# Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2A

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2A Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2A

Question 1.
If A = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right]$$ then reduce it to I3 by using column transformations.
Solution:
|A| = $$\left|\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right|$$
= 1(1 – 0) – 0 + 0 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right]$$
By C1 – 2C2, we get, A ~ $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 3 & 1 \end{array}\right]$$
By C1 + 3C3 and C2 – 3C3, we get,
A ~ $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ = I3.

Question 2.
If A = $$\left[\begin{array}{lll} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right]$$, then reduce it to I3 by using row transformations.
Solution:
|A| = $$\left|\begin{array}{lll} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right|$$
= 2 (0 – 1) – 1(1 – 1) + 3 (1 – 0)
= -2 – 0 + 3 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = $$\left[\begin{array}{lll} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right]$$
By R1 – R2, we get,

By R1 – R3 and By R2 – R3, we get
A ~ $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ = I3.

Question 3.
Check whether the following matrices are invertible or not:
(i) $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|$$ = 1 – 0 = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A-1 exists.

(ii) $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|$$ = 1 – 1 = 0.
∴ A is a singular matrix.
Hence, A-1 does not exist.

(iii) $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right|$$ = 3 – 6 = -3 ≠ 0.
∴ A is a non-singular matrix.
Hence, A-1 exist.

(iv) $$\left[\begin{array}{ll} 2 & 3 \\ 10 & 15 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 2 & 3 \\ 10 & 15 \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{ll} 2 & 3 \\ 10 & 15 \end{array}\right|$$ = 30 – 30 = 0.
∴ A is a singular matrix.
Hence, A-1 does not exist.

(v) $$\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{cc} \sec \theta & \tan \theta \\ \tan \theta & \sec \theta \end{array}\right|$$
= sec2θ – tan2θ = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A-1 exist.

(vii) $$\left[\begin{array}{lll} 3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{array}\right]$$
Solution:
let A = $$\left[\begin{array}{lll} 3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{lll} 3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{array}\right|$$
= 3(5 – 0) – 4(5 – 0) + 3(4 – 1)
= 15 – 20 + 9 = 4 ≠ 0
∴ A is a non-singular matrix.
Hence, A-1 exist.

(viii) $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & -1 & 3 \\ 1 & 2 & 3 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & -1 & 3 \\ 1 & 2 & 3 \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{lll} 1 & 2 & 3 \\ 2 & -1 & 3 \\ 1 & 2 & 3 \end{array}\right|$$
= 1 (-3 -6) – 2 (6 – 3) + 3 (4 + 1)
= -9 – 6 + 15 = 0
∴ A is a singular matrix.
Hence, A-1 does not exist.

(ix) $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 6 & 8 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 6 & 8 \end{array}\right]$$
Then, |A| = $$\left|\begin{array}{lll} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 6 & 8 \end{array}\right|$$
= 1(32 – 30) – 2(24 – 20) + 3(18 – 16)
= 2 – 8 + 6 = 0
∴ A is a singular matrix.
Hence, A-1 does not exist.

Question 4.
Find AB, if A = $$\left[\begin{array}{ccc} 1 & 2 & 3 \\ 1 & -2 & -3 \end{array}\right]$$ and B = $$\left[\begin{array}{cc} 1 & -1 \\ 1 & 2 \\ 1 & -2 \end{array}\right]$$ Examine whether AB has inverse or not.
Solution:

∴ A is a non-singular matrix.
Hence, (AB)-1 exist.

Question 5.
If A = $$\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]$$ is a nonsingular matrix then find A-1 by elementary row transformations.
Hence, find the inverse of $$\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$$
Solution:
Since A is a non-singular matrix, then find A-1 by using elementary row transformations.
We write AA-1 = I

Comparing $$\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$$ with $$\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]$$,
we get, x = 2, y = 1, z = -1
∴ $$\frac{1}{x}$$ = $$\frac{1}{2}$$, $$\frac{1}{y}$$ = $$\frac{1}{1}$$ = 1, $$\frac{1}{z}$$ = $$\frac{1}{-1}$$ = -1
$$\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$$ is $$\left(\begin{array}{rrr} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right)$$.

Question 6.
if A = $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$$ and X is a 2 × 2 matrix such that AX = I , then find X.
Solution:
We will reduce the matrix A to the identity matrix by using row transformations. During this pro¬cess, I will be converted to the matrix X.
We have AX = I.

Question 7.
Find the inverse of each of the following matrices (if they exist).
(i) $$\left[\begin{array}{ll} 1 & -1 \\ 2 & 3 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 1 & -1 \\ 2 & 3 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 1 & -1 \\ 2 & 3 \end{array}\right|$$ = 3 + 2 = 5 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(ii) $$\left[\begin{array}{ll} 2 & 1 \\ 1 & -1 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 2 & 1 \\ 1 & -1 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 2 & 1 \\ 1 & -1 \end{array}\right|$$ = -2 – 1 = -3 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(iii) $$\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right|$$ = 7 – 6 = 1 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(iv) $$\left[\begin{array}{ll} 2 & -3 \\ 5 & 7 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 2 & -3 \\ 5 & 7 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 2 & -3 \\ 5 & 7 \end{array}\right|$$ = 14 + 15 = 29 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(v) $$\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right|$$ = 8 – 7 = 1 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(vi) $$\left[\begin{array}{ll} 3 & -10 \\ 2 & -7 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{ll} 3 & -10 \\ 2 & -7 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 3 & -10 \\ 2 & -7 \end{array}\right|$$ = -21 + 20 = -1 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(vii) $$\left[\begin{array}{lll} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{lll} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{lll} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right|$$
= 2(4 + 6) +3(4 – 9) + 3(-4 – 6)
= 20 – 15 – 30 = -25 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(viii) $$\left[\begin{array}{lll} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{lll} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right|$$
= 1(0 + 25) + 3(0 + 10) + 2(-15 – 0)
= 25 + 30 -30
= 25 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

(ix) $$\left[\begin{array}{lll} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]$$
Solution:
Let A =$$\left[\begin{array}{lll} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{lll} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right|$$
= 2(3 – 0) – 0 – 1(5 – 0)
= 6 – 0 – 5 = 1 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

∴ A-1 = $$\left[\begin{array}{lll} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$$

(x) $$\left[\begin{array}{lll} 1 & 2 & -2 \\ 0 & -2 & 1 \\ -1 & 3 & 0 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 2 & -2 \\ 0 & -2 & 1 \\ -1 & 3 & 0 \end{array}\right]$$
∴ A-1 = $$\left[\begin{array}{lll} 1 & 2 & -2 \\ 0 & -2 & 1 \\ -1 & 3 & 0 \end{array}\right]$$
= 1$$\left|\begin{array}{ll} -2 & 1 \\ 3 & 0 \end{array}\right|$$ – 2$$\left|\begin{array}{ll} 0 & 1 \\ -1 & 1 \end{array}\right|$$ – 2$$\left|\begin{array}{ll} 0 & -2 \\ -1 & 3 \end{array}\right|$$
|A| = 1(0 – 3) – 2(0 + 1) – 2(0 – 2)
= -3 – 2 + 4
= -1 ≠ 0
∴ A-1 exists.
We have
AA-1 = I

∴ A-1 = $$\left[\begin{array}{lll} 3 & 6 & 2 \\ 1 & 2 & 1 \\ 2 & 5 & 2 \end{array}\right]$$

Question 8.
Find the inverse of A = $$\left[\begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right]$$ by
(i) elementary row transformations
Solution:
|A| = $$\left|\begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right|$$
= cosθ (cosθ – 0) + sinθ (sinθ – 0) + 0
= cos2θ + sin2θ = 1 ≠ 0
∴ A-1 exists.
(i) Consider AA-1 = I

(ii) elementary column transformations
Solution:
Consider A-1A = I

Question 9.
If A = $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]$$, B = $$\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right]$$ find AB and (AB)-1. Verify that (AB)-1 = B-1A-1
Solution:
AB = $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]$$ $$\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right]$$

From (1) and (2), (AB)-1 = B-1 ∙ A-1.

Question 10.
If A = $$\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]$$, then show that A-1 = $$\frac{1}{6}$$(A – 5I)
Solution:
|A| = $$\left|\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right|$$ = 4 – 10 = -6 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

Question 11.
Find matrix X such that AX = B, where A = $$\left[\begin{array}{ll} 1 & 2 \\ -1 & 3 \end{array}\right]$$ and B = $$\left[\begin{array}{ll} 0 & 1 \\ 2 & 4 \end{array}\right]$$
Solution:
AX = B

Question 12.
Find X, if AX = B where A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]$$ and B = $$\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$$.
Solution:
AX = B

Question 13.
If A = $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right]$$, B = $$\left[\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}\right]$$ and C = $$\left[\begin{array}{ll} 24 & 7 \\ 31 & 9 \end{array}\right]$$ then find matrix X such that AXB = C.
Solution:
AXB = C
∴ $$\left(\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right)(\mathrm{XB})$$ =$$\left[\begin{array}{ll} 24 & 7 \\ 31 & 9 \end{array}\right]$$
First we perform the row transformations.

Question 14.
Find the inverse of $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{array}\right]$$ by adjoint method.
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{array}\right|$$
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A-1 exists.
First we have to find the cofactor matrix
= [Aij]3×3 where Aij = (-1)i+jMij
Now, A11 = (-1)1+1M11 = $$\left|\begin{array}{ll} 1 & 5 \\ 4 & 7 \end{array}\right|$$ = 7 – 20 = -13
A12 = (-1)1+2M12 = $$\left|\begin{array}{ll} 1 & 5 \\ 2 & 7 \end{array}\right|$$ = -(7 – 10) = 3

Question 15.
Find the inverse of $$\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right]$$ by adjoint method.
Solution:
where A = $$\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right]$$
|A| = 1(2 – 6) – 0(0 – 3) + 1(0 – 2)
|A| = -4 – 2
|A| = -6 ≠ 0
∴ A-1 exists.
First we have to find the cofactor matrix
= [Aij]3×3, where Aij = (-1)i+jMij

Question 16.
Find A-1 by adjoint method and by elementary transformations if A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]$$
Solution:
|A| = $$\left|\begin{array}{lll} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right|$$
= 1(4 – 4) – 2(-4 – 2) + 3(-2 – 1)
= 0 + 12 – 9 = 3 ≠ 0
∴ A-1 exists.
We have to find the cofactor matrix
= [Aij]3×3, where Aij = (-1)i+j Mij

Question 17.
Find the inverse of A = $$\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right]$$ by elementary column transformations.
Solution:
|A| = $$\left|\begin{array}{lll} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right|$$
= 1 (2 – 6) – 0 + 1 (0 – 2)
= -4 – 2= -6 ≠ 0
∴ A-1 exists.
Consider A-1A = I
∴ A-1$$\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right]$$ = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
By C3 – C1, we get,

Question 18.
Find the inverse of $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{array}\right]$$ by elementary row transformations.
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{array}\right|$$
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A-1 exists.
Consider AA-1 = I

Question 19.
Show with usual notations that for any matrix A = [aij]3×3
(i) a11A21 + a12A22 + a13A23 = 0
Solution:
A = [aij]3×3 = $$\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]$$
(i) A21 = (-1)2+1M21 = $$-\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right|$$
= -(a12a33 – a13a32)
= -a12a33 + a13a32
A22 = (-1)2+2M22 = $$\left|\begin{array}{ll} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array}\right|$$
= a11a33 – a13a31
A23 = (-1)2+3M23 = $$-\left|\begin{array}{ll} a_{11} & a_{12} \\ a_{31} & a_{32} \end{array}\right|$$
= -(a11a32 – a12a31)
= -a11a32+ a12a31
∴ a11A21 + a12A22 + a13A23
= a11(-a1233 + a13a32) + a12(a11a33 – a13a31) + a13(-a11a32 + a12a31)
= -a11a12a33 + a11a13a32 + a11a12a33 – a12a13a31 – a11a13a32 + a12a13a31
= 0

(ii) a11A11 + a12A12 + a13A13 = |A|
Solution:

Question 20.
If A = $$\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right]$$ and B = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{array}\right]$$, then find a matrix X such that XA= B.
Solution:
Consider XA = B