NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise

Question 1.
Let A = \(\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\) show that (aI + bA)ⁿ = aⁿI + na^n-1 bA, where I is the identity matrix of order 2 and n ∈ N .
Solution:
Let P(n) : (aI + bA)ⁿ = aⁿI + na^n-1 bA
P(1) : (aI + bA)¹ = a¹I + 1a°bA
⇒ P(1) : aI + bA = aI + bA
∴ P(1) is true
Let us assume that P(k) is true.
i.e., P(k) : (aI + bA)^k = a^kI + ka^k-1bA
P(k+ 1) : (aI + bA)^k+1
= (aI + bA)^k.(aI + bA)
= (a^kI + ka^k-1A) (aI + bA)
= a^k.a.I.l + a^k.b. I.A + ka^k-1.b.a. AI + ka^k-1b.bA.A
= a^k+1I + a^kbA + ka^kbA + ka^k-1b²A²
= a^k+1I + (k + 1)a^kbA + ka^k-1b²A² … (1)
A² = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = 0
∴ (1) ⇒ P(k + 1) = a^k+1 I + (k + 1)a^kbA + kak^k-1b² x 0
= a^k+1 I + (k + 1)a^(k+1)-1Ab
which is true whenever P(L) is true.
Hence by the Principle of Mathematical Induction, the result is true for all n ∈ N .

Question 2.
If A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{ccc}
3^{n-1} & 3^{n-1} & 3^{n-1} \\
3^{n-1} & 3^{n-1} & 3^{n-1} \\
3^{n-1} & 3^{n-1} & 3^{n-1}
\end{array}\right]\), n ∈ N
Solution:

which is true
whenever P(k) is true.
Hence by the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Question 3.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), then prove that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), where n is any positive integer.
Solution:

∴ P(k + 1) is true, whenever P(k) is true.
Hence by the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Question 4.
If A and Bare symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Solution:
A and B are symmetric matrices
∴ A’ = A and B’ = B

(i) (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’ since (AB)’ = B’A’
= BA – AB since A’ = A and
B’ = B
= – (AB – BA)
∴ AB – BA is a skew-symmetric matrix.

(ii) Let AB = BA
Taking transpose on both sides,
(AB)’ = (BA)’
i.e.,(AB)’ = A’B’ since (BA)’ = A’B’
i.e., (AB)’ = AB, since A’ A and B’ = B
∴ AB ¡s a symmetric matrix.
Let AB be a symmetric matrix.
Then (AB)’ = AB
⇒ B’A’ = AB
⇒ BA = AB since B’ = B and A’ = A
∴ AB = BA

Question 5.
Show that the matrix B’AB is symmetric or skew symmetric according as A is sym¬metric or skew symmetric.
Solution:
Let A be a symmetric matrix.
Then A’ = A
∴ (B’AB)’ = (B’(AB))’
= (AB)’(B’)’
= B’A’B
= B’AB since A’ = A
∴ B’AB is a symmetric matrix
Let A be a skew-symmetric matrix.
Then A’ = – A
∴ (B’AB)’ (B’(AB))’
= (AB)’(B’)’
= B’A’B
= B’(-A)B since A’= – A
= – B’AB
∴ B’AB is a skew-symmetric matrix.

Question 6.
Find the values of x, y, z if the matrix A = \(\left[\begin{array}{ccc}
0 & 2 y & z \\
x & y & -z \\
x & -y & z
\end{array}\right]\) satisfied the equation A’A = I.
Solution:

Question 7.
For what values of x if
\(\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]\) = 0
Solution:

Question 8.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), show that A² – 5A + 7I = 0.
Solution:

Question 9.
Find x, if
\(\left[\begin{array}{lll}
x & -5 & -1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]\) = 0
Solution:

Question 10.
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

(i) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00 respectively, find the total revenue in each market with the help of matrix algebra.
(ii) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively, find the gross profit.
Solution:
(i) Let

represents annual sale of products in two markets and the column matrix
B = \(\left[\begin{array}{l|l}
2.50 \\
1.50 \\
1.00
\end{array}\right] \begin{aligned}
&x \\
&y \\
&z
\end{aligned}\) represents the unit sale price of the commodities x, y, z.
∴ The revenue collected by each market is given by AB.

The revenue in market I = ₹ 46,000 and
the revenue in market II = ₹ 53,000
Hence gross revenue in two markets
= ₹ 46,000 + 53,000 = ₹ 99,000

(ii) Let the column matrix C = \(\left[\begin{array}{c}
2.00 \\
1.00 \\
0.50
\end{array}\right]\)
Repre-sent the unit cost price of the commodities x, y and z.
∴ The cost price of the articles in the two markets is AC.

The cost price of articles in market I = ₹ 31,000 and the cost price of articles in market II = ₹ 36,000.
Hence the gross cost price
= ₹ 31,000 + 36,000 = ₹ 67,000
Gross profit = Gross revenue – Gross cost price
= ₹ 99,000 – 67, 000 = ₹ 32, 000.

Question 11.
Find the matrix X so that
X\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]\) = \(\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
Solution:
X\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]\) = \(\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
It is clear that the order of X is 2 x 2
since the product exists. Let X = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\).

Equating the corresponding elements
∴ a + 4b = – 7 … (1)
2a + 56 = – 8 … (2)
3a + 66 = – 9 … (3)
c + 4d = 2 … (4)
2c + 5d = 4 … (5)
3c + 6d = 6 … (6)
(1) + (2) ⇒ 3a + 96 = – 15
(3) ⇒ 3a + 6b = – 9
Subtracting we get 3b = – 6
⇒ 6 = – 2
∴ (1) ⇒ a = -7 + 8 = 1
(4) + (5) ⇒ 3c + 9d = 6
(6) ⇒ 3c + 6d = 6
Subtracting 3d = 0 ⇒ d = 0
(4) ⇒ c = 2
∴ X = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) = \(\left[\begin{array}{cc}
1 & -2 \\
2 & 0
\end{array}\right]\)

Question 12.
If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABⁿ = BⁿA. Further, prove that (AB)ⁿ = AⁿBⁿ for all n ∈ N.
Solution:
Let P(n) : ABⁿ = BⁿA be the given statement.
P(1): AB’ = B’A which is true.
Assume that P(&) is true, i.e., AB^k = B^kA
AB^k+1 = (AB)B^k = (BA)B^k
= B(AB^k)
= B(B^kA) since P(k) is true
= B^k+1A
Hence by the Principle of Mathematical In-duction, P(n) is true for all values of n ∈ N .
Let P(n) : (AB)ⁿ = AⁿBⁿ
P(1) : (AB)’ = A’B’ ⇒ AB = AB which is true.
Assume that P(k) is true i.e., (AB)^k = A^kB^k
(AB)^k+1 = (AB)^k.(AB) = A^k.B^k(AB)
= A^k(B^kA)B
= A^k(AB^k)B (since B^k A = AB^k)
= (A^k.A)(B^kB)
= A^k+1 B^k+1
∴ By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Choose the correct answer in the following questions.

Question 13.
If A = \(\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\) is such that A² = I, then
a. 1 + α² + βγ = 0
b. 1 – α² + βγ = 0
c. 1 – α² – βγ = 0
d. 1 + α² – βγ = 0
Solution:
c. 1 – α² – βγ = 0

Question 14.
If the matrix A is both symmetric and skew symmetric, then
a. A is a diagonal matrix
b. A is a zero matrix
c. A is a square matrix
d. None of these
Solution:
b. A is a zero matrix
For symmetric matrix a_ij = a_ji
For skew symmetric matrix a_ij = – a_ji
These conditions are satisfied only if a_ij = 0 for all i and j.
Hence the matrix is a zero matrix.

Question 15.
If A is square matrix such that A² = A, then (I + A)³ – 7A is equal to
a. A
b. I – A
c. I
d. 3A
Solution:
c. I
(I + A)³ – 7A = I³ + 3I²A + 3IA² + A³ – 7A
= I + 3A + 3A² + A²A – 7A
= I + 3A + 3A + A.A – 7A
= I + 6A + A – 7A
= I + 7A – 7A = I