Class 12 RD Sharma Solutions - Chapter 7 Adjoint and Inverse of a Matrix - Exercise 7.1 | Set 2
Last Updated : 21 Aug, 2024
Question 10. For the following parts of matrices verify that (AB)-1 = B-1A-1.
(i) A = \begin{bmatrix}3&2\\7&5\end{bmatrix} and B = \begin{bmatrix}4&6\\3&2\end{bmatrix}
Solution:
To prove (AB)-1= B-1A-1
We take LHS
AB = \begin{bmatrix}3&2\\7&5\end{bmatrix}\begin{bmatrix}4&6\\3&2\end{bmatrix}
= \begin{bmatrix}18&22\\43&52\end{bmatrix}
|AB| = 18 × 52 - 22 × 43
= 936 - 946 = -10
adj(AB) = \begin{bmatrix}52&-22\\-43&18\end{bmatrix}
AB-1= adj(AB)/|AB| = \frac{1}{(-10)}\begin{bmatrix}52&-22\\-43&18\end{bmatrix}
= \frac{1}{10}\begin{bmatrix}-52&22\\43&-18\end{bmatrix}
Now,
A = \begin{bmatrix}3&2\\7&5\end{bmatrix}
|A| = 15 - 14 = 1
adj A = \begin{bmatrix}5&-2\\-7&3\end{bmatrix}
Therefore, A-1 = adj A/|A| = \frac{1}{1}\begin{bmatrix}5&-2\\-7&3\end{bmatrix}
B = \begin{bmatrix}4&6\\3&2\end{bmatrix}
|B| = 8 - 18 = -10
adj B = \begin{bmatrix}2&-6\\-3&4\end{bmatrix}
Therefore, B-1= adj B/|B| = \frac{1}{-10}\begin{bmatrix}2&-6\\-3&4\end{bmatrix}
Now, we take RHS
B-1A-1 = \frac{-1}{10}\begin{bmatrix}2&-6\\-3&4\end{bmatrix}\begin{bmatrix}5&-2\\-7&3\end{bmatrix}
= \frac{-1}{10}\begin{bmatrix}52&-22\\-43&18\end{bmatrix}
= \frac{1}{10}\begin{bmatrix}-52&22\\43&-18\end{bmatrix}
LHS = RHS
Hence, Proved
(ii) A = \begin{bmatrix}2&1\\5&3\end{bmatrix} and B = \begin{bmatrix}4&5\\3&4\end{bmatrix}
Solution:
To prove (AB)-1 = B-1A-1
We take LHS
AB = \begin{bmatrix}2&1\\5&3\end{bmatrix}\begin{bmatrix}4&5\\3&4\end{bmatrix}
= \begin{bmatrix}11&14\\29&27\end{bmatrix}
|AB| = 11 × 27 - 29 × 14
= 407 - 406 = 1
adj(AB) = \begin{bmatrix}37&-14\\-29&11\end{bmatrix}
AB-1= adj(AB)/|AB| = \frac{1}{1}\begin{bmatrix}37&-14\\-29&11\end{bmatrix}
=\begin{bmatrix}37&-14\\-29&11\end{bmatrix}
Now,
A = \begin{bmatrix}2&1\\5&3\end{bmatrix}
|A| = 6 - 5 = 1
adj A= \begin{bmatrix}3&-1\\-5&2\end{bmatrix}
Therefore, A-1 = adj A/|A| = \frac{1}{1}\begin{bmatrix}3&-1\\-5&2\end{bmatrix}
B = \begin{bmatrix}4&5\\3&4\end{bmatrix}
|B| = 16 - 15 = 1
adj B = \begin{bmatrix}4&-5\\-3&4\end{bmatrix}
Therefore, B-1= adj B/|B| = \frac{1}{1}\begin{bmatrix}4&-5\\-3&4\end{bmatrix}
Now, we take RHS
B-1A-1 = \begin{bmatrix}4&-5\\-3&4\end{bmatrix}\begin{bmatrix}3&-1\\-5&2\end{bmatrix}
= \begin{bmatrix}37&-14\\-29&11\end{bmatrix}
LHS = RHS
Hence, Proved
Question 11. Let A = \begin{bmatrix}3&2\\7&5\end{bmatrix} and B = \begin{bmatrix}6&7\\8&9\end{bmatrix} . Find (AB)-1.
Solution:
AB = \begin{bmatrix}3&2\\7&5\end{bmatrix}\begin{bmatrix}4&6\\3&2\end{bmatrix}
=\begin{bmatrix}34&39\\82&94\end{bmatrix}
|AB| = 34 × 94 - 39 × 82 = -2
adj(AB) = \begin{bmatrix}94&-39\\-82&34\end{bmatrix}
AB-1 = adj(AB)/|AB| = \frac{-1}{2}\begin{bmatrix}94&-39\\-82&34\end{bmatrix}
= \begin{bmatrix}-47&39/2\\41&-17\end{bmatrix}
Question 12. Given A = \begin{bmatrix}2&-3\\-4&7\end{bmatrix}, Compute A-1 and show that 2A-1 = 9I - A.
Solution:
A = \begin{bmatrix}2&-3\\-4&7\end{bmatrix}
|A| = 14 - 12 = 2
adj A = \begin{bmatrix}7&3\\4&2\end{bmatrix}
Therefore, A-1 = adj A/|A| = \frac{1}{2}\begin{bmatrix}7&3\\4&2\end{bmatrix}
To show 2A-1 = 9I - A.
LHS = 2 × (1/2) \begin{bmatrix}7&3\\4&2\end{bmatrix}
= \begin{bmatrix}7&3\\4&2\end{bmatrix}
Now we take RHS
= 9I - A
= \begin{bmatrix}9&0\\0&9\end{bmatrix} - \begin{bmatrix}2&-3\\-4&7\end{bmatrix}
=\begin{bmatrix}7&3\\4&2\end{bmatrix}
LHS = RHS
Hence Proved
Question 13. If A = \begin{bmatrix}4&5\\2&1\end{bmatrix}, then show that A - 3I = 2(I + 3A-1).
Solution:
Here, A = \begin{bmatrix}4&5\\2&1\end{bmatrix}
|A| = 4 - 10 = -6
adj A = \begin{bmatrix}1&-5\\-2&4\end{bmatrix}
Therefore, A-1 = adj A/|A| = \frac{1}{(-6)}\begin{bmatrix}1&-2\\-5&4\end{bmatrix}
To show, A - 3I = 2(I + 3A-1)
Now we take LHS
= A - 3I
=\begin{bmatrix}4&5\\2&1\end{bmatrix} - 3\begin{bmatrix}1&0\\0&1\end{bmatrix}
=\begin{bmatrix}1&5\\2&-2\end{bmatrix}
Now we take RHS
= 2I + 6A-1
= 2\begin{bmatrix}1&0\\0&1\end{bmatrix} + 6 × (1/6)\begin{bmatrix}-1&5\\2&-4\end{bmatrix}
= \begin{bmatrix}1&5\\2&-2\end{bmatrix}
LHS = RHS
Hence Proved
Question 14. Find the inverse of the matrix A = \begin{bmatrix}a&b\\c&(1+bc)/a\end{bmatrix} and show that aA-1 = (a2 + bc + 1)I - aA.
Solution:
Here, A = \begin{bmatrix}a&b\\c&(1+bc)/a\end{bmatrix}
|A| = (a + abc)/a - bc = 1
Therefore, inverse of A exists
Cofactor of A are,
C11 = (1 + bc)/a C12 = -c
C21 = -b C22 = a
adj A = \begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T
= \begin{bmatrix}(1+bc)/a &-c\\-b&a\end{bmatrix}^T
= \begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}
A-1 = 1/|A|. adj A
= 1/1 \begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}
= \begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}
To show that
aA-1 = (a2 + bc + 1)I - aA.
LHS = aA-1
= a\begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}
= \begin{bmatrix}1+bc &-ab\\-ac&a^2\end{bmatrix}
RHS = (a2 + bc + 1)I - aA
= \begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1\end{bmatrix} - a\begin{bmatrix}a&b\\c&(1+bc)/a\end{bmatrix}
= \begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1\end{bmatrix} - \begin{bmatrix}a^2&ab\\ac&1+bc\end{bmatrix}
= \begin{bmatrix}1+bc&-ab\\-ac&a^2\end{bmatrix}
LHS = RHS
Hence Proved
Question 15. Given A = \begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix}, B-1 = \begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix} , Compute (AB)-1.
Solution:
We know (AB)-1 = B-1A-1
Here, A = \begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix}
|A| = 5(3 - 4) + 4(4 - 3) = -5 + 4 = -1
Co-factors of A are:
C11 = -1 C12 = 0 C13 = 1
C21 = 8 C22 = 1 C23 = -10
C31 = -12 C32 = -2 C33 = 15
adj A = \begin{bmatrix}-1&8&-12\\0&1&-2\\1&-10&15\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{(-1)}\begin{bmatrix}-1&8&-12\\0&1&-2\\1&-10&15\end{bmatrix}
= \begin{bmatrix}1&-8&12\\0&-1&2\\-1&10&-15\end{bmatrix}
(AB)-1 = B-1A-1
= \begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}\begin{bmatrix}1&-8&12\\0&-1&2\\-1&10&-15\end{bmatrix}
= \begin{bmatrix}-2&19&-27\\-2&18&-25\\-3&29&-42\end{bmatrix}
Question 16. Let F(α) = \begin{bmatrix}cosα&-sinα&0\\sinα&cosα&0\\0&0&1\end{bmatrix} and G(β) = \begin{bmatrix}cosβ&0&sinβ\\0&1&0\\-sinβ&0&cosβ\end{bmatrix} , Show that
(i) [F(α)]-1 = F(-α)
Solution:
We have F(α) = \begin{bmatrix}cosα&-sinα&0\\sinα&cosα&0\\0&0&1\end{bmatrix}
|F(α)| = cos2α + sin2α = 1
Therefore, inverse of F(α) exists
Cofactors of F(α) are:
C11 = cosα C12 = -sinα C13 = 0
C21 = sinα C22 = cosα C23 = 0
C31 = 0 C32 = 0 C33 = 1
Adj F(α) = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
=\begin{bmatrix}cosα&-sinα&0\\sinα&cosα&0\\0&0&1\end{bmatrix}^T
=\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}
[F(α)]-1 = 1/|F(α)|. adj F(α)
Hence, [F(α)]-1 = 1/1\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}
=\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}
Now, F(-α) = \begin{bmatrix}cos(-α)&-sin(-α)&0\\sin(-α)&cos(-α)&0\\0&0&1\end{bmatrix}
=\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}
So, [F(α)]-1 = F(-α)
Hence, Proved
(ii) [G(β)]-1 = G(-β)
Solution:
We have G(β) = \begin{bmatrix}cosβ&0&sinβ\\0&1&0\\-sinβ&0&cosβ\end{bmatrix}
|G(β)| = cos2β + sin2β = 1
Therefore, inverse of G(β) exists
Cofactors of G(β) are:
C11 = cosβ C12 = 0 C13 = sinβ
C21 = 0 C22 = 1 C23 = 0
C31 = -sinβ C32 = 0 C33 = sinβ
Adj G(β) = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
=\begin{bmatrix}cosβ&0&sinβ\\0&1&0\\-sinβ&0&cosβ\end{bmatrix}^T
=\begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}
[G(β)]-1 = 1/|G(β)|. adj G(β)
Hence, [G(β)]-1 = 1/1\begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}
= \begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}
Now, G(-β) =\begin{bmatrix}cos(-β)&0&sin(-β)\\0&1&0\\-sin(-β)&0&cos(-β)\end{bmatrix}
= \begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}
So, [G(β)]-1 = G(-β)
Hence, Proved
(iii) [F(α)G(β)]-1 = F(-α)G(-β)
Solution:
We already know that S[G(β)]-1 = G(-β)
[F(α)]-1 = F(-α)
Taking LHS = [F(α)G(β)]-1
= [F(α)]-1[G(β)]-1
= F(-α)G(-β) = RHS
Hence, Proved
Question 17. If A = \begin{bmatrix}2&3\\1&2\end{bmatrix} , Verify that A2 - 4A + I = O, where I = \begin{bmatrix}1&0\\0&1\end{bmatrix} and O = \begin{bmatrix}0&0\\0&0\end{bmatrix} , Hence, find A-1.
Solution:
Here, A = \begin{bmatrix}2&3\\1&2\end{bmatrix}
A2 = \begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}
= \begin{bmatrix}7&12\\4&7\end{bmatrix}
4A = 4\begin{bmatrix}2&3\\1&2\end{bmatrix}
= \begin{bmatrix}8&12\\4&8\end{bmatrix}
A2 - 4A + I = O
= \begin{bmatrix}7&12\\4&7\end{bmatrix} -\begin{bmatrix}8&12\\4&8\end{bmatrix} + \begin{bmatrix}1&0\\0&1\end{bmatrix}
= \begin{bmatrix}7-8+1&12-2+0\\4-4+0&7-8+1\end{bmatrix}
Hence, = \begin{bmatrix}0&0\\0&0\end{bmatrix}
Now, A2 - 4A + I = O
A2 - 4A = -I
Multiplying both side by A-1 both sides we get
A.A(A-1) - 4AA-1 = -IA-1
AI - 4I = -A-1
A-1 = 4I - AI
= \begin{bmatrix}4&0\\0&4\end{bmatrix} - \begin{bmatrix}2&3\\1&2\end{bmatrix}
= \begin{bmatrix}2&-3\\-1&2\end{bmatrix}
Question 18. Show that A = \begin{bmatrix}-8&5\\2&4\end{bmatrix} satisfies the equation A2 + 4A - 42I = O. Hence, Find A-1.
Solution:
Here, A = \begin{bmatrix}-8&5\\2&4\end{bmatrix}
A2 = \begin{bmatrix}-8&5\\2&4\end{bmatrix}\begin{bmatrix}-8&5\\2&4\end{bmatrix}
=\begin{bmatrix}64+10&-40+20\\-16+8&10+16\end{bmatrix}
=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}
4A = 4\begin{bmatrix}-8&5\\2&4\end{bmatrix}
= \begin{bmatrix}-32&20\\8&16\end{bmatrix}
A2 + 4A - 42I = \begin{bmatrix}74&-20\\-8&26\end{bmatrix} + \begin{bmatrix}-32&20\\8&16\end{bmatrix} - \begin{bmatrix}42&0\\0&42\end{bmatrix}
=\begin{bmatrix}74-74&-20+20\\-8+8&42-42\end{bmatrix}
Hence, \begin{bmatrix}0&0\\0&0\end{bmatrix}
Now, A2 + 4A - 42I = 0
⇒ A-1A.A + 4A-1.A - 42A-1I = 0
⇒ IA + 4I - 42A-1 = 0
⇒ A-1 = 1/42 [A + 4I]
⇒ A-1 = \frac{1}{42}\begin{bmatrix}-4&5\\2&8\end{bmatrix}
Question 19. If A = \begin{bmatrix}3&1\\-1&2\end{bmatrix} , show that A2 - 5A + 7I = O. Hence find A-1.
Solution:
Here, A = \begin{bmatrix}3&1\\-1&2\end{bmatrix}
A2 = \begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}
=\begin{bmatrix}8&5\\-5&3\end{bmatrix}
Now, A2 - 5A + 7I = \begin{bmatrix}8&5\\-5&3\end{bmatrix} + 5\begin{bmatrix}3&1\\-1&2\end{bmatrix} + 7\begin{bmatrix}1&0\\0&1\end{bmatrix}
=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}
=\begin{bmatrix}0&0\\0&0\end{bmatrix}
Now, A2 - 5A + 7I = O
Multiplying by A-1 both sides
⇒ A-1AA + 5AA - 1 + 7IA-1 = 0
⇒ A-1 = 1/7[5I - A]
⇒ A-1 = \frac{1}{7}\begin{bmatrix}5&0\\0&5\end{bmatrix}-\begin{bmatrix}3&1\\-1&2\end{bmatrix}
⇒ A-1 = \frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}
Question 20. If A = \begin{bmatrix}4&3\\2&5\end{bmatrix} , find x and y such that A2 - xA + yI = O. Hence, evaluate A-1.
Solution:
Here, A = \begin{bmatrix}4&3\\2&5\end{bmatrix}
A2 = \begin{bmatrix}4&3\\2&5\end{bmatrix}\begin{bmatrix}4&3\\2&5\end{bmatrix}
= \begin{bmatrix}22&27\\18&31\end{bmatrix}
Now, A2 - xA + yI = O
⇒ \begin{bmatrix}22&27\\18&31\end{bmatrix} - \begin{bmatrix}4x&3x\\2x&5x\end{bmatrix} + \begin{bmatrix}y&0\\0&y\end{bmatrix}
= \begin{bmatrix}0&0\\0&0\end{bmatrix}
⇒ 22 - 4x + y = 0
⇒ 4x - y = 22 .........(i)
or
18 - 2x = 0
⇒ x = 9
Putting x = 9 in eq (i)
⇒ y = 14
A2 - 9A + 14I = 0
⇒ 9A = A2 + 14I
⇒ 9A-1A = A-1AA + 14A-1
⇒ 9I = IA + 14A-1
⇒ A-1 = 1/14[9I - A] = 1/14(\begin{bmatrix}9&0\\0&9\end{bmatrix}-\begin{bmatrix}4&3\\2&4\end{bmatrix} )
⇒ A-1= \frac{1}{14}\begin{bmatrix}5&-3\\-2&4\end{bmatrix}
Question 21. If A = \begin{bmatrix}3&-2\\4&-2\end{bmatrix} , find the value of λ so that A2 = λA - 2I. Hence, find A-1.
Solution:
Here, A = \begin{bmatrix}3&-2\\4&-2\end{bmatrix}
A2 = \begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}
= \begin{bmatrix}1&-2\\4&-4\end{bmatrix}
If A2 = λA - 2I
λA = A2 + 2I
⇒ λ \begin{bmatrix}3&-2\\4&-2\end{bmatrix} = \begin{bmatrix}1&-2\\4&-4\end{bmatrix} + \begin{bmatrix}2&0\\0&2\end{bmatrix}
⇒ λ \begin{bmatrix}3&-2\\4&-2\end{bmatrix} = \begin{bmatrix}3&-2\\4&-2\end{bmatrix}
⇒ λ = 1
Now, A2 = λA - 2I
Multiplying both side A-1
⇒ A-1AA = A-1A - 2A-1I
⇒ A = I - 2A-1
⇒ 2A-1 = I - A = \begin{bmatrix}1&0\\0&1\end{bmatrix} - \begin{bmatrix}3&-2\\4&-2\end{bmatrix}
A-1 = \frac{1}{2}\begin{bmatrix}-2&2\\-4&3\end{bmatrix}
Question 22. Show that A = \begin{bmatrix}5&3\\-1&-2\end{bmatrix} satisfies the equation x2 - 3x - 7 = 0. Thus, find A-1.
Solution:
Here, A = \begin{bmatrix}5&3\\-1&-2\end{bmatrix}
A2 = \begin{bmatrix}5&3\\-1&-2\end{bmatrix}\begin{bmatrix}5&3\\-1&-2\end{bmatrix} = \begin{bmatrix}22&9\\-3&-1\end{bmatrix}
Now, A2 - 3A - 7= \begin{bmatrix}22&9\\-3&-1\end{bmatrix}-\begin{bmatrix}15&9\\-3&-6\end{bmatrix}-\begin{bmatrix}7&0\\0&7\end{bmatrix}
=\begin{bmatrix}0&0\\0&0\end{bmatrix}
We have, A2 - 3A - 7 = 0
⇒ A-1AA - 3A-1A - 7A-1 = 0
⇒ A-3I - 7A-1 = 0
⇒ 7A-1 = A - 3I
⇒ 7A-1 = \begin{bmatrix}5&3\\-1&-2\end{bmatrix} - \begin{bmatrix}3&0\\0&3\end{bmatrix}
A-1 = \begin{bmatrix}2/7&3/7\\-1/7&-5/7\end{bmatrix}
Question 23. Show that A = \begin{bmatrix}6&5\\7&6\end{bmatrix} satisfies the equation x2 - 12x + 1 = 0. Thus, find A-1.
Solution:
Here, A = \begin{bmatrix}6&5\\7&6\end{bmatrix}
A2 = \begin{bmatrix}6&5\\7&6\end{bmatrix}\begin{bmatrix}6&5\\7&6\end{bmatrix}
= \begin{bmatrix}71&60\\84&71\end{bmatrix}
Now, A2 - 12A + I = \begin{bmatrix}71&60\\84&71\end{bmatrix} - \begin{bmatrix}6&5\\7&6\end{bmatrix} + \begin{bmatrix}1&0\\0&1\end{bmatrix}
= \begin{bmatrix}0&0\\0&0\end{bmatrix}
We have, A2 - 12A + I = 0
⇒ A - 12I + A-1 = 0
⇒ A-1 = 12I - A
⇒ A-1 = \begin{bmatrix}12&0\\0&12\end{bmatrix}-\begin{bmatrix}6&5\\7&6\end{bmatrix}
⇒ A-1 = \begin{bmatrix}6&-5\\-7&6\end{bmatrix}
Question 24. For the matrix A =\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} show that A3 - 6A2 + 5A + 11I3 = O. Hence, find A-1.
Solution:
Here, A = \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
A2 = \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
= \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}
A3 = \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix} \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
= \begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}
A3 - 6A2 + 5A + 11I
= \begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix} - 6 \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}+ 5 \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}+11 \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
=\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix} - \begin{bmatrix}24&12&6\\-18&48&-84\\42&-18&84\end{bmatrix}+ \begin{bmatrix}5&5&5\\5&10&-15\\10&-5&15\end{bmatrix}+\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}
= \begin{bmatrix}24-24&12-12&6-6\\-18+18&48-48&-84+84\\42-42&-18+18&84-84\end{bmatrix}
= \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
We have, A3 - 6A2 + 5A + 11I = O.
⇒ A-1(AAA) - 6A-1(AA) + 5A-1A + 11IA-1 = 0
⇒ A2 - 6A + 5I = -11A-1
⇒ -11A-1 = (A2 - 6A + 5I)
=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}-6 \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}+5 \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}-\begin{bmatrix}6&6&6\\6&12&-18\\22&-6&18\end{bmatrix}+\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}
=\begin{bmatrix}4-6+5&2-6&1-6\\-3-6&8-12+5&-14+18\\7-12&-3+6&14-18+5\end{bmatrix}
=\begin{bmatrix}3&-4&-5\\-9&1&4\\-5&3&1\end{bmatrix}
Therefore, A-1 = \frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}
Summary
Chapter 7, Exercise 7.1 | Set 2 typically builds upon the concepts introduced in Set 1, focusing on more advanced applications of adjoint and inverse matrices. The main topics covered usually include:
- Complex problems involving adjoint and inverse matrices
- Properties and theorems related to adjoint and inverse matrices
- Application of adjoint and inverse matrices in solving systems of equations
- Relationship between determinants, adjoint, and inverse matrices
- Matrix equations and their solutions using adjoint and inverse matrices
- Special matrices and their adjoints/inverses
- Proofs involving adjoint and inverse matrices
This exercise set emphasizes problem-solving skills and the ability to apply adjoint and inverse matrix concepts in diverse mathematical contexts.
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Chapter 1: Relations
Class 12 RD Sharma Solutions - Chapter 1 Relations - Exercise 1.1 | Set 1In mathematics, understanding the properties of relations is very fundamental to forming a strong basic. Reflexive, symmetric, and transitive relations form the cornerstone of relational algebra, which finds a huge application in various fields such as computer science, logic finding, and data scien
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Class 12 RD Sharma Solutions - Chapter 1 Relations - Exercise 1.1 | Set 2Question 11. Is it true every relation which is symmetric and transitive is also reflexive? Give reasons.Solution:We will verify this by taking an example.Consider a set A = {1, 2, 3} and a relation R on A such that R = { (1, 2), (2,1), (2,3), (1,3) }The relation R over the set A is symmetric and tr
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Class 12 RD Sharma Solutions - Chapter 1 Relations - Exercise 1.2 | Set 1Question 1. Show that the relation R = {(a,b): a-b is divisible by 3;, a, b ∈ Z} is an equivalence relation. Solution: According to question, relation R = {(a,b): a-b is divisible by 3;, a, b ∈ Z} We have to show that R is an equivalence relation. (i) reflexibity: let a = z => a - a = 0 => 0 i
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Class 12 RD Sharma Solutions - Chapter 1 Relations - Exercise 1.2 | Set 2Question 11. Let O be the origin . We define a relation between two points P and Q in a plane if OP = OQ . Show that the relation, so defined is an equivalence relation.Solution:Let A be the set of points on planeand let R = {(P, Q): OP = OQ} be a relation on A where O is the origin.now (i) reflexib
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Chapter 2: Functions
Class 12 RD Sharma Mathematics Solutions - Chapter 2 Functions - Exercise 2.1 | Set 1Chapter 2 of RD Sharma's Class 12 Mathematics textbook delves into the concept of functions which is foundational in the higher mathematics. Exercise 2.1 | Set 1 introduces students to the basic concepts and types of functions laying the groundwork for the more advanced topics. Understanding functio
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Class 12 RD Sharma Solutions - Chapter 2 Functions - Exercise 2.1 | Set 2Question 12: Show that the exponential function f: R → R, given by f(x) = ex is one one but not onto. What happens if the codomain is replaced by Ro+.Solution:We have f: R → R, given by f(x) = ex.Let x,y ϵ R, such that=> f(x) = f(y)=> ex = ey=> e(x-y) = 1 = e0=> x - y = 0=> x = yHence
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Class 12 RD Sharma Solutions - Chapter 2 Functions - Exercise 2.2Question 1(i). Find g o f and f o g when f: R -> R and g: R -> R is defined by f(x) = 2x + 3 and g(x) = x2 + 5Solution:f: R -> R and g: R -> RTherefore, f o g: R -> R and g o f: R -> RNow, f(x) = 2x + 3 and g(x) = x2 + 5g o f(x) = g(2x + 3) =(2x + 3)2 + 5=> g o f(x) = 4x2 + 12x
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Class 12 RD Sharma Solutions - Chapter 2 Functions - Exercise 2.3Question 1. Find fog and gof, if(i) f (x) = ex,g (x) = \log_exSolution:Let f: R → (0, ∞); and g: (0, ∞) → RClearly, the range of g is a subset of the domain of f.So, fog: (0, ∞) → R and we know, (fog)(x) = f(g(x))= f(\log_ex)= e^{log_ex}(fog)(x) = xClearly, the range of f is a subset of the domain o
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Chapter 3: Binary Operations
Class 12 RD Sharma Solutions - Chapter 3 Binary Operations - Exercise 3.1Question 1. Determine whether the following operation define a binary operation on the given set or not:(i) ‘*’ on N defined by a * b = ab for all a, b ∈ N.(ii) ‘O’ on Z defined by a O b = ab for all a, b ∈ Z.(iii) ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N(iv) ‘×6‘ on S = {1, 2, 3, 4, 5
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Class 12 RD Sharma Solutions - Chapter 3 Binary Operations - Exercise 3.2Question 1. Let ‘*’ be a binary operation on N defined by a * b = 1.c.m. (a, b) for all a, b ∈ N(i) Find 2 * 4, 3 * 5, 1 * 6Solution:We are given that a * b = L.C.M. (a, b) ⇒ 2 * 4 = L.C.M. (2, 4) = 4and, 3 * 5 = L.C.M. (3, 5) = 15now, 1 * 6 = L.C.M. (1, 6) = 6Hence, 2 * 4 = 4, 3 * 5 = 15 and 1 * 6
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Class 12 RD Sharma Solutions- Chapter 3 Binary Operations - Exercise 3.3Binary operations are fundamental concepts in mathematics and computer science that involve operations on pairs of elements. In Chapter 3 of RD Sharma’s Class 12 textbook, we explore the binary operations in detail focusing on their properties and applications. Exercise 3.3 covers practical problems
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Class 12 RD Sharma Solutions - Chapter 3 Binary Operations - Exercise 3.4In this article, we will explore the solutions to Exercise 3.4 from Chapter 3 of RD Sharma's Class 12 Mathematics textbook which focuses on "Binary Operations". This chapter is essential for understanding how operations can be performed on pairs of elements within a set a fundamental concept in alge
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Class 12 RD Sharma Solutions - Chapter 3 Binary Operations - Exercise 3.5Exercise 3.5 in Chapter 3 of RD Sharma's Class 12 Mathematics textbook focuses on binary operations. This exercise likely covers advanced concepts related to binary operations, including their properties, applications, and problem-solving techniques. Students will be challenged to apply their unders
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Chapter 4: Inverse Trigonometric Functions
Chapter 5: Algebra of Matrices
Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.1 | Set 1Chapter 5 of RD Sharma's Class 12 Mathematics textbook focuses on the Algebra of Matrices, with Exercise 5.1 Set 1 introducing fundamental concepts and operations of matrices. This exercise covers matrix notation, types of matrices, and basic matrix operations. Students will learn to represent data
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Class 12 RD Sharma Solutions- Chapter 5 Algebra of Matrices - Exercise 5.1 | Set 2Chapter 5 of RD Sharma's Class 12 Mathematics textbook continues exploring the Algebra of Matrices, with Exercise 5.1 Set 2 building upon the foundational concepts introduced in Set 1. This set likely delves deeper into matrix operations, properties, and applications. Students will encounter more co
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Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.2 | Set 1Chapter 5 of RD Sharma's Class 12 Mathematics textbook focuses on the Algebra of Matrices. Exercise 5.2 specifically deals with operations on matrices, including addition, subtraction, and multiplication. This exercise helps students understand how to perform these operations and apply them to solve
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Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.2 | Set 2Chapter 5 of RD Sharma's Class 12 Mathematics textbook continues to explore the Algebra of Matrices. Exercise 5.2 Set 2 typically delves deeper into matrix operations, focusing on more complex problems involving matrix addition, subtraction, multiplication, and related concepts. This set often inclu
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Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.3 | Set 1Chapter 5 of RD Sharma's Class 12 Mathematics textbook on the Algebra of Matrices continues with Exercise 5.3. This exercise typically focuses on more advanced matrix operations and properties, including determinants, adjoint matrices, and inverse matrices. Set 1 of this exercise often introduces th
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Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.3 | Set 2Exercise 5.3 Set 2 in Chapter 5 of RD Sharma's Class 12 Mathematics textbook continues the exploration of advanced matrix concepts. This set typically delves deeper into the properties of determinants, adjoint matrices, and inverse matrices. It often includes more complex problems that require stude
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Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.3 | Set 3Exercise 5.3 Set 3 in Chapter 5 of RD Sharma's Class 12 Mathematics textbook further expands on the concepts of determinants, adjoint matrices, and inverse matrices. This set typically includes more challenging problems that require a deeper understanding of matrix properties and their interrelation
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Class 12 RD Sharma Solutions- Chapter 5 Algebra of Matrices - Exercise 5.4Chapter 5 of RD Sharma's Class 12 Mathematics textbook focuses on the Algebra of Matrices. Exercise 5.4 specifically deals with elementary operations on matrices and their properties. This exercise is crucial for understanding how matrices can be manipulated and transformed, which is fundamental in
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Class 12 RD Sharma Solutions- Chapter 5 Algebra of Matrices - Exercise 5.5Exercise 5.5 in Chapter 5 of RD Sharma's Class 12 Mathematics focuses on the concept of rank of a matrix. This is a fundamental concept in linear algebra that helps in understanding the structure and properties of matrices, as well as their applications in solving systems of linear equations.Questio
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Chapter 6: Determinants
Class 12 RD Sharma Solutions - Chapter 6 - Exercise 6.1In this article, we will be going to solve the entire exercise 6.1 of our RD Sharma textbook. In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix and the linear transformations it represent
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Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.2 | Set 1In Chapter 6 of RD Sharma's Class 12 Mathematics textbook determinants play a crucial role in the solving systems of linear equations and various matrix-related problems. Exercise 6.2 | Set 1 provides the practice problems designed to deepen students' understanding of the determinants and their prop
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Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.2 | Set 2Prove the following identities:Question 18. \begin{vmatrix}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix} = -2Solution:Considering the determinant, we have\triangle = \begin{vmatrix}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 &
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Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.2 | Set 3Prove the following identities:Question 35. \begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix} = 4xyzSolution:Considering the determinant, we have\triangle = \begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix}R1
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Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.3Question 1. Find the area of the triangle with vertices at the points:(i) (3, 8), (−4, 2) and (5, −1)Solution:Given (3, 8), (−4, 2) and (5, −1) are the vertices of the triangle. We know that, if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,A =
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Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.4 | Set 1Question 1. Solve the following system of linear equations by Cramer’s rule.x – 2y = 4−3x + 5y = −7Solution:Using Cramer's Rule, we get,D=\left|\begin{array}{cc} 1 & -2 \\ -3 & 5 \end{array} \right|= 5 − 6 = −1Also, we get,D_1=\left|\begin{array}{cc} 4 & -2 \\ -7 & 5 \end{array} \rig
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Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.4 | Set 2Question 17. Solve the following system of the linear equations by Cramer's rule.2x − 3y − 4z = 29−2x + 5y − z = −153x − y + 5z = −11Solution:Using Cramer's Rule, we get,D=\left|\begin{array}{cc} 2 & -3 & -4 \\ -2 & 5 & -1 \\ 3 & -1 & 5 \end{array} \right|Expanding along R1,
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Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.5Question 1. Solve each of the following system of homogeneous linear equations:x + y - 2z = 02x + y - 3z =05x + 4y - 9z = 0Solution:Given: x + y - 2z = 02x + y - 3z =05x + 4y - 9z = 0This system of equations can be expressed in the form of a matrix AX = BNow find the determinant,D=\begin{vmatrix}1
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Chapter 7: Adjoint and Inverse of a Matrix
Chapter 8: Solutions of Simultaneous Linear Equations
Chapter 9: Continuity
Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.1 | Set 1Question 1. Test the continuity of the following function at the origin:Â Â f(x)= \begin{cases}\frac{x}{|x|},& Â x \neq 0 \\1,& x=0\end{cases}Solution:Given thatf(x)= \begin{cases}\frac{x}{|x|},& Â x\neq0 \\1,& x=0\end{cases}Â Â Â Now, let us consider LHL at x = 0\lim_{x\to0^-}f(x) =\lim_{h
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Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.1 | Set 2Question 16. Discuss the continuity of the function f(x)=\begin{cases}x,&0\leq x<(\frac{1}{2}) \\(\frac{1}{2}),&x=(\frac{1}{2}) \\1-x,&(\frac{1}{2})<x\leq1\end{cases} at the point x = 1/2.Solution:Given that, f(x)=\begin{cases}x,&0\leq x<(\frac{1}{2}) \\(\frac{1}{2}),&x=
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Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.1 | Set 3Question 31. If f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases} is continuous at x = 2, find k.Solution: Given that,f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases} Also, f(x) is continuous at
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Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.2 | Set 1Question 1. Prove that the function f(x)=\begin{cases}\frac{sinx}{x} \ \ \ \ ,x≤0\\x+1\ \ \ ,x≤0\end{cases} is continuous everywhere.Solution:We know sin x/ x is continuous everywhere since it is the composite function of the functions sin x and x which are continuous.When x > 0, we have f(x) = x
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Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.2 | Set 2Question 8. If f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x} for x ≠π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].Solution:If x ≠π/4, tan (π/4 - x) and cot2x are continuous in [0, π/2]. T
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