begin{array}{cc} 1 & -2 \ 2 & 1 \end{array} \right| + 5 \times \left| \begin{array}{cc} 1 & 1 \ 2 & 3 \end{array} \right|
= 3(1 - (-6)) + 4(1 - (-4)) + 5(3 - 2) = 3(7) + 4(5) + 5(1) = 21 + 20 + 5 = 46.

(iii) $\left| \begin{array}{lll}0 & 1 & 2\ -1 & 0 & -3\ -2 & 3 & 0 \end{array} \right|$
= 0 \times \left| \begin{array}{cc} 0 & -3 \ 3 & 0 \end{array} \right| - 1 \times \left| \begin{array}{cc} -1 & -3 \ -2 & 0 \end{array} \right| + 2 \times \left| \begin{array}{cc} -1 & 0 \ -2 & 3 \end{array} \right|
= 0 - 1((-1)(0) - (-3)(-2)) + 2((-1)(3) - 0(-2)) = 0 - 1(0 - 6) + 2(-3 - 0) = 0 - (-6) + (-6) = 6 - 6 = 0.

(iv) $\left| \begin{array}{rrr}2 & -1 & -2\ 0 & 2 & -1\ 3 & -5 & 0 \end{array} \right|$
= 2 \times \left| \begin{array}{cc} 2 & -1 \ -5 & 0 \end{array} \right| - (-1) \times \left| \begin{array}{cc} 0 & -1 \ 3 & 0 \end{array} \right| + (-2) \times \left| \begin{array}{cc} 0 & 2 \ 3 & -5 \end{array} \right|
= 2(0 - (-5)(-1)) + 1(0 - (-3)) - 2(0 - 6) = 2(0 - 5) + 3 + 12 = 2(-5) + 3 + 12 = -10 + 3 + 12 = 5.

6. $\mathrm{A} = \left[ \begin{array}{rrr}1 & 1 & -2\ 2 & 1 & -3\ 5 & 4 & -9 \end{array} \right]$
Calculate determinant:
= 1 \times \left| \begin{array}{cc} 1 & -3 \ 4 & -9 \end{array} \right| - 1 \times \left| \begin{array}{cc} 2 & -3 \ 5 & -9 \end{array} \right| + (-2) \times \left| \begin{array}{cc} 2 & 1 \ 5 & 4 \end{array} \right|
= 1(1 \times -9 - (-3) \times 4) - 1(2 \times -9 - (-3) \times 5) - 2(2 \times 4 - 1 \times 5)
= 1(-9 + 12) - 1(-18 + 15) - 2(8 - 5)
= 1(3) - 1(-3) - 2(3) = 3 + 3 - 6 = 0.

7. Find values of $x$ such that:
(i) $\left| \begin{array}{ll}2 & 4 \ 5 & 1 \end{array} \right| = \left| \begin{array}{ll}2x & 4 \ 6 & x \end{array} \right|$
Calculate left determinant: $2 \times 1 - 4 \times 5 = 2 - 20 = -18.$
Calculate right determinant: $2x \times x - 4 \times 6 = 2x^{2} - 24.$
Equate: $2x^{2} - 24 = -18 \Rightarrow 2x^{2} = 6 \Rightarrow x^{2} = 3 \Rightarrow x = \pm \sqrt{3}.$

(ii) $\left| \begin{array}{ll}2 & 3 \ 4 & 5 \end{array} \right| = \left| \begin{array}{ll}x & 3 \ 2x & 5 \end{array} \right|$
Left determinant: $2 \times 5 - 3 \times 4 = 10 - 12 = -2.$
Right determinant: $x \times 5 - 3 \times 2x = 5x - 6x = -x.$
Equate: $-x = -2 \Rightarrow x = 2.$

8. If $\left| \begin{array}{ll}x & 2 \ 18 & x \end{array} \right| = \left| \begin{array}{ll}6 & 2 \ 18 & 6 \end{array} \right|$, then find $x$.
Calculate right determinant: $6 \times 6 - 2 \times 18 = 36 - 36 = 0.$
Calculate left determinant: $x \times x - 2 \times 18 = x^{2} - 36.$
Equate: $x^{2} - 36 = 0 \Rightarrow x^{2} = 36 \Rightarrow x = \pm 6.$

Hence correct option is (B) $\pm 6$. — Step-by-step evaluation of each determinant and verification of properties:
- For 2x2 determinants, use formula $ad - bc$.
- For 3x3 determinants, use expansion by minors.
- For matrix scalar multiplication, use property $|kA| = k^{n} |A|$ where $n$ is order of matrix.
- For equations involving determinants, equate given determinants and solve for unknowns.
- For MCQ, calculate both sides and compare to find correct option.

Question 5

Find area of the triangle with vertices at the point given in each of the following:
(i) (1,0), (6,0), (4,3)
(ii) (2,7), (1,1), (10,8)
(iii) (-2, -3), (3,2), (-1, -8)

Accepted Answer

Solution:
The area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3) is given by:
Area = (1/2) * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

(i) Vertices: (1,0), (6,0), (4,3)
Area = (1/2) * |1(0 - 3) + 6(3 - 0) + 4(0 - 0)|
= (1/2) * |1*(-3) + 6*3 + 4*0|
= (1/2) * |-3 + 18 + 0| = (1/2)*15 = 7.5 sq. units

(ii) Vertices: (2,7), (1,1), (10,8)
Area = (1/2) * |2(1 - 8) + 1(8 - 7) + 10(7 - 1)|
= (1/2) * |2*(-7) + 1*1 + 10*6|
= (1/2) * |-14 + 1 + 60| = (1/2)*47 = 23.5 sq. units

(iii) Vertices: (-2, -3), (3,2), (-1, -8)
Area = (1/2) * |-2(2 + 8) + 3(-8 + 3) + (-1)(-3 - 2)|
= (1/2) * |-2*10 + 3*(-5) + (-1)*(-5)|
= (1/2) * |-20 -15 + 5| = (1/2)*(-30) = 15 sq. units (area is positive)

Hence, areas are:
(i) 7.5 sq. units
(ii) 23.5 sq. units
(iii) 15 sq. units — Used the formula for area of triangle using determinant method or coordinate geometry formula for area of triangle given vertices. Calculated stepwise for each set of points.

Question 6

Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Accepted Answer

Solution:
To show points A, B, C are collinear, the area of triangle formed by them must be zero.
Using the area formula:
Area = (1/2) * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Let A = (a, b + c), B = (b, c + a), C = (c, a + b)

Grouping terms:
= (1/2) * |(a c - c a) + (b a - a b) + (c b - b c)|
= (1/2) * |0 + 0 + 0| = 0

Since area is zero, points A, B, C are collinear. — Used the area formula for triangle and simplified the expression to show it equals zero, proving collinearity.

Question 7

Find values of k if area of triangle is 4 sq. units and vertices are
(i) (k,0), (4,0), (0,2)
(ii) (-2,0), (0,4), (0,k)

Accepted Answer

Solution:
Area of triangle with vertices (x1,y1), (x2,y2), (x3,y3) is:
Area = (1/2) * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Given area = 4 sq. units

(i) Vertices: (k,0), (4,0), (0,2)
Area = (1/2) * |k(0 - 2) + 4(2 - 0) + 0(0 - 0)| = 4
=> (1/2) * |-2k + 8 + 0| = 4
=> |-2k + 8| = 8

Case 1: -2k + 8 = 8
=> -2k = 0
=> k = 0

Case 2: -2k + 8 = -8
=> -2k = -16
=> k = 8

So, k = 0 or 8

(ii) Vertices: (-2,0), (0,4), (0,k)
Area = (1/2) * |-2(4 - k) + 0(k - 0) + 0(0 - 4)| = 4
=> (1/2) * |-2(4 - k)| = 4
=> |-2(4 - k)| = 8
=> | -8 + 2k | = 8

Case 1: -8 + 2k = 8
=> 2k = 16
=> k = 8

Case 2: -8 + 2k = -8
=> 2k = 0
=> k = 0

So, k = 0 or 8 — Used the area formula and set equal to 4, then solved the resulting absolute value equations to find possible values of k.

Question 8

(i) Find equation of line joining (1,2) and (3,6) using determinants.
(ii) Find equation of line joining (3,1) and (9,3) using determinants.

Accepted Answer

Solution:
(i) Equation of line joining points (x1,y1) and (x2,y2) using determinant:
\begin{vmatrix} x & y & 1 \ x_1 & y_1 & 1 \ x_2 & y_2 & 1 \end{vmatrix} = 0

For points (1,2) and (3,6):
\begin{vmatrix} x & y & 1 \ 1 & 2 & 1 \ 3 & 6 & 1 \end{vmatrix} = 0

Expanding:
(x)(2*1 - 6*1) - (y)(1*1 - 3*1) + 1(1*6 - 3*2) = 0
=> x(2 - 6) - y(1 - 3) + (6 - 6) = 0
=> x(-4) - y(-2) + 0 = 0
=> -4x + 2y = 0
=> 2y = 4x
=> y = 2x

(ii) For points (3,1) and (9,3):
\begin{vmatrix} x & y & 1 \ 3 & 1 & 1 \ 9 & 3 & 1 \end{vmatrix} = 0

Expanding:
(x)(1*1 - 3*1) - (y)(3*1 - 9*1) + 1(3*3 - 9*1) = 0
=> x(1 - 3) - y(3 - 9) + (9 - 9) = 0
=> x(-2) - y(-6) + 0 = 0
=> -2x + 6y = 0
=> 6y = 2x
=> y = (1/3)x — Used determinant form of line equation passing through two points and expanded determinant to get linear equation in x and y.

Question 9

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

Accepted Answer

Solution:
Area of triangle with vertices (x1,y1), (x2,y2), (x3,y3) is:
Area = (1/2) * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Given area = 35 sq. units
Vertices: (2, -6), (5, 4), (k, 4)

Calculate:
Area = (1/2) * |2(4 - 4) + 5(4 + 6) + k(-6 - 4)| = 35
=> (1/2) * |2*0 + 5*10 + k*(-10)| = 35
=> (1/2) * |0 + 50 - 10k| = 35
=> |50 - 10k| = 70

Case 1: 50 - 10k = 70
=> -10k = 20
=> k = -2

Case 2: 50 - 10k = -70
=> -10k = -120
=> k = 12

Therefore, k = 12 or k = -2

Correct option is (D) 12, -2 — Used area formula and solved absolute value equation for k. Both values satisfy the area condition.

Question 10

Write Minors and Cofactors of the elements of following determinants:

1. (i) $ \left| \begin{array}{cc} 2 & -4 \ 0 & 3 \end{array} \right| $ (ii) $ \left| \begin{array}{cc} a & c \ b & d \end{array} \right| $

2. (i) $ \left| \begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array} \right| $ (ii) $ \left| \begin{array}{ccc} 1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2 \end{array} \right| $

3. Using Cofactors of elements of second row, evaluate $ \Delta = \left| \begin{array}{lll} 5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3 \end{array} \right| $.

4. Using Cofactors of elements of third column, evaluate $ \Delta = \left| \begin{array}{ccc} 1 & x & yz \ 1 & y & zx \ 1 & z & xy \end{array} \right| $.

5. If $ \Delta = \left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{array} \right| $ and $ A_{ij} $ is Cofactors 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} $

Accepted Answer

1. (i) For $ \left| \begin{array}{cc} 2 & -4 \ 0 & 3 \end{array} \right| $:
- Minors:
  - Minor of 2 (element a11) is determinant of matrix formed by deleting row 1 and column 1, which is 3.
  - Minor of -4 (a12) is 0.
  - Minor of 0 (a21) is -4.
  - Minor of 3 (a22) is 2.
- Cofactors:
  - Cofactor of 2: $ (+1)^{1+1} \times 3 = 3 $
  - Cofactor of -4: $ (-1)^{1+2} \times 0 = 0 $
  - Cofactor of 0: $ (-1)^{2+1} \times (-4) = 4 $
  - Cofactor of 3: $ (+1)^{2+2} \times 2 = 2 $

(ii) For $ \left| \begin{array}{cc} a & c \ b & d \end{array} \right| $:
- Minors:
  - Minor of a (a11) is d.
  - Minor of c (a12) is b.
  - Minor of b (a21) is c.
  - Minor of d (a22) is a.
- Cofactors:
  - Cofactor of a: $ (+1)^{1+1} \times d = d $
  - Cofactor of c: $ (-1)^{1+2} \times b = -b $
  - Cofactor of b: $ (-1)^{2+1} \times c = -c $
  - Cofactor of d: $ (+1)^{2+2} \times a = a $

2. (i) For identity matrix $ I_3 $:
- Minors and cofactors of diagonal elements are 1, others are 0.
- Cofactors matrix is same as identity matrix.

(ii) For $ \left| \begin{array}{ccc} 1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2 \end{array} \right| $:
- Calculate minors and cofactors for each element by deleting corresponding row and column and finding determinant of 2x2 matrix.

3. Evaluate $ \Delta = \left| \begin{array}{lll} 5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3 \end{array} \right| $ using cofactors of second row:
- Elements of second row: 2, 0, 1
- Cofactors:
  - Cofactor of 2 (a21): $ (-1)^{2+1} \times \left| \begin{array}{cc} 3 & 8 \ 2 & 3 \end{array} \right| = - (3*3 - 8*2) = - (9 - 16) = 7 $
  - Cofactor of 0 (a22): $ (-1)^{2+2} \times \left| \begin{array}{cc} 5 & 8 \ 1 & 3 \end{array} \right| = + (5*3 - 8*1) = 7 $
  - Cofactor of 1 (a23): $ (-1)^{2+3} \times \left| \begin{array}{cc} 5 & 3 \ 1 & 2 \end{array} \right| = - (5*2 - 3*1) = - (10 - 3) = -7 $
- $ \Delta = 2 \times 7 + 0 \times 7 + 1 \times (-7) = 14 - 7 = 7 $

4. Evaluate $ \Delta = \left| \begin{array}{ccc} 1 & x & yz \ 1 & y & zx \ 1 & z & xy \end{array} \right| $ using cofactors of third column:
- Elements of third column: yz, zx, xy
- Cofactors:
  - Cofactor of yz (a13): $ (-1)^{1+3} \times \left| \begin{array}{cc} 1 & x \ 1 & y \end{array} \right| = + (1*y - 1*x) = y - x $
  - Cofactor of zx (a23): $ (-1)^{2+3} \times \left| \begin{array}{cc} 1 & x \ 1 & z \end{array} \right| = - (1*z - 1*x) = -(z - x) = x - z $
  - Cofactor of xy (a33): $ (-1)^{3+3} \times \left| \begin{array}{cc} 1 & x \ 1 & y \end{array} \right| = + (1*y - 1*x) = y - x $
- $ \Delta = yz (y - x) + zx (x - z) + xy (y - x) $
- Simplify to find the value.

5. The correct value of $ \Delta $ in terms of cofactors is given by expansion along the first row:
$ \Delta = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} $
- None of the options exactly match this, but option (D) is $ a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31} $ which is expansion along first column.
- The correct formula for expansion along first row is not given exactly, but (B) is $ a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31} $ which is incorrect because cofactors indices do not match.
- Hence, none of the options is exactly correct for expansion along first row.
- However, the standard formula is $ \Delta = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} $.

Hence correct answer is none of the above options as given, but from the options, (D) corresponds to expansion along first column, which is a valid expansion formula.

Therefore, correct answer is (D). — Step-by-step solutions:

1. For each element, minor is determinant of matrix after deleting row and column of that element. Cofactor is minor multiplied by $ (-1)^{i+j} $.

2. For 3x3 matrices, minors are 2x2 determinants. Cofactors include sign changes.

3. Use cofactor expansion along second row: multiply each element of second row by its cofactor and sum.

4. Use cofactor expansion along third column similarly.

5. Expansion of determinant along first row is $ \Delta = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} $. Options given do not exactly match this; (D) corresponds to expansion along first column.

Hence, detailed calculations and sign considerations are essential.

Question 11

Verify A  $(adj\mathrm{A}) = (adj\mathrm{A})\mathrm{A} = |\mathrm{A}|\mathrm{I}$  in 
1. $\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$
2. $\begin{bmatrix} 1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1 \end{bmatrix}$

Accepted Answer

Solution:

Let us verify the relation $A (\mathrm{adj} A) = (\mathrm{adj} A) A = |A| I$ for each matrix.

1. For $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$:

Step 1: Find $|A|$:
$|A| = (1)(4) - (2)(3) = 4 - 6 = -2$

Step 2: Find $\mathrm{adj} A$:
The adjoint of a 2x2 matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$

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

Step 3: Compute $A (\mathrm{adj} A)$:
$= \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix} = \begin{bmatrix} (1)(4)+(2)(-3) & (1)(-2)+(2)(1) \ (3)(4)+(4)(-3) & (3)(-2)+(4)(1) \end{bmatrix} = \begin{bmatrix} 4 - 6 & -2 + 2 \ 12 - 12 & -6 + 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 0 & -2 \end{bmatrix}$

Step 4: Compute $|A| I$:
$|A| I = -2 	imes \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 0 & -2 \end{bmatrix}$

Hence, $A (\mathrm{adj} A) = |A| I$.

Similarly, $(\mathrm{adj} A) A = |A| I$ can be verified.

2. For $A = \begin{bmatrix} 1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1 \end{bmatrix}$:

Step 1: Find $|A|$:
Calculate determinant by expansion:

$|A| = 1 	imes \begin{vmatrix} 3 & 5 \ 0 & 1 \end{vmatrix} - (-1) 	imes \begin{vmatrix} 2 & 5 \ -2 & 1 \end{vmatrix} + 2 	imes \begin{vmatrix} 2 & 3 \ -2 & 0 \end{vmatrix}$

Calculate minors:
$\begin{vmatrix} 3 & 5 \ 0 & 1 \end{vmatrix} = (3)(1) - (5)(0) = 3$

$\begin{vmatrix} 2 & 5 \ -2 & 1 \end{vmatrix} = (2)(1) - (5)(-2) = 2 + 10 = 12$

$\begin{vmatrix} 2 & 3 \ -2 & 0 \end{vmatrix} = (2)(0) - (3)(-2) = 0 + 6 = 6$

So,
$|A| = 1 	imes 3 + 1 	imes 12 + 2 	imes 6 = 3 + 12 + 12 = 27$

Step 2: Find $\mathrm{adj} A$:
Calculate cofactors matrix and then transpose.

Cofactors:

$C_{11} = + \begin{vmatrix} 3 & 5 \ 0 & 1 \end{vmatrix} = 3$

$C_{12} = - \begin{vmatrix} 2 & 5 \ -2 & 1 \end{vmatrix} = -12$

$C_{13} = + \begin{vmatrix} 2 & 3 \ -2 & 0 \end{vmatrix} = 6$

$C_{21} = - \begin{vmatrix} -1 & 2 \ 0 & 1 \end{vmatrix} = -((-1)(1) - (2)(0)) = -(-1) = 1$

$C_{22} = + \begin{vmatrix} 1 & 2 \ -2 & 1 \end{vmatrix} = (1)(1) - (2)(-2) = 1 + 4 = 5$

$C_{23} = - \begin{vmatrix} 1 & -1 \ -2 & 0 \end{vmatrix} = -((1)(0) - (-1)(-2)) = -(0 - 2) = -(-2) = 2$

$C_{31} = + \begin{vmatrix} -1 & 2 \ 3 & 5 \end{vmatrix} = (-1)(5) - (2)(3) = -5 - 6 = -11$

$C_{32} = - \begin{vmatrix} 1 & 2 \ 2 & 5 \end{vmatrix} = -((1)(5) - (2)(2)) = -(5 - 4) = -1$

$C_{33} = + \begin{vmatrix} 1 & -1 \ 2 & 3 \end{vmatrix} = (1)(3) - (-1)(2) = 3 + 2 = 5$

Cofactor matrix:
$\begin{bmatrix} 3 & -12 & 6 \ 1 & 5 & 2 \ -11 & -1 & 5 \end{bmatrix}$

Adjoint is transpose:
$\mathrm{adj} A = \begin{bmatrix} 3 & 1 & -11 \ -12 & 5 & -1 \ 6 & 2 & 5 \end{bmatrix}$

Step 3: Compute $A (\mathrm{adj} A)$:
Multiply $A$ and $\mathrm{adj} A$:

$= \begin{bmatrix} 1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 1 & -11 \ -12 & 5 & -1 \ 6 & 2 & 5 \end{bmatrix}$

Calculate each element:

Row 1:
$(1)(3) + (-1)(-12) + (2)(6) = 3 + 12 + 12 = 27$

$(1)(1) + (-1)(5) + (2)(2) = 1 - 5 + 4 = 0$

$(1)(-11) + (-1)(-1) + (2)(5) = -11 + 1 + 10 = 0$

Row 2:
$(2)(3) + (3)(-12) + (5)(6) = 6 - 36 + 30 = 0$

$(2)(1) + (3)(5) + (5)(2) = 2 + 15 + 10 = 27$

$(2)(-11) + (3)(-1) + (5)(5) = -22 - 3 + 25 = 0$

Row 3:
$(-2)(3) + (0)(-12) + (1)(6) = -6 + 0 + 6 = 0$

$(-2)(1) + (0)(5) + (1)(2) = -2 + 0 + 2 = 0$

$(-2)(-11) + (0)(-1) + (1)(5) = 22 + 0 + 5 = 27$

So,
$A (\mathrm{adj} A) = \begin{bmatrix} 27 & 0 & 0 \ 0 & 27 & 0 \ 0 & 0 & 27 \end{bmatrix} = 27 I$

Step 4: Since $|A| = 27$, $A (\mathrm{adj} A) = |A| I$.

Similarly, $(\mathrm{adj} A) A = |A| I$ can be verified.

Hence, the relation is verified for both matrices. — The problem asks to verify the matrix identity $A (\mathrm{adj} A) = (\mathrm{adj} A) A = |A| I$ for two given matrices.

For each matrix, we:
1. Calculate the determinant $|A|$.
2. Find the adjoint matrix $\mathrm{adj} A$ by computing cofactors and transposing.
3. Multiply $A$ by $\mathrm{adj} A$ and check if the product equals $|A|$ times the identity matrix.

The step-by-step calculations for determinants, cofactors, adjoints, and matrix multiplications confirm the identity holds in both cases.

Question 12

Find the inverse of each of the matrices (if it exists) given in
3.  $\begin{bmatrix} 2 & 3 \ -4 & -6 \end{bmatrix}$
4.  $\begin{bmatrix} 1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3 \end{bmatrix}$

Accepted Answer

3. Matrix A = $\begin{bmatrix} 2 & 3 \ -4 & -6 \end{bmatrix}$
Calculate determinant:
$\det(A) = (2)(-6) - (3)(-4) = -12 + 12 = 0$
Since determinant is zero, inverse does not exist.

4. Matrix B = $\begin{bmatrix} 1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3 \end{bmatrix}$
Calculate determinant:
$\det(B) = 1 \times \begin{vmatrix} 0 & -2 \ 0 & 3 \end{vmatrix} - (-1) \times \begin{vmatrix} 3 & -2 \ 1 & 3 \end{vmatrix} + 2 \times \begin{vmatrix} 3 & 0 \ 1 & 0 \end{vmatrix}$
$= 1(0 \times 3 - 0 \times (-2)) + 1(3 \times 3 - 1 \times (-2)) + 2(3 \times 0 - 1 \times 0)$
$= 0 + 1(9 + 2) + 2(0) = 11$
Since determinant $\neq 0$, inverse exists.

Find adjoint matrix and then inverse:
Calculate cofactors:
Cofactor matrix =
$\begin{bmatrix}
\begin{vmatrix} 0 & -2 \ 0 & 3 \end{vmatrix} & -\begin{vmatrix} 3 & -2 \ 1 & 3 \end{vmatrix} & \begin{vmatrix} 3 & 0 \ 1 & 0 \end{vmatrix} \
-\begin{vmatrix} -1 & 2 \ 0 & 3 \end{vmatrix} & \begin{vmatrix} 1 & 2 \ 1 & 3 \end{vmatrix} & -\begin{vmatrix} 1 & -1 \ 1 & 0 \end{vmatrix} \
\begin{vmatrix} -1 & 2 \ 0 & -2 \end{vmatrix} & -\begin{vmatrix} 1 & 2 \ 3 & -2 \end{vmatrix} & \begin{vmatrix} 1 & -1 \ 3 & 0 \end{vmatrix}
\end{bmatrix}$
Calculate each minor:
$C_{11} = (0)(3) - (0)(-2) = 0$
$C_{12} = -[(3)(3) - (1)(-2)] = -[9 + 2] = -11$
$C_{13} = (3)(0) - (1)(0) = 0$
$C_{21} = -[(-1)(3) - (0)(2)] = -[-3 - 0] = 3$
$C_{22} = (1)(3) - (1)(2) = 3 - 2 = 1$
$C_{23} = -[(1)(0) - (1)(-1)] = -[0 + 1] = -1$
$C_{31} = (-1)(-2) - (0)(2) = 2 - 0 = 2$
$C_{32} = -[(1)(-2) - (3)(2)] = -[-2 - 6] = 8$
$C_{33} = (1)(0) - (3)(-1) = 0 + 3 = 3$
Cofactor matrix =
$\begin{bmatrix} 0 & -11 & 0 \ 3 & 1 & -1 \ 2 & 8 & 3 \end{bmatrix}$
Transpose to get adjoint:
$\text{adj}(B) = \begin{bmatrix} 0 & 3 & 2 \ -11 & 1 & 8 \ 0 & -1 & 3 \end{bmatrix}$
Inverse:
$B^{-1} = \frac{1}{11} \times \text{adj}(B) = \frac{1}{11} \begin{bmatrix} 0 & 3 & 2 \ -11 & 1 & 8 \ 0 & -1 & 3 \end{bmatrix}$ — Step 1: Calculate determinant to check if inverse exists.
Step 2: For matrix 3, determinant is zero, so inverse does not exist.
Step 3: For matrix 4, determinant is 11, so inverse exists.
Step 4: Calculate cofactors for each element.
Step 5: Form cofactor matrix and transpose it to get adjoint.
Step 6: Multiply adjoint by 1/determinant to get inverse.

Question 13

Find the determinant of the matrix $ \begin{bmatrix} 2 & -2 \ 4 & 3 \end{bmatrix} $.

Accepted Answer

The determinant of a 2x2 matrix $ \begin{bmatrix} a & b \ c & d \end{bmatrix} $ is given by $ ad - bc $.

Here, $ a=2, b=-2, c=4, d=3 $.

So, determinant = $ (2)(3) - (4)(-2) = 6 + 8 = 14 $.

Therefore, the determinant is 14. — Using the formula for determinant of 2x2 matrix, calculate $ ad - bc $ with given values.

Question 14

Find the determinant of the matrix $ \begin{bmatrix} -1 & 5 \ -3 & 2 \end{bmatrix} $.

Accepted Answer

Using the determinant formula for 2x2 matrix:

$ a = -1, b = 5, c = -3, d = 2 $

Determinant = $ (-1)(2) - (-3)(5) = -2 + 15 = 13 $.

Hence, the determinant is 13. — Calculate $ ad - bc $ for the given matrix elements.

Question 15

Find the determinant of the matrix $ \begin{bmatrix} 1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5 \end{bmatrix} $.

Accepted Answer

The given matrix is upper triangular. The determinant of a triangular matrix is the product of the diagonal elements.

Diagonal elements are 1, 2, and 5.

Determinant = $ 1 \times 2 \times 5 = 10 $.

Therefore, the determinant is 10. — Use the property that determinant of triangular matrix equals product of diagonal entries.

Determinants

Determinants — Study Notes

4.1 Introduction

4.2 Determinant

4.2.3 Determinant of a matrix of order 3 × 3

Practice Questions — Determinants

All 6 Chapters in Mathematics Part-I