If A , B , C are matrices of order 2 × 2 each and 2 A + B + C = 1 2 3 0 , A + B + C = 0 1 2 1 and A + B - C = 1 2 1 0 , the find A , B and C .

Given, 2 A + B + C = 1 2 3 0 → i A + B + C = 0 1 2 1 → i i A + B - C = 1 2 1 0 → i i i To find : A , B , and C Subtracting i i i from i i , we have A + B + C - A + B - C = 0 1 2 1 - 1 2 1 0 ⇒ 2 C = 0 - 1 1 - 2 2 - 1 1 - 0 ⇒ 2 C = - 1 - 1 1 1 ⇒ C = 1 2 - 1 - 1 1 1 ⇒ C = - 1 2 - 1 2 1 2 1 2 Subtracting i i from i , we have 2 A + B + C - A + B + C = 1 2 3 0 - 0 1 2 1 ⇒ A = 1 - 0 2 - 1 3 - 2 0 - 1 ⇒ A = 1 1 1 - 1 Since A + B + C = 0 1 2 1 ⇒ B = 0 1 2 1 - A - C ⇒ B = 0 1 2 1 - 1 1 1 - 1 - - 1 2 - 1 2 1 2 1 2 ⇒ B = 0 - 1 + 1 2 1 - 1 + 1 2 2 - 1 - 1 2 1 + 1 - 1 2 ⇒ B = - 1 2 1 2 1 2 3 2 Hence, A = 1 1 1 - 1 , B = - 1 2 1 2 1 2 3 2 , and C = - 1 2 - 1 2 1 2 1 2 .

If A = 1 2 3 2 3 1 and B = 3 - 1 3 - 1 0 2 then find 2 A - B .

Given, A = 1 2 3 2 3 1 , and B = 3 - 1 3 - 1 0 2 To find: 2 A - B 2 A = 2 1 2 3 2 3 1 = 2 4 6 4 6 2 Now, 2 A - B = 2 4 6 4 6 2 - 3 - 1 3 - 1 0 2 = 2 - 3 4 + 1 6 - 3 4 + 1 6 - 0 2 - 2 = - 1 5 3 5 6 0 Hence, 2 A - B = - 1 5 3 5 6 0

If A = 1 2 − 3 5 0 2 1 − 1 1 , B = 3 − 1 2 4 2 5 2 0 3 and C = 4 1 2 0 3 2 1 − 2 3 then compute A + B and B - C . Also verify that A + B - C = A + B - C

Given that, A = 1 2 - 3 5 0 2 1 - 1 1 and B = 3 - 1 2 4 2 5 2 0 3 , C = 4 1 2 0 3 2 1 - 1 3 To find : A + B and B - C To proof : A + B - C = A + B - C Now, A + B = 1 2 - 3 5 0 2 1 - 1 1 + 3 - 1 2 4 2 5 2 0 3 = 1 + 3 2 - 1 - 3 + 2 5 + 4 0 + 2 2 + 5 1 + 2 - 1 + 0 1 + 3 = 4 1 - 1 9 2 7 3 - 1 4 B - C = 3 - 1 2 4 2 5 2 0 3 - 4 1 2 0 3 2 1 - 1 3 = 3 - 4 - 1 - 1 2 - 2 4 - 0 2 - 3 5 - 2 2 - 1 0 - ( - 2 ) 3 - 3 = - 1 - 2 0 4 - 1 3 1 2 0 Now, A + B - C = 1 2 - 3 5 0 2 1 - 1 1 + - 1 - 2 0 4 - 1 3 1 2 0 = 0 0 - 3 9 - 1 5 2 1 1 . . . . 1 A + B - C = 4 1 - 1 9 2 7 3 - 1 4 - 4 1 2 0 3 2 1 - 1 3 = 0 0 - 3 9 - 1 5 2 1 1 . . . . . 2 From equation 1 and 2 , A + B - C = A + B - C Hence, proved.

Let A = 2 4 3 2 , B = 1 3 - 2 5 , find 2 A + B .

Given that, A = 2 4 3 2 , B = 1 3 - 2 5 . Consider, 2 A + B = 2 2 4 3 2 + 1 3 - 2 5 = 4 8 6 4 + 1 3 - 2 5 = 5 11 4 9 Thus the value of 2 A + B is 5 11 4 9 .

Simplify: sin A sin A - cos A cos A sin A + cos A cos A sin A - sin A cos A

Given: sin A sin A - cos A cos A sin A + cos A cos A sin A - sin A cos A Multiplying the term outside with each term of the matrices, we get: ⇒ sin 2 A - sin A cos A sinA cos A sin 2 A + cos 2 A cos A sin A - cos A sin A cos 2 A Now, adding the matrices, we get: ⇒ sin 2 A + cos 2 A 0 0 sin 2 A + cos 2 A ⇒ 1 0 0 1 Hence, sin A sin A - cos A cos A sin A + cos A cos A sin A - sin A cos A = 1 0 0 1 .

If x + y = 2 1 1 2 , 2 x - y = 1 2 2 1 , then find the value of x .

Given x + y = 2 1 1 2 . . . . i , 2 x - y = 1 2 2 1 . . . . . . . ii Adding both equations, ⇒ 3 x = 2 1 1 2 + 1 2 2 1 ⇒ 3 x = 3 3 3 3 ⇒ x = 1 3 3 3 3 3 ⇒ x = 1 1 1 1

If A = 1 5 6 - 6 7 0 and B = 1 - 5 7 8 - 7 7 , then find the value of A - B .

Given that, the matrices are A = 1 5 6 - 6 7 0 and B = 1 - 5 7 8 - 7 7 . Consider, A - B = 1 5 6 - 6 7 0 - 1 - 5 7 8 - 7 7 = 0 10 - 1 - 14 14 - 7

If A = 0 - tan α 2 tan α 2 0 , then show that I + A = I - A cos α - sin α sin α cos α , where I is the identity matrix of order 2 .

Given, A = 0 - tan α 2 tan α 2 0 LHS = I + A = 1 0 0 1 + 0 - tan α 2 tan α 2 0 = 1 - tan α 2 tan α 2 1 ...(i) RHS = ( I - A ) cos α - sin α sin α cos α = 1 0 0 1 - 0 - tan α 2 tan α 2 0 cos α - sin α sin α cos α = 1 tan α 2 - tan α 2 1 cos α - sin α sin α cos α = cos α + sin α tan α 2 - sin α + cos α tan α 2 - cos α tan α 2 + sin α sin α tan α 2 + cos α = 1 - 2 sin 2 α 2 + 2 sin α 2 cos α 2 tan α 2 - 2 sin α 2 cos α 2 + 2 cos 2 α 2 - 1 tan α 2 - 2 cos 2 α 2 - 1 tan α 2 + 2 sin α 2 cos α 2 2 sin α 2 cos α 2 tan α 2 + 1 - 2 sin 2 α 2 = 1 - 2 sin 2 α 2 + 2 sin 2 α 2 - 2 sin α 2 cos α 2 + 2 sin α 2 cos α 2 - tan α 2 - 2 sin α 2 cos α 2 + tan α 2 + 2 sin α 2 cos α 2 2 sin 2 α 2 + 1 - 2 sin 2 α 2 = 1 - tan α 2 tan α 2 1 ...(ii) From (i) and (ii), LHS = RHS

If A = 1 3 - 2 5 , B = - 2 5 3 4 , then find the value of A - B .

Given that, A = 1 3 - 2 5 , B = - 2 5 3 4 Consider A - B = 1 3 - 2 5 - - 2 5 3 4 = 3 - 2 - 5 1 Thus the value of A - B is 3 - 2 - 5 1 .

Find x and y if x 2 3 + y - 1 1 = 10 5 .

Given, x 2 3 + y - 1 1 = 10 5 ⇒ 2 x 3 x + - y y = 10 5 ⇒ 2 x - y 3 x + y = 10 5 Equating corresponding elements, we have 2 x - y = 10 ....(i) 3 x + y = 5 ....(ii) Adding (i) and (ii), we get 5 x = 15 ⇒ x = 3 Substituting x = 3 in (i) we get, 6 - y = 10 ⇒ y = - 4 ∴ x = 3 , y = - 4

If A 3 0 5 1 and B - 4 2 1 0 Find A 2 - 2 A B + B 2

Given, If A 3 0 5 1 and B - 4 2 1 0 Now, A 2 = A × A = 3 0 5 1 × 3 0 5 1 = 9 + 0 0 + 0 15 + 5 0 + 1 = 9 0 20 1 A B = A × B = 3 0 5 1 × - 4 2 1 0 = 12 + 0 6 + 0 - 20 + 1 10 + 0 = - 12 6 - 19 10 B 2 = B × B = - 4 2 1 0 × - 4 2 1 0 = 16 + 2 - 8 + 0 - 4 + 0 2 + 0 = 18 - 8 - 4 2 Now, A 2 - 2 A B + B 2 = 9 0 20 1 - 2 - 12 6 - 19 10 + 18 - 8 - 4 2 = 9 0 20 1 + 24 - 12 38 - 20 + 18 - 8 - 4 2 = 9 + 24 + 18 0 - 12 - 8 20 + 38 - 4 1 - 20 + 2 = 51 - 20 54 - 17 Thus, the value of A 2 - 2 A B + B 2 is 51 - 20 54 - 17

Find A , if 2 A - 3 - 1 1 2 0 5 = 1 5 3 2 0 1 .

Given that, 2 A - 3 - 1 1 2 0 5 = 1 5 3 2 0 1 ⇒ 2 A = 1 5 3 2 0 1 + 3 - 1 1 2 0 5 = 1 + 3 5 - 1 3 + 1 2 + 2 0 + 0 1 + 5 ⇒ 2 A = 4 4 4 4 0 6 ⇒ A = 1 2 4 4 4 4 0 6 ∴ A = 2 2 2 2 0 3

Find X and Y , if X + Y = 7 0 2 5 and X - Y = 3 0 0 3

Given: X + Y = 7 0 2 5 ⋯ ⋯ ( 1 ) and X - Y = 3 0 0 3 ⋯ ⋯ ( 2 ) Adding equation ( 1 ) and ( 2 ) , we get X + Y + X - Y = 7 0 2 5 + 3 0 0 3 ⇒ 2 X = 7 + 3 0 + 0 2 + 0 5 + 3 2 X = 10 0 2 8 ⇒ X = 1 2 10 0 2 8 ⇒ X = 10 2 0 2 2 2 8 2 ⇒ X = 5 0 1 4 Putting the value of X in equation ( 1 ) , we get 5 0 1 4 + Y = 7 0 2 5 Y = 7 0 2 5 - 5 0 1 4 ⇒ Y = 7 - 5 0 - 0 2 - 1 5 - 4 ⇒ Y = 2 0 1 1 Therefore, X = 5 0 1 4 and Y = 2 0 1 1 .

If A = 1 3 - 2 5 and B = - 2 5 3 4 , then find A - 4 B .

Given, A = 1 3 - 2 5 , and B = - 2 5 3 4 To find: A - 4 B Now, 4 B = 4 - 2 5 3 4 = - 8 20 12 16 Thus, A - 4 B = 1 3 - 2 5 - - 8 20 12 16 = 1 - - 8 3 - 20 - 2 - 12 5 - 16 = 9 - 17 - 14 - 11 Hence, A - 4 B = 9 - 17 - 14 - 11

Find the value of x and y if: 2 x 7 9 y - 5 + 6 - 7 4 5 = 10 7 22 15

Given, 2 x 7 9 y - 5 + 6 - 7 4 5 = 10 7 22 15 2 x 14 18 2 y - 10 + 6 - 7 4 5 = 10 7 22 15 2 x + 6 14 - 7 18 + 4 2 y - 10 + 5 = 10 7 22 15 2 x + 6 7 22 2 y - 5 = 10 7 22 15 2 x + 6 = 10 2 x = 10 - 6 2 x = 4 x = 4 2 = 2 2 y - 5 = 15 2 y = 15 + 5 2 y = 20 y = 20 2 = 10 Hence, the value of x = 2 , y = 10

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If $A$ and $B$ are square matrices of order $3$ , then

37.5% studentsanswered this correctly

Important Questions on Matrices and Determinants

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A, B, C

are matrices of order

2 \times 2

each and

2 A + B + C = [\begin{array}{l} 1 & 2 \\ 3 & 0 \end{array}], A + B + C = [\begin{array}{l} 0 & 1 \\ 2 & 1 \end{array}]

and

A + B - C = [\begin{array}{l} 1 & 2 \\ 1 & 0 \end{array}]

, the find

A, B and C

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{array}{l} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}]

and

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

then find

2 A - B .

HARD

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{array}{ccc} 1 & 2 & - 3 \\ 5 & 0 & 2 \\ 1 & - 1 & 1 \end{array}], B = [\begin{array}{ccc} 3 & - 1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{array}] and C = [\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & - 2 & 3 \end{array}]

then compute

(A + B)

and

(B - C)

. Also verify that

A + (B - C) = (A + B) - C

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

Let

A = [\begin{array}{l} 2 & 4 \\ 3 & 2 \end{array}]

B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]

, find

2 A + B

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

Simplify:

$\sin A [\begin{matrix} \sin A & - \cos A \\ \cos A & \sin A \end{matrix}] + \cos A [\begin{matrix} \cos A & \sin A \\ - \sin A & \cos A \end{matrix}]$

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

x + y = [\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}]

2 x - y = [\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}]

, then find the value of

x

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{array}{l} 4 & 2 & 3 \\ 1 & 5 & 7 \end{array}]

and

B = [\begin{array}{l} 1 & 3 & 7 \\ 0 & 4 & 1 \end{array}]

, then

2 A + B

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{matrix} 1 & 5 & 6 \\ - 6 & 7 & 0 \end{matrix}]

and

B = [\begin{array}{l} 1 & - 5 & 7 \\ 8 & - 7 & 7 \end{array}]

, then find the value of

A - B

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{matrix} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{matrix}],

then show that

I + A = (I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]

, where

I

is the identity matrix of order

2

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

If $A = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}], B = [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$ , then find the value of $A - B$ .

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

Find

x

and

y

x [\begin{array}{l} 2 \\ 3 \end{array}] + y [\begin{matrix} - 1 \\ 1 \end{matrix}] = [\begin{matrix} 10 \\ 5 \end{matrix}]

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

If $A [\begin{matrix} 3 & 0 \\ 5 & 1 \end{matrix}]$ and $B [\begin{matrix} - 4 & 2 \\ 1 & 0 \end{matrix}]$ Find $A^{2} - 2 A B + B^{2}$

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

The value of

\cos θ [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] + \sin θ [\begin{matrix} \sin θ & \cos θ \\ - \cos θ & \sin θ \end{matrix}]

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

Find

A

, if

2 A - [\begin{matrix} 3 & - 1 \\ 1 & 2 \\ 0 & 5 \end{matrix}] = [\begin{matrix} 1 & 5 \\ 3 & 2 \\ 0 & 1 \end{matrix}]

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

If $X + Y = [\begin{array}{l} 5 & 2 \\ 0 & 9 \end{array}]$ and $X - Y = [\begin{matrix} 3 & 6 \\ - 2 & 1 \end{matrix}]$ , then matrix $X$ is :

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]

, and

A + A^{'} = I

, then the value of

α

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

Find $X$ and $Y$ , if $X + Y = [\begin{array}{l} 7 & 0 \\ 2 & 5 \end{array}]$ and $X - Y = [\begin{array}{l} 3 & 0 \\ 0 & 3 \end{array}]$

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]

and

B = [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]

, then find

A - 4 B

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

Find the value of $x$ and $y$ if:

$2 [\begin{matrix} x & 7 \\ 9 & y - 5 \end{matrix}] + [\begin{matrix} 6 & - 7 \\ 4 & 5 \end{matrix}] = [\begin{matrix} 10 & 7 \\ 22 & 15 \end{matrix}]$

EASY

Mathematics>Algebra>Matrices and Determinants>Algebra of Matrices

A = [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}], B = [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}], C = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]

then

If A and B are square matrices of order 3, then

Important Questions on Matrices and Determinants

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If $A$ and $B$ are square matrices of order $3$ , then