Chapter 4 Determinants and Matrices Ex 4.4 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.4

ii. aij= i – 3j
∴ a11= 1 – 3(1) = 1 – 3 = -2,
a12= 1 – 3(2) = 1 – 6 = -5,
a21= 2 – 3(1) = 2 – 3 =-1,
a22= 2 – 3(2) = 2 – 6 = – 4
a31= 3 – 3(1) = 3-3 = 0,
a32= 3 – 3(2) = 3 – 6 = -3
∴ A = ≤ft[{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
{array}

iii. a_ij=

ii. ≤ft[{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
{array}
Solution:
Let A = ≤ft[{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
{array}
∴ AT= ≤ft[{array}{ccc}
0 & -4 & -7 \\
4 & 0 & 3 \\
7 & -3 & 0
{array}
∴ AT= -≤ft[{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
{array}
∴ AT= -A, i.e., A = -AT
∴ A is a skew-symmetric matrix.

iii. ≤ft[{array}{c}
5 \\
4 \\
-3
{array}
Solution:
Let A = ≤ft[{array}{c}
5 \\
4 \\
-3
{array}
∴ As matrix A has only one column.
∴ A is a column matrix.

iv. ≤ft[{array}{lll}
9 & & -3
{array}
Solution:
Let A = ≤ft[{array}{lll}
9 & & -3
{array}
As matrix A has only one row.
∴ A is a row matrix.

v. ≤ft[{array}{ll}
6 & 0 \\
0 & 6
{array}
Solution:
Let A = ≤ft[{array}{ll}
6 & 0 \\
0 & 6
{array}
As matrix A has all its non-diagonal elements zero and diagonal elements same.
∴ A is a scalar matrix.

vi. ≤ft[{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
{array}
Solution:
Let A = ≤ft[{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
{array}
As every element above the diagonal is zero in matrix A.
∴ A is a lower triangular matrix.

vii. ≤ft[{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 &
{array}
Solution:
Let A = ≤ft[{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 &
{array}
As matrix A has all its non-diagonal elements zero.
∴ A is a diagonal matrix.

viii. ≤ft[{array}{ccc}
10 & -15 & 27 \\
-15 & 0 & \\
27 & &
{array}
Solution:
∴ AT= A, i/e., A = AT
∴ A is a symmetric matrix.

ix. ≤ft[{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
{array}
Solution:
A = ≤ft[{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
{array}
In matrix A, all the non-diagonal elements are zero and diagonal elements are one.
∴ A is a unit (identity) matrix.

x. ≤ft[{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
{array}
Solution:
Let A = ≤ft[{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
{array}
∴ AT= A, i/e., A = AT
∴ A is a symmetric matrix.
∴ A is a symmetric matrix.

ii. Let A = ≤ft[{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
{array}
∴ |A| = ≤ft|{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
{array}
Applying C2→ C2+ C1
|A| = ≤ft|{array}{ccc}
5 & 5 & 5 \\
1 & 100 & 100 \\
6 & 105 & 105
{array}
= 0 … [∵ C2and C3are identical]
∴ A is a singular matrix.

iii. Let A = ≤ft[{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
{array}
∴ |A| = ≤ft|{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
{array}
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49 = 52 ≠ 0
∴ A is a non-singular matrix.

iv. Let A = ≤ft[{array}{cc}
7 & 5 \\
-4 & 7
{array}
∴ |A| = ≤ft[{array}{cc}
7 & 5 \\
-4 & 7
{array} = 49 + 20 = 69 ≠ 0

ii. Let A = ≤ft[{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
{array}
Since A is a singular matrix,
|A|= 0
∴ ≤ft|{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
{array} = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6

iii. Let A = ≤ft[{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
{array}
Since A is a singular matrix
|A| = 0
∴ ≤ft|{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
{array}
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3 (-6 – 1) = 0
∴ 8k-8-20-21 =0
∴ 8k = 49
∴ k = 49/8

ii.

∴ A ≠ A^T, i.e., A ≠ -A^T
∴ A is neither a symmetric nor skew-symmetric matrix.

iii.

∴ A^T= -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.