Maharashtra State Board Class 12 (HSC Board) Mathematics & Statistics Solutions Chapter 2 Matrices Ex 2.2 - Digest Answers | Maharashtra Board Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Ex 2.2

Question 1 Maharashtra Board Solution

Find the co-factors of the elements of the following matrices (i)

Solution & Step-by-Step Answer:

Let A = Here, a11 = -11, M11 = 4 ∴ A11 = (-1)1+1(4) = 4 a12 = 2, M12 = -3 ∴ A12 = (-1)1+2(- 3) = 3 a21 = – 3, M21 = -2 ∴ A21 = (- 1)2+1(2) = -2 a22 = 4, M22 = -1 ∴ A22 = (-1)2+2(-1) = -1.

(ii) ≤ft[{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
{array}
Solution:
Let A = ≤ft[{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
{array}
The co-factor of aijis given by Aij= (-1)i+jMij

Question 2 Maharashtra Board Solution

Find the matrix of co-factors for the following matrices (i)

Solution & Step-by-Step Answer:

Let A = Here, a11 = 1, M11 = -1 ∴ A11 = (-1)1+1(-1) = -1 a12 = 3, M12 = 4 ∴ A12 = (-1)1+2(4) = -4 a21 = 4, M21 = 3 ∴ A21 = (-1)2+1(3) = -3 a22 = -1, M22 = 1 ∴ A22 = (-1)2+1(1) = 1 ∴ the co-factor matrix = =

(ii) ≤ft[{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
{array}
Solution:
Let A = ≤ft[{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
{array}
A11= -14, A12= -10, A13= -6,
A21= 6, A22= -5, A23= -3,
A31= -2, A32= -7, A33= 1.
∴ the co-factor matrix
= ≤ft[{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
{array}] = ≤ft[{array}{rrr}
-14 & -10 & -6 \\
6 & -5 & -3 \\
-2 & -7 & 1
{array}

Question 3 Maharashtra Board Solution

Find the adjoint of the following matrices. (i)

Solution & Step-by-Step Answer:

Let A = Here, a11 = 2, M11= 5 ∴ A11 = (-1)1+1(5) = 5 a12 = -3, M12 = 3 ∴ A12 = (-1)1+2(3) = -3 a21 = 3, M21 = -3 ∴ A A21 = (-1)2+1(-3) = 3 a22 = 5, M22 = 2 ∴ A22 = (-1)2+1 = 2 ∴ the co-factor matrix = = ∴ adj A =

(ii) ≤ft[{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
{array}
Solution:
A11= -3, A12= -12, A13= 6,
A21= -1, A22= 3, A23= 2,
A31= -11, A32= -9, A33= 1
∴ the co-factor matrix = ≤ft[{array}{lll}
A_{11} & ~A_{12} & ~A_{15} \\
~A_{21} & ~A_{22} & ~A_{23} \\
~A_{31} & ~A_{32} & ~A_{33}
{array}
= ≤ft[{array}{rrr}
-3 & -12 & 6 \\
-1 & 3 & 2 \\
-11 & -9 & 1
{array}
∴ adj A = ≤ft[{array}{rrr}
-3 & -1 & -11 \\
-12 & 3 & -9 \\
6 & 2 & 1
{array}

Question 4 Maharashtra Board Solution

If A = , verify that A (adj A) = (adj A) A = | A | ∙ I

Solution & Step-by-Step Answer:

A = From (1), (2) and (3), we get, A(adj A) = (adj A)A = |A|∙I. Note: This relation is valid for any non-singular matrix A.

Question 5 Maharashtra Board Solution

Find the inverse of the following matrices by the adjoint method (i)

Solution & Step-by-Step Answer:

Let A = ∴ |A| = = -2 + 15 = 13 ≠ 0 ∴ A-1 exists. First we have to find the co-factor matrix = [Aij]2×2, where Aij = (-1)i+jMij Now, A11 = (-1)1+1M11 = 2 A12 = (-1)1+2M12 = -(-3) = 3 A21 = (-1)2+1M21 = -5 A22 = (-1)2+2M22 = -1 Hence, the co-factor matrix

(ii) ≤ft[{array}{cc}
2 & -2 \\
4 & 3
{array}
Solution:
Let A = ≤ft[{array}{cc}
2 & -2 \\
4 & 3
{array}
|A| = = 6 + 8 = 14 ≠ 0
∴ A-1exist
First we have to find the co-factor matrix
= [Aij]2×2where Aij= (-1)i+jMij
Now, A11= (-1)1+1M11= 3
A12= (-1)1+2M = -4
A21= (-2)2+1M21= (-2) = 2
A22= (-1)2+2M22= 2
Hence the co-factor matrix
= ≤ft[{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
{array}] = ≤ft[{array}{cc}
3 & -4 \\
2 & 2
{array}
∴ adj A = ≤ft[{array}{cc}
3 & 2 \\
-4 & 2
{array}
∴ A-1= (adj A) = ≤ft({array}{cc}
3 & 2 \\
-4 & 2
{array}

(iii) ≤ft[{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
{array}
Solution:
Let A = ≤ft[{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
{array}
∴ A-1= ≤ft[{array}{rrr}
3 & 0 & 0 \\
-3 & 1 & 0 \\
9 & 2 & -3
{array}

(iv) ≤ft[{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
{array}
Solution:
Let A = ≤ft[{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
{array}
∴ |A| = ≤ft[{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
{array}
= 1(10 – 0) – 0 + 0
= 1(10) – 0 + 0
= 10 ≠ 0
∴ A-1exists.
First we have to find the co-factor matrix
∴ A-1= (adj A)
= ≤ft({array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
{array}
∴ A-1= ≤ft({array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
{array}

Question 6 Maharashtra Board Solution

Find the inverse of the following matrices (i)

Solution & Step-by-Step Answer:

Let A =

(ii) ≤ft[{array}{cc}
2 & -3 \\
-1 & 2
{array}
Solution:
Let A = ≤ft[{array}{cc}
2 & -3 \\
-1 & 2
{array}
∴ A-1= ≤ft({array}{ll}
2 & 3 \\
1 & 2
{array}

(iii) ≤ft[{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
{array}
Solution:
Let A = ≤ft[{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
{array}

(iv) ≤ft[{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
{array}
Solution:
Let A = ≤ft[{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
{array}

Chapter 2 Matrices Ex 2.2 Solutions