For a 2 × 2 matrix A = a i j , whose elements are given by a i j = i j , Find matrix A and write the value of a 12 .

Given A is a matrix of order 2 × 2 . The general matrix of order 2 × 2 is given by A = a 11 a 12 a 21 a 22 , where a i j = i j for 1 ≤ i ≤ 2 , and 1 ≤ j ≤ 2 Putting, i = 1 , j = 1 ⇒ a 11 = 1 i = 2 , j = 1 ⇒ a 21 = 2 i = 1 , j = 2 ⇒ a 12 = 1 2 i = 2 , j = 2 ⇒ a 22 = 1 So, the matrix is A = 1 1 2 2 1 , and a 12 = 1 2

For a 2 × 2 matrix A = a i j , whose elements are given by a i j = i j , Find matrix A and write the value of a 12 .

Given A is a matrix of order 2 × 2 . The general matrix of order 2 × 2 is given by A = a 11 a 12 a 21 a 22 , where a i j = i j for 1 ≤ i ≤ 2 , and 1 ≤ j ≤ 2 Putting, i = 1 , j = 1 ⇒ a 11 = 1 i = 2 , j = 1 ⇒ a 21 = 2 i = 1 , j = 2 ⇒ a 12 = 1 2 i = 2 , j = 2 ⇒ a 22 = 1 So, the matrix is A = 1 1 2 2 1 , and a 12 = 1 2

If A = 3 2 0 2 , B = − 2 1 0 3 and 2 A − B + X = O , then find the matrix X .

Given, A = 3 2 0 2 , B = - 2 1 0 3 , and 2 A - B + X = O We need to find the matrix X Consider 2 A - B + X = O ⇒ 2 3 2 0 2 - - 2 1 0 3 + X = 0 0 0 0 We know that in scalar multiplication, each entry of the matrix is multiplied by the given scalar. ⇒ 6 4 0 4 - - 2 1 0 3 + X = 0 0 0 0 We know that to add or subtract two matrices, we add or subtract their corresponding elements. ⇒ 6 - - 2 4 - 1 0 - 0 4 - 3 + X = 0 0 0 0 ⇒ 8 3 0 1 + X = 0 0 0 0 ⇒ X = 0 0 0 0 - 8 3 0 1 = 0 - 8 0 - 3 0 - 0 0 - 1 = - 8 - 3 0 - 1 Hence, the required matrix X is - 8 - 3 0 - 1

Find the values of x and y when 2 1 3 0 x + y 0 1 2 = 5 6 1 8

Given that: 2 1 3 0 x + y 0 1 2 = 5 6 1 8 We have to find the values of x and y 2 1 + y 3 1 x + 2 = 5 6 1 8 ⇒ 2 1 + y 6 2 2 x + 2 = 5 6 1 8 We know that, Two matrices are equal if they have the same dimension or order and the corresponding elements are identical. Thus, we have 2 1 + y = 5 and 2 x + 2 = 8 ⇒ y = 3 2 and x = 2 Hence, the required value of x and y are 2 and 3 2 .

Find the values of x and y when: x y 3 y x 1 2 = 3 5

Given, x y 3 y x 1 2 = 3 5 ⇒ x × 1 + y × 2 3 y × 1 + x × 2 = 3 5 ⇒ x + 2 y 3 y + 2 x = 3 5 We know that two matrices are equal if and only if they are of the same order, and their corresponding elements are equal. Thus, we have x + 2 y = 3 → i and 2 x + 3 y = 5 → i i Now, applying 2 i - i i , we get 2 x + 4 y - 2 x + 3 y = 6 - 5 ⇒ y = 1 Thus, x + 2 1 = 3 ⇒ x = 1 Hence, x = 1 , and y = 1

Find A + B and A - B in the following case: A = 1 5 2 0 , B = 7 2 − 3 5

Given A = 1 5 2 0 and B = 7 2 − 3 5 Now, A + B = 1 5 2 0 + 7 2 - 3 5 = 1 + 7 5 + 2 2 - 3 0 + 5 = 8 7 − 1 5 And A - B = 1 5 2 0 - 7 2 - 3 5 = 1 - 7 5 - 2 2 + 3 0 - 5 = − 6 3 5 − 5 .

Find A + B and A - B in the following case: A = x 1 y 1 x 2 y 2 x 3 y 3 , B = u 1 v 1 u 2 v 2 u 3 v 3

Given A = x 1 y 1 x 2 y 2 x 3 y 3 and B = u 1 v 1 u 2 v 2 u 3 v 3 Now, A + B = x 1 y 1 x 2 y 2 x 3 y 3 + u 1 v 1 u 2 v 2 u 3 v 3 = x 1 + u 1 y 1 + v 1 x 2 + u 2 y 2 + v 2 x 3 + u 3 y 3 + v 3 And A − B = x 1 y 1 x 2 y 2 x 3 y 3 - u 1 v 1 u 2 v 2 u 3 v 3 = x 1 − u 1 y 1 − v 1 x 2 − u 2 y 2 − v 2 x 3 − u 3 y 3 − v 3

Find the values of x , y , z , t so that, A = B , where A = x − y y − t z + t x + z , B = y − z 2 − z 2 + t 3 + y

Given matrices are A = x − y y − t z + t x + z , B = y − z 2 − z 2 + t 3 + y And also given that A = B x − y y − t z + t x + z = y − z 2 − z 2 + t 3 + y From the definition of equality of matrices, we have x - y = y - z , y - t = 2 - z , z + t = 2 + t and x + z = 3 + y x + z = 2 y , y + z - t = 2 , z = 2 and x + z - y = 3 Substituting z = 2 in y + z - t = 2 , we have y = t Substituting z = 2 in x + z - y = 3 , we have x - y = 1 . . . . i Substituting z = 2 in x + z = 2 y , we have x - 2 y = - 2 . . . i i Subtracting i and i i , we get y = 3 and substituting it in i , we get x = 4 Therefore, x = 4 , y = 3 , z = 2 , t = 3 .

Find the values of x , y , z , t so that, 3 x y z t = x 6 − 1 2 t + 4 x + y z + t 3

Given 3 x y z t = x 6 − 1 2 t + 4 x + y z + t 3 ⇒ 3 x 3 y 3 z 3 t = x + 4 6 + x + y - 1 + z + t 2 t + 3 From the definition of equality of matrices, we have 3 x = x + 4 , 3 y = 6 + x + y , 3 z = - 1 + z + t and 3 t = 2 t + 3 x = 2 , 2 y - x = 6 , 2 z - t = - 1 and t = 3 Substituting x = 2 in 2 y - x = 6 , we get y = 4 Substituting t = 3 in 2 z - t = - 1 , we get z = 1 Therefore, x = 2 , y = 4 , z = 1 , t = 3 .

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 15

EASY

12th West Bengal Board

IMPORTANT

Earn 100

For a $2 \times 2$ matrix $A = [a_{i j}]$ , whose elements are given by $a_{i j} = \frac{i}{j}$ , Find matrix $A$ and write the value of $a_{12} .$

Important Questions on Elementary Theory of Matrices

EASY

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 16

If

A = [\begin{array}{ll} 3 2 \\ 0 2 \end{array}]

,

B = [\begin{array}{cc} - 2 & 1 \\ 0 & 3 \end{array}]

and

2 A - B + X = O

, then find the matrix

X

.

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 17

Find the values of

x

and

y

when

2 [\begin{array}{ll} 1 3 \\ 0 x \end{array}] + [\begin{array}{ll} y 0 \\ 1 2 \end{array}] = [\begin{array}{ll} 5 6 \\ 1 8 \end{array}]

EASY

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 18

Find the values of

x

and

y

when:

[\begin{array}{cc} x & y \\ 3 y & x \end{array}] [\begin{array}{l} 1 \\ 2 \end{array}] = [\begin{array}{l} 3 \\ 5 \end{array}]

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 19

If

A = (\begin{matrix} 3 & 1 \\ 7 & 5 \end{matrix})

and

A^{2} = - x I + y A

, then find the values of

x

and

y

. Here

I

is the identity matrix of order

2 \times 2

.

EASY

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 1

Find $A + B$ and $A - B$ in the following case:

$A = (\begin{array}{ll} 1 5 \\ 2 0 \end{array})$ , $B = (\begin{array}{cc} 7 & 2 \\ - 3 & 5 \end{array})$

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 1

Find $A + B$ and $A - B$ in the following case:

$A = (\begin{array}{ll} x_{1} y_{1} \\ x_{2} y_{2} \\ x_{3} y_{3} \end{array})$ , $B = (\begin{array}{ll} u_{1} v_{1} \\ u_{2} v_{2} \\ u_{3} v_{3} \end{array})$

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 2

Find the values of $x$ , $y$ , $z$ , $t$ so that,

$A = B$ , where $A = (\begin{array}{cc} x - y & y - t \\ z + t & x + z \end{array})$ , $B = (\begin{array}{cc} y - z & 2 - z \\ 2 + t & 3 + y \end{array})$

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 3 - Elementary Theory of Matrices>EXERCISE>Q 2

Find the values of $x$ , $y$ , $z$ , $t$ so that,

$3 (\begin{array}{ll} x y \\ z t \end{array}) = (\begin{array}{cc} x & 6 \\ - 1 & 2 t \end{array}) + (\begin{array}{cc} 4 & x + y \\ z + t & 3 \end{array})$