NCERT Solutions Matrices Exercise 3.3 Class 12 Math Chapter 3 free PDF Download

Chapter 3 of the Class 12 NCERT Mathematics textbook, titled “Matrices,” delves into the fundamental concepts of matrices, including their types, operations, and applications. Exercise 3.3 focuses on practical problems involving matrix operations, such as addition, subtraction, and multiplication of matrices. This exercise helps students apply their understanding of matrices to solve various mathematical problems.

NCERT Solutions for Mathematics – Chapter 3 Matrices – Exercise 3.3

This section provides detailed solutions for Exercise 3.3 from Chapter 3 of the Class 12 NCERT Mathematics textbook. The exercise includes a variety of problems related to matrix operations, offering step-by-step explanations to ensure students can effectively solve these problems and grasp the underlying concepts of matrix algebra.

Question 1. Find the transpose of each of the following matrices:

(i) [Tex]

⎡ ⎣ ⎢ 5 1 2 - 1 ⎤ ⎦ ⎥

[/Tex]

(ii) [Tex]

[12 - 1 3]

[/Tex]

(iii) [Tex]

⎡ ⎣ ⎢ - 1 3 - \sqrt 2 553 66 - 1 ⎤ ⎦ ⎥

[/Tex]

Solution:

(i) Let A =[Tex]
$⎡ ⎣ ⎢ 5 1 2 - 1 ⎤ ⎦ ⎥$ [/Tex]
∴Transpose of A = A’ = AT = [Tex]
$[5 1 2 - 1]$ [/Tex]
(ii) Let A =[Tex]
$[12 - 1 3]$ [/Tex]
∴Transpose of A = A’ = AT =[Tex]
$[1 - 1 23]$ [/Tex]
(iii) Let A =[Tex]
$⎡ ⎣ ⎢ - 1 3 - \sqrt 2 553 66 - 1 ⎤ ⎦ ⎥$ [/Tex]
∴Transpose of A = A’ = AT =[Tex]
$⎡ ⎣ ⎢ - 1 56 3 - \sqrt 56 23 - 1 ⎤ ⎦ ⎥$ [/Tex]

Question 2. If A =[Tex] $⎡ ⎣ ⎢ - 1 5 - 2 271391 ⎤ ⎦ ⎥$ [/Tex] and B = [Tex] $⎡ ⎣ ⎢ - 4 11 223 - 5 01 ⎤ ⎦ ⎥$ [/Tex] then verify that:

(i) (A+B)’ = A’+B’

(ii) (A-B)’ = A’- B’

Solution:

(i) A+B =[Tex]
$⎡ ⎣ ⎢ - 1 5 - 2 271391 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ - 4 11 123 - 5 01 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ - 1 - 4 5 + 1 - 2 + 1 2 = 1 7 + 2 1 + 3 3 - 5 9 + 0 1 + 1 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ - 5 6 - 1 394 - 2 92 ⎤ ⎦ ⎥$ [/Tex]
L.H.S. = (A+B)’ = [Tex]
$⎡ ⎣ ⎢ - 5 3 - 2 699 - 1 42 ⎤ ⎦ ⎥$ [/Tex]
R.H.S. = A’+B’ = [Tex]
$⎡ ⎣ ⎢ - 1 2 - 2 571 - 2 11 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ - 4 1 - 5 120131 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ - 1 - 4 2 + 1 - 2 - 5 5 + 1 7 + 2 1 + 0 - 2 + 1 1 + 3 1 + 1 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ - 5 3 - 2 699 - 1 42 ⎤ ⎦ ⎥$ [/Tex]
∴L.H.S = R.H.S.
Hence, proved.
(ii) A-B = [Tex]
$⎡ ⎣ ⎢ - 1 5 - 2 271391 ⎤ ⎦ ⎥$ - $⎡ ⎣ ⎢ - 4 11 223 - 5 01 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 34 - 3 15 - 2 890 ⎤ ⎦ ⎥$ [/Tex]
L.H.S. = (A-B)’[Tex]=
$⎡ ⎣ ⎢ 318459 - 3 - 2 0 ⎤ ⎦ ⎥$ [/Tex]
R.H.S. = A’-B’ =[Tex]
$⎡ ⎣ ⎢ - 1 2 - 2 571 - 2 11 ⎤ ⎦ ⎥$ - $⎡ ⎣ ⎢ - 4 1 - 5 120131 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 318459 - 3 - 2 0 ⎤ ⎦ ⎥$ [/Tex]
∴ L.H.S. = R.H.S.
Hence, proved.

Question 3. If A’ =[Tex] $⎡ ⎣ ⎢ 3 - 1 0 421 ⎤ ⎦ ⎥$ [/Tex] and B = [Tex] $[- 1 1 2213]$ [/Tex], then verify that:

(i) (A+B)’=A’+B’

(ii) (A-B)’=A’-B’

Solution:

Given A’=[Tex]
$⎡ ⎣ ⎢ 3 - 1 0 421 ⎤ ⎦ ⎥$ [/Tex]and B=[Tex] $[- 1 1 2213]$ [/Tex]
then, (A’)’ = A =[Tex]
$[34 - 1 2 01]$ [/Tex]
(i) A+B =[Tex]
$[34 - 1 2 01]$ + $[- 1 1 2213]$ = $[251414]$ [/Tex]
∴ L.H.S. = (A+B)’=[Tex]
$⎡ ⎣ ⎢ 211544 ⎤ ⎦ ⎥$ [/Tex]
R.H.S.= A’+B’ = [Tex]
$⎡ ⎣ ⎢ 3 - 1 0 421 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ - 1 21 123 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 211544 ⎤ ⎦ ⎥$ [/Tex]
∴ L.H.S. = R.H.S.
Hence, proved.
(ii) A-B = [Tex]
$[34 - 1 2 01]$ - $[- 1 1 2213]$ = $[43 - 3 0 - 1 - 2]$ [/Tex]
∴ L.H.S. = (A-B)’=[Tex]
$⎡ ⎣ ⎢ 4 - 3 - 1 30 - 2 ⎤ ⎦ ⎥$ [/Tex]
R.H.S.= A’-B’ = [Tex]
$⎡ ⎣ ⎢ 3 - 1 0 421 ⎤ ⎦ ⎥$ - $⎡ ⎣ ⎢ - 1 21 123 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 4 - 3 - 1 30 - 2 ⎤ ⎦ ⎥$ [/Tex]
∴ L.H.S. = R.H.S.
Hence, proved.

Question 4. If A’ = [Tex] $[- 2 1 32]$ [/Tex]and B = [Tex] $[- 1 1 02]$ [/Tex]then find (A+2B)’.

Solution:

Given: A’ =[Tex]
$[- 2 1 32]$ [/Tex]and B =[Tex] $[- 1 1 02]$ [/Tex]
then (A’)’ =A=[Tex]
$[- 2 1 32]$ [/Tex]
Now, A+2B = [Tex]
$[- 2 1 32]$ +2 $[- 1 1 02]$ = $[- 2 1 32]$ + $[- 2 2 04]$ = $[- 2 - 2 3 + 2 1 + 0 2 + 4]$ = $[- 4 5 16]$ [/Tex]
∴(A+2B)’ = [Tex]
$[- 4 1 56]$ [/Tex]

Question 5. For the matrices A and B, verify that (AB)′ = B′A′, where

(i) A =[Tex]

⎡ ⎣ ⎢ 1 - 4 3 ⎤ ⎦ ⎥

[/Tex] and B = [Tex]

[- 1 21]

[/Tex]

(ii) A =[Tex]

⎡ ⎣ ⎢ 012 ⎤ ⎦ ⎥

[/Tex] and B =[Tex]

[157]

[/Tex]

Solution:

(i) AB = =[Tex]
$⎡ ⎣ ⎢ 1 - 4 3 ⎤ ⎦ ⎥$ $[- 1 21]$ = $⎡ ⎣ ⎢ - 1 4 - 3 2 - 8 6 1 - 4 3 ⎤ ⎦ ⎥$ [/Tex]
∴ L.H.S. = (AB)′ =[Tex]
$⎡ ⎣ ⎢ - 1 21 4 - 8 - 4 - 3 63 ⎤ ⎦ ⎥$ [/Tex]
R.H.S.= B′A’ = [Tex]
$⎡ ⎣ ⎢ - 1 21 ⎤ ⎦ ⎥$ $[1 - 4 3]$ = $⎡ ⎣ ⎢ - 1 21 4 - 8 - 4 - 3 63 ⎤ ⎦ ⎥$ [/Tex]
∴ L.H.S. = R.H.S.
Hence, proved.
(ii) AB =[Tex]
$⎡ ⎣ ⎢ 012 ⎤ ⎦ ⎥$ $[157]$ = $⎡ ⎣ ⎢ 01205100714 ⎤ ⎦ ⎥$ [/Tex]
∴ L.H.S. = (AB)′ =[Tex]
$⎡ ⎣ ⎢ 00015721014 ⎤ ⎦ ⎥$ [/Tex]
Now, R.H.S.=B’A’ = [Tex]
$⎡ ⎣ ⎢ 157 ⎤ ⎦ ⎥$ $[012]$ = $⎡ ⎣ ⎢ 0001572714 ⎤ ⎦ ⎥$ [/Tex]
∴ L.H.S. = R.H.S.
Hence, proved.

Question 6. If (i) A =[Tex] $[c o s α - s i n α s i n α c o s α]$ [/Tex] , then verify that A′ A = I.

(ii) A =[Tex] $[s i n α - c o s α c o s α s i n α]$ [/Tex] ,then verify that A′ A = I.

Solution:

(i) [Tex]
$[c o s α s i n α - s i n α c o s α]$ $[c o s α - s i n α s i n α c o s α]$ = $[c o s 2 α + s i n 2 α s i n α c o s α - c o s α s i n α c o s α s i n α - s i n α c o s α s i n 2 α + c o s 2 α]$ = $[1001]$ [/Tex]
= I = R.H.S.
∴ L.H.S. = R.H.S.
(ii) [Tex]
$[s i n α c o s α - c o s α s i n α]$ $[s i n α - c o s α c o s α s i n α]$ = $[s i n 2 α + s i n 2 α c o s α s i n α - s i n α c o s α s i n α c o s α - c o s α s i n α c o s 2 α + s i n 2 α]$ = $[1001]$ [/Tex]
= I = R.H.S.
∴ L.H.S. = R.H.S.

Question 7. (i) Show that the matrix A[Tex] $⎡ ⎣ ⎢ 1 - 1 5 - 1 21 513 ⎤ ⎦ ⎥$ [/Tex] = is a symmetric matrix.

(ii) Show that the matrix A[Tex] $⎡ ⎣ ⎢ 0 - 1 1 10 - 1 - 1 10 ⎤ ⎦ ⎥$ [/Tex] = is a symmetric matrix.

(i) Given: A =[Tex]
$⎡ ⎣ ⎢ 1 - 1 5 - 1 21 513 ⎤ ⎦ ⎥$ [/Tex]
Now, A’=[Tex]
$⎡ ⎣ ⎢ 1 - 1 5 - 1 21 513 ⎤ ⎦ ⎥$ ’= $⎡ ⎣ ⎢ 1 - 1 5 - 1 21 513 ⎤ ⎦ ⎥$ [/Tex]
∵ A = A’
∴ A is a symmetric matrix.
(ii) Given: A = [Tex]
$⎡ ⎣ ⎢ 0 - 1 1 10 - 1 - 1 10 ⎤ ⎦ ⎥$ [/Tex]
Now, A’=[Tex]
$⎡ ⎣ ⎢ 0 - 1 1 10 - 1 - 1 10 ⎤ ⎦ ⎥$ ’= $⎡ ⎣ ⎢ 0 - 1 1 10 - 1 - 1 10 ⎤ ⎦ ⎥$ [/Tex]
∵ A = A’
∴ A is a symmetric matrix.

Question 8. For the matrix A =[Tex] $[1657]$ [/Tex], verify that:

(i) (A + A′) is a symmetric matrix

(ii) (A – A′) is a skew symmetric matrix

Solution:

(i) Given: A =[Tex]
$[1657]$ [/Tex]
Let B = (A+A’) = [Tex]
$[1657]$ + $[1567]$ = $[1 + 1 6 + 5 5 + 6 7 + 7]$ = $[2111114]$ [/Tex]
Now, B’ = (A+A’)’ = [Tex]
$[2111114]$ ’= $[2111114]$ [/Tex]
∵ B = B’
∴ B=(A+A’) is a symmetric matrix.
(ii) Given: A =[Tex]
$[1657]$ [/Tex]
Let B = (A-A’) =[Tex]
$[1657]$ - $[1567]$ = $[1 - 1 6 - 5 5 - 6 7 - 7]$ = $[01 - 1 0]$ [/Tex]
Now, B’ = (A-A’)’ =[Tex]
$[01 - 1 0]$ ’= $[0 - 1 10]$ =- $[01 - 1 0]$ [/Tex]
∵ -B = B’
∴ B=(A-A’) is a skew symmetric matrix.

Question 9. Find 1/2(A+A’) and 1/2(A-A’) ,when A =[Tex] $⎡ ⎣ ⎢ 0 - a - b a 0 - c b c 0 ⎤ ⎦ ⎥$ [/Tex].

Solution:

Given: A = [Tex]
$⎡ ⎣ ⎢ 0 - a - b a 0 - c b c 0 ⎤ ⎦ ⎥$ [/Tex]
∴ A’ = [Tex]
$⎡ ⎣ ⎢ 0 - a - b a 0 - c b c 0 ⎤ ⎦ ⎥$ ’= $⎡ ⎣ ⎢ 0 a b - a 0 c - b - c 0 ⎤ ⎦ ⎥$ [/Tex]
Now, A+A’ = +[Tex]\frac{1}{2}\{
$⎡ ⎣ ⎢ 0 - a - b a 0 - c b c 0 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ 0 a b - a 0 c - b - c 0 ⎤ ⎦ ⎥$ \}=\frac{1}{2} $⎡ ⎣ ⎢ 0 + 0 - a + a - b + b a - a 0 + 0 - c + c b - b c - c 0 + 0 ⎤ ⎦ ⎥$ =\frac{1}{2} $⎡ ⎣ ⎢ 000000000 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 000000000 ⎤ ⎦ ⎥$ [/Tex]
Now, A-A’ =[Tex]\frac{1}{2}\{
$⎡ ⎣ ⎢ 0 - a - b a 0 - c b c 0 ⎤ ⎦ ⎥$ - $⎡ ⎣ ⎢ 0 a b - a 0 c - b - c 0 ⎤ ⎦ ⎥$ \}=\frac{1}{2} $⎡ ⎣ ⎢ 0 - 0 - a - a - b - b a + a 0 - 0 - c - c b + b c + c 0 - 0 ⎤ ⎦ ⎥$ =\frac{1}{2} $⎡ ⎣ ⎢ 0 - 2 a - 2 b 2 a 0 - 2 c 2 b 2 c 0 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 0 - a - b a 0 - c b c 0 ⎤ ⎦ ⎥$ [/Tex]

Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) [Tex]

[31 5 - 1]

[/Tex]

(ii) [Tex]

⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥

[/Tex]

(iii) [Tex]

⎡ ⎣ ⎢ 3 - 2 - 4 3 - 2 - 5 - 1 12 ⎤ ⎦ ⎥

[/Tex]

(iv) [Tex]

[1 - 1 52]

[/Tex]

Solution:

(i) Given : A =[Tex]
$[31 5 - 1]$ [/Tex]
⇒ A’=[Tex]
$[35 1 - 1]$ [/Tex]
Let P = [Tex]\frac{1}{2}(A+A’)[/Tex]
and Q = [Tex]\frac{1}{2}(A-A’)[/Tex]
Now, P =[Tex]\frac{1}{2} \{
$[31 5 - 1]$ + $[35 1 - 1]$ \}=\frac{1}{2} $[66 6 - 2]$ = $[33 3 - 1]$     [/Tex]…..(1)
& P’ = [Tex]
$[35 1 - 1]$ ’= $[35 1 - 1]$ [/Tex]
∵ P=P’
∴ P is a symmetric matrix.
Now, Q =[Tex]\frac{1}{2}(A-A’)=\frac{1}{2} \{
$[31 5 - 1]$ - $[35 1 - 1]$ \}=\frac{1}{2} $[0 - 4 40]$ = $[0 - 2 20]$     [/Tex]…..(2)
& Q’ = [Tex]
$[0 - 2 20]$ ’=- $[02 - 2 0]$ [/Tex]
∵ -Q=Q’
∴ Q is a skew symmetric matrix.
By adding (1) and (2), we get,
[Tex]
$[35 1 - 1]$ + $[0 - 2 20]$ = $[31 5 - 1]$ [/Tex]
Therefore, A =P + Q
(ii) Given : [Tex]
$⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ [/Tex]
⇒ A’=[Tex]
$⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ [/Tex]
P = [Tex]\frac{1}{2}(A+A’)=\frac{1}{2}(
$⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ )=\frac{1}{2} $⎡ ⎣ ⎢ 12 - 4 4 - 4 6 - 2 4 - 2 6 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ [/Tex]
…..(1)
Q = [Tex]\frac{1}{2}(A-A’)=\frac{1}{2}(
$⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ - $⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ )=\frac{1}{2} $⎡ ⎣ ⎢ 000000000 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 000000000 ⎤ ⎦ ⎥$ [/Tex]
……(2)
By adding (1) and (2), we get,
$⎡ ⎣ ⎢ 000000000 ⎤ ⎦ ⎥$ [Tex] $⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ 000000000 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 6 - 2 2 - 2 3 - 1 2 - 1 3 ⎤ ⎦ ⎥$ [/Tex]
Therefore, A =P + Q
(iii) Given: A =[Tex]
$⎡ ⎣ ⎢ 3 - 2 - 4 3 - 2 - 5 - 1 12 ⎤ ⎦ ⎥$ [/Tex]
⇒ A’=[Tex]
$⎡ ⎣ ⎢ 33 - 1 - 2 - 2 1 - 4 - 5 2 ⎤ ⎦ ⎥$ [/Tex]
P = }[Tex]\frac{1}{2}(A+A’)=\frac{1}{2}(
$⎡ ⎣ ⎢ 3 - 2 - 4 3 - 2 - 5 - 1 12 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ 33 - 1 - 2 - 2 1 - 4 - 5 2 ⎤ ⎦ ⎥$ )= $⎡ ⎣ ⎢ 61 - 5 1 - 4 - 4 - 5 - 4 4 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 3 1 / 2 - 5 / 2 1 / 2 - 2 - 2 - 5 / 2 - 2 2 ⎤ ⎦ ⎥$     [/Tex]…..(1)
Q = [Tex]\frac{1}{2}(A-A’)=\frac{1}{2}(
$⎡ ⎣ ⎢ 3 - 2 - 4 3 - 2 - 5 - 1 12 ⎤ ⎦ ⎥$ - $⎡ ⎣ ⎢ 33 - 1 - 2 - 2 1 - 4 - 5 2 ⎤ ⎦ ⎥$ )=\frac{1}{2} $⎡ ⎣ ⎢ 0 - 5 - 3 50 - 6 360 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 0 - 5 / 2 - 3 / 2 5 / 2 0 - 3 3 / 2 30 ⎤ ⎦ ⎥$     [/Tex]……(2)
By adding (1) and (2), we get
}[Tex]
$⎡ ⎣ ⎢ 3 1 / 2 - 5 / 2 1 / 2 - 2 - 2 - 5 / 2 - 2 2 ⎤ ⎦ ⎥$ + $⎡ ⎣ ⎢ 0 - 5 / 2 - 3 / 2 5 / 2 0 - 3 3 / 2 30 ⎤ ⎦ ⎥$ = $⎡ ⎣ ⎢ 3 - 2 - 4 3 - 2 - 5 - 1 12 ⎤ ⎦ ⎥$ [/Tex]
Therefore, A =P + Q
(iv) Given: A = [Tex]
$[1 - 1 52]$ [/Tex]
⇒ A’= [Tex]
$[15 - 1 2]$ [/Tex]
P =[Tex]\frac{1}{2} \{
$[1 - 1 52]$ + $[15 - 1 2]$ \}=\frac{1}{2} $[2444]$ = $[1222]$ [/Tex]
…..(1)
Q = [Tex]\frac{1}{2}(A-A’)=\frac{1}{2} \{
$[1 - 1 52]$ - $[15 - 1 2]$ \}=\frac{1}{2} $[15 - 1 2]$ = $[0 - 3 30]$ [/Tex]
…..(2)
By adding (1) and (2), we get
[Tex]
$[1222]$ + $[0 - 3 30]$ = $[1 - 1 52]$ [/Tex]
Therefore, A =P + Q

Question 11. If A, B are symmetric matrices of same order, then AB – BA is a

(A) Skew symmetric matrix (B) Symmetric matrix

Solution:

Given: A and B are symmetric matrices.
⇒ A=A’
⇒ B=B’
Now, ( AB – BA)’ =(AB)’-(BA)’ [∵ (X-Y)’=X’-Y’]
=B’A’-A’B’ [∵ (XY)’=Y’X’]
=BA-AB [∵ Given]
= -(AB-BA)
∴(AB-BA) is a skew symmetric matrix.
∴ The option (A) is correct.

Question 12. If A =[Tex] $[c o s α s i n α - s i n α c o s α]$ [/Tex], and A + A′ = I, then the value of α is

(A)π/6 (B) π/3

Solution:

[Tex]
$[c o s α s i n α - s i n α c o s α]$ + $[c o s α - s i n α s i n α c o s α]$ = $[1001]$ [/Tex]
[Tex]
$[2 c o s α 0 0 2 c o s α]$ = $[1001]$ [/Tex]
On comparing both sides, we get
2cosα = 1
⇒ cosα = [Tex]\frac{1}{2}[/Tex]
⇒ cosα = cos[Tex]\frac{π}{3}[/Tex]
⇒ α = [Tex]\frac{π}{3}[/Tex]
∴ The option (B) is correct.

Summary

Chapter 3 of the Class 12 NCERT Mathematics textbook, “Matrices,” explores essential concepts such as matrix operations and their applications. Exercise 3.3 focuses on practical problems involving matrix addition, subtraction, and multiplication. This exercise provides step-by-step solutions to help students understand and apply matrix operations effectively. Key topics include matrix addition and subtraction, matrix multiplication, determinants, and finding matrix inverses.

FAQs on Matrices

What are matrices, and why are they important in mathematics?

Matrices are rectangular arrays of numbers arranged in rows and columns. They are important in mathematics because they provide a systematic way to handle and solve systems of linear equations, perform linear transformations, and represent data in various applications.

Can matrices be multiplied if their dimensions do not match?

Matrices can only be multiplied if the number of columns in the first matrix is equal to the number of rows in the second matrix. If this condition is not met, matrix multiplication is not possible.

How do you find the inverse of a matrix, and when does a matrix have an inverse?

To find the inverse of a matrix A, the matrix must be square (same number of rows and columns) and have a non-zero determinant. The inverse of A can be found using various methods, such as the adjoint method or Gaussian elimination. The matrix A has an inverse if and only if its determinant is non-zero.

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