If the matrix A is involutary, show that 1 2 ( I + A ) and 1 2 ( I - A ) are idempotent and 1 2 ( I + A ) . 1 2 ( I − A ) = 0 .

Given, A is involutary, ∴ A 2 = I To show that 1 2 ( I + A ) and 1 2 ( I - A ) are idempotent ⇒ 1 2 ( I + A ) × 1 2 ( I + A ) = 1 4 I 2 + I A + A I + A 2 ⇒ 1 2 I + A × 1 2 I + A = 1 4 I + A + A + I ∵ I 2 = I & any matrix multiplied by an identity matrix is that same matrix ⇒ 1 2 I + A × 1 2 I + A = 1 4 2 I + 2 A ⇒ 1 2 I + A × 1 2 I + A = 1 2 ( I + A ) Hence, 1 2 ( I + A ) is idempotent. Similarly 1 2 ( I − A ) will also be idempotent. To prove, 1 2 ( I + A ) . 1 2 ( I − A ) = 0 1 2 ( I + A ) . 1 2 ( I − A ) = 1 4 I 2 - I A + A I - A 2 = 1 4 I - A + A - I = 1 4 × 0 ⇒ 1 2 ( I + A ) . 1 2 ( I − A ) = 0 hence proved.

If the matrix A is involutary, show that 1 2 ( I + A ) and 1 2 ( I - A ) are idempotent and 1 2 ( I + A ) . 1 2 ( I − A ) = 0 .

Given, A is involutary, ∴ A 2 = I To show that 1 2 ( I + A ) and 1 2 ( I - A ) are idempotent ⇒ 1 2 ( I + A ) × 1 2 ( I + A ) = 1 4 I 2 + I A + A I + A 2 ⇒ 1 2 I + A × 1 2 I + A = 1 4 I + A + A + I ∵ I 2 = I & any matrix multiplied by an identity matrix is that same matrix ⇒ 1 2 I + A × 1 2 I + A = 1 4 2 I + 2 A ⇒ 1 2 I + A × 1 2 I + A = 1 2 ( I + A ) Hence, 1 2 ( I + A ) is idempotent. Similarly 1 2 ( I − A ) will also be idempotent. To prove, 1 2 ( I + A ) . 1 2 ( I − A ) = 0 1 2 ( I + A ) . 1 2 ( I − A ) = 1 4 I 2 - I A + A I - A 2 = 1 4 I - A + A - I = 1 4 × 0 ⇒ 1 2 ( I + A ) . 1 2 ( I − A ) = 0 hence proved.

Given A = 1 1 1 2 4 1 2 3 1 , B = 2 3 3 4 . Find P such that B P A = 1 0 1 0 1 0 .

B P A = 1 0 1 0 1 0 ⇒ B − 1 B P A = B − 1 1 0 1 0 1 0 B − 1 = a d j · B | B | Now, a d j . B = 4 - 3 - 3 + 2 T = 4 - 3 - 3 + 2 | B | = 8 − 9 = − 1 B − 1 = 1 - 1 4 - 3 - 3 + 2 = 4 3 3 - 2 B - 1 1 0 1 0 1 0 = - 4 + 3 3 - 2 1 0 1 0 1 0 B - 1 1 0 1 0 1 0 = - 4 + 3 - 4 3 - 2 3 ∴ P A = - 4 + 3 - 4 3 - 2 3 Now, P A A − 1 = - 4 + 3 - 4 3 - 2 3 A − 1 A - 1 = adj . A | A | Now, a d j . A = 1 0 - 2 2 - 1 - 2 - 3 1 2 T = 1 2 - 3 0 - 1 1 - 2 - 1 2 | A | = 1 ( 4 - 3 ) - 1 ( 2 - 2 ) + 1 ( 6 - 8 ) = 1 - 2 = - 1 A - 1 = 1 - 1 1 2 - 3 0 - 1 1 - 2 - 1 2 = - 1 - 2 3 0 1 - 1 2 1 - 2 P I = - 4 + 3 - 4 3 - 2 3 - 1 - 2 3 0 1 - 1 2 1 - 2 P = 4 − 8 8 + 3 − 4 − 12 − 3 + 8 - 3 + 6 - 6 - 2 + 3 9 + 2 - 6 ⇒ P = - 4 7 - 7 3 - 5 5

Show that 1 - tan θ 2 tan θ 2 1 1 tan θ 2 - tan θ 2 1 - 1 = cosθ - sinθ sinθ cosθ

We have to prove that 1 - tan θ 2 tan θ 2 1 1 tan θ 2 - tan θ 2 1 - 1 = cosθ - sinθ sinθ cosθ L . H . S . = 1 - tan θ 2 tan θ 2 1 1 tan θ 2 - tan θ 2 1 - 1 . . . . . . . . ( i ) Let A = 1 tan θ 2 - tan θ 2 1 ∴ A = 1 - tan θ 2 ( - tan θ 2 ) = 1 + tan 2 θ 2 And Adj . A = 1 - tan θ 2 tan θ 2 1 [ ∵ adj a b c d = d - b - c a ] ∴ A - 1 = 1 A adjA = 1 1 + tan 2 θ 2 1 - tan θ 2 tan θ 2 1 Put this value in equation in ( i ) , L . H . S . = 1 - tan θ 2 tan θ 2 1 1 1 + tan 2 θ 2 1 - tan θ 2 tan θ 2 1 = 1 1 + tan 2 θ 2 1 - tan 2 θ 2 - tan θ 2 - tan θ 2 tan θ 2 + tan θ 2 - tan 2 θ 2 + 1 = 1 - tan 2 θ 2 1 + tan 2 θ 2 - 2 tan θ 2 1 + tan 2 θ 2 2 tan θ 2 1 + tan 2 θ 2 1 - tan 2 θ 2 1 + tan 2 θ 2 = cosθ - sinθ sinθ cosθ = R . H . S . Hence, proved.

If F x = cos x − sin x 0 sin x cos x 0 0 0 1 then show that F x . F y = F x + y . Hence prove that F x − 1 = F − x .

Given, F x = cos x - sin x 0 sin x cos x 0 0 0 1 To proof:- F x . F y = F x + y L H S = cos x - sin x 0 sin x cos x 0 0 0 1 cos y - sin y 0 sin y cos y 0 0 0 1 = cos x . cos y - sin x sin y - cos x sin y - sin x cos y 0 sin x cos y + cos x sin y - sin x sin y + cos x cos y 0 0 0 1 = cos x + y - sin x + y 0 sin x + y cos x + y 0 0 0 1 R H S = F x + y = cos x + y - sin x + y 0 sin x + y cos x + y 0 0 0 1 L H S = R H S ⇒ F x . F y = F x + y Proved. Now to proof:- F x - 1 = F - x F - x = cos - x - sin - x 0 sin - x cos - x 0 0 0 1 = - cos x sin x 0 sin x cos x 0 0 0 1 F x = cos x - sin x 0 sin x cos x 0 0 0 1 a d j . F x = cos x - sin x 0 sin x cos x 0 0 0 cos 2 x + sin 2 x T = cos x - sin x 0 sin x cos x 0 0 0 1 T = cos x sin x 0 - sin x cos x 0 0 0 1 F x = cos 2 x + sin 2 x = 1 F x - 1 = cos x sin x 0 - sin x cos x 0 0 0 1 = F - x Proved.

Use matrix to solve the following system of equations:- x + y + z = 3 x + 2 y + 3 z = 4 x + 4 y + 9 z = 6

The equations x + y + z = 3 , x + 2 y + 3 z = 4 ,and x + 4 y + 9 z = 0 can be written as, 1 1 1 1 2 3 1 4 9 x y z = 3 4 6 Or, A X = B where A = 1 1 1 1 2 3 1 4 9 , X = x y z , B = 3 4 6 A = 1 1 1 1 2 3 1 4 9 = 1 18 - 12 - 1 9 - 3 + 1 4 - 2 = 2 Co-factor of matrix A = 6 - 6 2 - 5 8 - 3 1 - 2 - 2 a d j . A = 6 - 6 2 - 5 8 - 3 1 - 2 1 T = 6 - 5 1 - 6 8 - 2 2 - 3 1 A - 1 = 1 A a d j . A = 1 2 6 - 5 1 - 6 8 - 2 2 - 3 1 As, A X = B , ⇒ X = A - 1 B x y z = 1 2 6 - 5 1 - 6 8 - 2 2 - 3 1 3 4 6 x y z = 1 2 6 × 3 - 5 × 4 + 1 × 6 - 6 × 3 + 8 × 4 - 2 × 6 2 × 3 - 3 × 4 + 1 × 6 = 1 2 4 2 0 = 2 1 0 ⇒ x = 2 , y = 1 , z = 0

Use matrix to solve the following system of equations (c) x + y + z = 3 x + 2 y + 3 z = 4 2 x + 3 y + 4 z = 7

(a) Given, x + y + z = 3 … … … 1 x + 2 y + 3 z = 4 … … … 2 2 x + 3 y + 4 z = 7 … … … 3 A X = B Where A = 1 1 1 1 2 3 2 3 4 , X = x y z & B = 3 4 7 A = 1 8 - 9 - 1 4 - 6 + 1 3 - 4 = 0 Now, a d j . A = - 1 2 - 1 - 1 2 - 1 1 - 2 1 T = - 1 - 1 1 2 2 - 2 - 1 - 1 1 a d j . A B = - 1 - 1 1 2 2 - 2 - 1 - 1 1 3 4 7 = - 3 - 4 + 7 6 + 8 - 14 - 3 - 3 + 7 = 0 0 0 Thus the system of equation has infinitely many solution. Let z = k where k ∈ R , Substituting value in equation 1 , x + y = 3 − k Substituting value of z in equation 2 we get, x + 2 y = 4 - 3 k Solving these two equations we get, x = 2 + k , y = 1 − 2 k , z = k

Use matrix to solve the following system of equations (d) x + y + z = 3 x + 2 y + 3 z = 4 2 x + 3 y + 4 z = 9

(d) x + y + z = 3 x + 2 y + 3 z = 4 2 x + 3 y + 4 z = 9 ⇒ ∆ = 1 1 1 1 2 3 2 3 4 = 1 ( 8 − 9 ) − 1 ( 4 − 6 ) + 1 ( 3 − 4 ) = 0 Since, ∆ = 0 hence an entire row or column is zero.Therefore the system of equations is inconsistent, hence no solution. (An inconsistant system of equations has no solution).

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 26

EASY

JEE Main/Advance

IMPORTANT

Earn 100

If the matrix A is involutary, show that $\frac{1}{2} (I + A)$ and $\frac{1}{2} (I - A)$ are idempotent and $\frac{1}{2} (I + A) . \frac{1}{2} (I - A) = 0$ .

Important Questions on Matrices and Determinants

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 27

Express the matrix

[\begin{matrix} 1 & 2 & 5 \\ 2 & 3 & - 6 \\ - 1 & 0 & 4 \end{matrix}]

as a sum of a lower triangular matrix & an upper triangular matrix with zero in leading diagonal of upper triangular matrix. Also express the matrix as a sum of a symmetric and a skew symmetric matrix.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 28

Given

A = [\begin{array}{l} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 2 & 3 & 1 \end{array}], B = [\begin{array}{l} 2 & 3 \\ 3 & 4 \end{array}] .

Find P such that

B P A = [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}]

.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 29

Show that $[\begin{matrix} 1 & - \tan \frac{θ}{2} \\ \tan \frac{θ}{2} & 1 \end{matrix}] {[\begin{matrix} 1 & \tan \frac{θ}{2} \\ - \tan \frac{θ}{2} & 1 \end{matrix}]}^{- 1} = [\begin{matrix} cosθ & - sinθ \\ sinθ & cosθ \end{matrix}]$