10th Class Mathematics Unit 3 Matrices and Determinants Numericals with Solutions

3.5.1.Find the values of each of the determinant. (i) |10 5; 4 6| (ii) |-5 8; -3 -7| (iii) |3 8; 0 2|

(i) |A|=(10)(6)-(5)(4)=60-20=40

(ii) |A|=(-5)(-7)-(8)(-3)=35+24=59

(iii) |A|=(3)(2)-(8)(0)=6-0=6

Result

Answers (i) 40quad (ii) 59quad (iii) 6

3.5.2.Find whether the following matrices are singular or non-singular. (i) A=[[5,3],[3,2]] (ii) B=[[7,21],[2,6]] (iii) C=[[13,5],[7,3]] (iv) D=[[2,0],[0,0]]

(i) |A|=(5)(2)-(3)(3)=10-9=1neq0 ⇒ non-singular

(ii) |B|=(7)(6)-(21)(2)=42-42=0 ⇒ singular

(iii) |C|=(13)(3)-(5)(7)=39-35=4neq0 ⇒ non-singular

(iv) |D|=(2)(0)-(0)(0)=0 ⇒ singular

A:non-singular; B:singular; C:non-singular; D:singular

3.5.3.Find the value of x when A=[[x,6],[5,15]] is a singular matrix.

Formula

For singular matrix |A|=0

Working

(x)(15)-(6)(5)=15x-30

15x-30=0 ⇒ 15x=30

Result

Answer x=2

3.5.4.Find the adjoint of the following matrices: (i) A=[[a,b],[c,d]] (ii) B=[[5,-2],[3,7]] (iii) C=[[-3,5],[-3,2]]

Formula

For 2x2 matrix adjbegin{bmatrix}a&bc&dend{bmatrix}=begin{bmatrix}d&-b-c&aend{bmatrix}

Result

(ii) adj B=begin{bmatrix}7&2-3&5end{bmatrix}

(iii) adj C=begin{bmatrix}2&-53&-3end{bmatrix}

3.5.5.Find multiplicative inverse of the following matrices: (i) A=[[5,0],[0,5]] (ii) A=[[-4,8],[7,2]] (iii) A=[[40,8],[5,2]] (iv) A=[[3,5],[5,-3]] (v) A=[[10,8],[3,3]] (vi) A=[[-2,-3],[4,5]]

(i) |A|=25 ⇒ A^-1=begin{bmatrix}1/5&00&1/5end{bmatrix}

(ii) |A|=(-4)(2)-(8)(7)=-64 ⇒ A^-1=begin{bmatrix}-1/32&1/87/64&1/16end{bmatrix}

(iii) |A|=80-40=40 ⇒ A^-1=begin{bmatrix}1/20&-1/5-1/8&1end{bmatrix}

(iv) |A|=-9-25=-34 ⇒ A^-1=begin{bmatrix}3/34&5/345/34&-3/34end{bmatrix}

(v) |A|=30-24=6 ⇒ A^-1=begin{bmatrix}1/2&-4/3-1/2&5/3end{bmatrix}

(vi) |A|=-10+12=2 ⇒ A^-1=begin{bmatrix}5/2&3/2-2&-1end{bmatrix}

(i)begin{bmatrix}1/5&00&1/5end{bmatrix} (ii)begin{bmatrix}-1/32&1/87/64&1/16end{bmatrix} (iii)begin{bmatrix}1/20&-1/5-1/8&1end{bmatrix} (iv)begin{bmatrix}3/34&5/345/34&-3/34end{bmatrix} (v)begin{bmatrix}1/2&-4/3-1/2&5/3end{bmatrix} (vi)begin{bmatrix}5/2&3/2-2&-1end{bmatrix}

3.5.6.If A=[[5,-3],[2,-1]], then find A^-1 and prove that AA^-1 = A^-1A = I.

Determinant |A|=(5)(-1)-(-3)(2)=-5+6=1neq0

Formula

Inverse A^-1=1|A|begin{bmatrix}d&-b-c&aend{bmatrix}=begin{bmatrix}-1&3-2&5end{bmatrix}

AA^-1 AA^-1=begin{bmatrix}5(-1)+(-3)(-2)&5(3)+(-3)(5)2(-1)+(-1)(-2)&2(3)+(-1)(5)end{bmatrix}=begin{bmatrix}1&00&1end{bmatrix}

A^-1A A^-1A=begin{bmatrix}-1(5)+3(2)&-1(-3)+3(-1)-2(5)+5(2)&-2(-3)+5(-1)end{bmatrix}=begin{bmatrix}1&00&1end{bmatrix}

Result

Conclusion AA^-1=A^-1A=I

3.5.7.Show that the following matrices are multiplicative inverse of each other. (i) A=[[2,-1],[-3,2]], B=[[2,1],[3,2]]

AB AB=begin{bmatrix}2(2)+(-1)(3)&2(1)+(-1)(2)-3(2)+2(3)&-3(1)+2(2)end{bmatrix}=begin{bmatrix}1&00&1end{bmatrix}=I

BA BA=begin{bmatrix}2(2)+1(-3)&2(-1)+1(2)3(2)+2(-3)&3(-1)+2(2)end{bmatrix}=begin{bmatrix}1&00&1end{bmatrix}=I

Result

Conclusion A and B are multiplicative inverses

3.5.8.Prove that (AB)^-1 = B^-1 A^-1, if (i) A=[[-3,-2],[5,6]], B=[[2,-1],[-3,2]]

Step 1: A^-1, B^-1 |A|=-18+10=-8 ⇒ A^-1=begin{bmatrix}-3/4&-1/45/8&3/8end{bmatrix};quad |B|=4-3=1 ⇒ B^-1=begin{bmatrix}2&13&2end{bmatrix}

Step 2: (AB)^-1 AB=begin{bmatrix}0&-1-8&7end{bmatrix}; |AB|=-8 ⇒ (AB)^-1=begin{bmatrix}-7/8&-1/8-1&0end{bmatrix}

Step 3: B^-1 A^-1 B^-1A^-1=begin{bmatrix}-7/8&-1/8-1&0end{bmatrix}

Result

Conclusion (AB)^-1=B^-1A^-1

Unit 3: Matrices and Determinants — Exercise 3 5