12th NCERT/CBSE Matrices Miscellaneous Exercise Questions 15

Do or do not
There is no try

Question (1)

Let $A = \left[ {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right]$, show that (aI + bA)ⁿ = aⁿI + na^n-1bA, where, I is the identity matrix of order 2 and n ∈N

Solution

(aI + bA)ⁿ = aⁿI + na^n-1bA Let us prove for n=1
\[LHS = {\left( {aI + bA} \right)^n} = aI + bA\] \[RHS = {a^n}I + n{a^{n - 1}}bA\] \[RHS = aI + 1{a^0}bA = aI + bA = LHS\] It is true for n=1
Let us assue it is true for n=k
∴ (aI+nA)^k = a^kI + ka^k-1bbA
Let us prove for n = k + 1, then we have to prove
(aI + bA)^k+1 = a^k+1 I + (k+1)a^kbA
\[{\rm{LHS = }}{\left( {aI + bA} \right)^{k + 1}}\] \[{\rm{ = }}{\left( {aI + bA} \right)^k} \cdot \left( {aI + bA} \right)\] \[ = \left( {{a^k}I + k{a^{k - 1}}bA} \right)\left( {aI + bA} \right)\] \[ = {a^{k + 1}}{I^2} + {a^k}bIA + k{a^k}bAI + {a^{k - 1}}{b^2}{A^2}\] \[ = {a^{k + 1}}I + \left( {k + 1} \right){a^k}bA = RHS\] ∴ It is true for n = k+1
∴ By PMI, it is true for ∀ n ∈ N

Question (2)

If $A = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}} \right]$, prove that
${A^n} = \left[ {\begin{array}{*{20}{c}}{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\end{array}} \right],n \in N$

Solution

$A = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}} \right]$,
${A^n} = \left[ {\begin{array}{*{20}{c}}{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\end{array}} \right],n \in N$ Let us prove for n=1
\[{A^1} = \left[ {\begin{array}{*{20}{c}}{{3^{1 - 1}}}&{{3^{1 - 1}}}&{{3^{1 - 1}}}\\{{3^{1 - 1}}}&{{3^{1 - 1}}}&{{3^{1 - 1}}}\\{{3^{1 - 1}}}&{{3^{1 - 1}}}&{{3^{1 - 1}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}} \right]\] ∴ It is true for n = 1
Let us assume it is true for n ∈k
\[{A^k} = \left[ {\begin{array}{*{20}{c}}{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\end{array}} \right]\] Let us prove for n = k+1, to prove that we will prove \[{A^{k + 1}} = \left[ {\begin{array}{*{20}{c}}{{3^k}}&{{3^k}}&{{3^k}}\\{{3^k}}&{{3^k}}&{{3^k}}\\{{3^k}}&{{3^k}}&{{3^k}}\end{array}} \right]\] LHS = A^k+1
LHS= A^k A
\[LHS = \left[ {\begin{array}{*{20}{c}}{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}} \right]\] \[LHS = \left[ {\begin{array}{*{20}{c}}{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}\\{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}\\{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}\end{array}} \right]\] \[LHS = \left[ {\begin{array}{*{20}{c}}{{3^k}}&{{3^k}}&{{3^k}}\\{{3^k}}&{{3^k}}&{{3^k}}\\{{3^k}}&{{3^k}}&{{3^k}}\end{array}} \right] = RHS\] ∴ It is true for n = k+1
∴ By PMI, it is true ∀n ∈ N

Question (3)

If $A = \left[ {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right]$, then prove that ${A^n} = \left[ {\begin{array}{*{20}{c}}{1 + 2n}&{ - 4n}\\n&{1 - 2n}\end{array}} \right]$, where n is any positive integer.

Solution

$A = \left[ {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right]$
${A^n} = \left[ {\begin{array}{*{20}{c}}{1 + 2n}&{ - 4n}\\n&{1 - 2n}\end{array}} \right]$
Let us prove for n = 1
\[\therefore {A^1} = \left[ {\begin{array}{*{20}{c}}{1 + 2}&{ - 4}\\n&{1 - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right] = A\] ∴ It is true for n=1
Let us assue it is true for n = k
\[ \therefore {A^k} = \left[ {\begin{array}{*{20}{c}}{1 + 2k}&{ - 4k}\\k&{1 - 2k}\end{array}} \right]\] \[{A^k} = \left[ {\begin{array}{*{20}{c}}{1 + 2k}&{ - 4k}\\k&{1 - 2k}\end{array}} \right]\] Let us prove for n = k+1
so we will prove that
\[{A^{k + 1}} = \left[ {\begin{array}{*{20}{c}}{2k + 3}&{ - 4k - 4}\\{k + 1}&{ - 2k - 1}\end{array}} \right]\] \[LHS = {A^{k + 1}}\] \[LHS = {A^k} \cdot A\] \[LHS = \left[ {\begin{array}{*{20}{c}}{1 + 2k}&{ - 4k}\\k&{1 - 2k}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}{3 + 6k - 4k}&{ - 4 - 8k + 4k}\\{3k + 1 - 2k}&{ - 4k - 1 + 2k}\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}{2k + 3}&{ - 4k - 4}\\{k + 1}&{ - 2k - 1}\end{array}} \right] = RHS\] ∴ It is true for n = k +1
∴ By PMI it is true for ∀ n ∈ N

Question (4)

If A and B are symmetric matrices, prove that AB - BA is a skew symmetric matrix.

Solution

A and B are symmetric matrix
∴ A' = A and B' = B
Now (AB - BA)' = (AB)' - (BA)'
= B'A' - A'B'
= BA - AB
= -(AB - BA)
since transpose matrix is negative of matrix
∴ (AB - BA) is skew symmetric matrix

Question (5)

Show that the matrix B'AB is symmetric or skew symmetric according as A is symmetric or skew symmetric

Solution

BAB
Let A is symmetric matrix
A' = A
(B'AB)' = [B'(AB)]'
= (AB)' (B')' ∵ (B')' = B
=(B'A'). B ∵ A' = A
=B'A.B
(B'AB)' = B'AB'
⇒ B'AB is symmetric mattery
If A is skew symmetric
A' = A
(B'AB)' = [B'(AB)]¹
= (AB)' (B')'
= B'A'B
= B'(-A)BB
=- B'AB
⇒ B'ABB is skew symmetric
∴ B'AB is symetric or skew dependsupon matrix 'A'

Question (6)

Find the values of x, y, z if the matrix $A = \left[ {\begin{array}{*{20}{c}}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right]$ satisfy the equation A'A = I

Solution

\[A = \left[ {\begin{array}{*{20}{c}}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right],A' = \left[ {\begin{array}{*{20}{c}}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right]\] \[A'A = I\] \[ \Rightarrow \left[ {\begin{array}{*{20}{c}}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[ \Rightarrow \left[ {\begin{array}{*{20}{c}}{2{x^2}}&{xy - xy}&{ - xz + xz}\\{xy - xy}&{6{y^2}}&{zyz - yz - yz}\\{ - xz + xz}&{2yz - yz - yz}&{3{z^2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[ \Rightarrow \left[ {\begin{array}{*{20}{c}}{2{x^2}}&0&0\\0&{6{y^2}}&0\\0&0&{3{z^2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] \[ \Rightarrow 2{x^2} = 1\] \[x = \pm \frac{1}{{\sqrt 2 }}\] \[6{y^2} = 1\] \[y = \pm \frac{1}{{\sqrt 6 }}\] \[3{z^2} = 1\] \[z = \pm \frac{1}{{\sqrt 3 }}\]

Question (7)

For what values of $x:\left[ {\begin{array}{*{20}{c}}1&2&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&2&0\\2&0&1\\1&0&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\2\\x\end{array}} \right] = 0?$

Solution

\[{\left[ {\begin{array}{*{20}{c}}1&2&1\end{array}} \right]_{1 \times 1}}{\left[ {\begin{array}{*{20}{c}}1&2&0\\2&0&1\\1&0&2\end{array}} \right]_{3 \times 3}}\left[ {\begin{array}{*{20}{c}}0\\2\\x\end{array}} \right] = 0\] \[ \Rightarrow \left[ {\begin{array}{*{20}{c}}{1 + 4 + 1}&{2 + 0 + 0}&{0 + 2 + 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\2\\x\end{array}} \right] = 0\] \[ \Rightarrow \left[ {\begin{array}{*{20}{c}}6&2&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\2\\x\end{array}} \right] = 0\] \[\left[ {0 + 4 + 4x} \right] = 0\] \[x = - 1\]

Question (8)

If $A = \left[ {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right]$ show that A² -5A + 7I = 0

Solution

\[{A^2} = A \cdot A\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right]\] \[{A^2} = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right]\] \[LHS = {A^2} - 5A + 7I\] \[ = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right] = 0\] \[ = RHS\]

Question (9)

Findx, if $\left[ {x - 5 - 1} \right]\left[ {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\1\end{array}} \right] = 0$

Solution

\[\left[ {\begin{array}{*{20}{c}}x&{ - 5}&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\1\end{array}} \right] = 0\] \[\left[ {\begin{array}{*{20}{c}}{x + 0 - 2}&{0 - 10 + 0}&{2x - 5 - 3}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\1\end{array}} \right] = 0\] \[\left[ {\begin{array}{*{20}{c}}{x - 2}&{ - 10}&{2x - 8}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\1\end{array}} \right] = 0\] \[\left[ {{x^2} - 2x - 40 + 2x - 8} \right] = 0\] \[\left[ {{x^2} - 48} \right] = 0\] \[{x^2} = 48\] \[x = \pm \sqrt {48} \] \[x = \pm 4\sqrt 3 \]

Question (10)

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

Market	Product
I	10,000	2,000	18,000
II	6000	20,000	8000

(a) If unit sale price of x, y, and z are Rs. 2.50, Rs1.50 and Rs1.00, respectively, find the total revenue in each market with the help of matrix algebra
(b) If the unit costs of the above three commodities are Rs. 2.00, Rs. 1.00 and 50 paise respectively. Find the gross profit

Solution

Total revenue
\[{\left[ {\begin{array}{*{20}{c}}{10,000}&{2000}&{18,000}\\{6,000}&{20,000}&{8,000}\end{array}} \right]_{2 \times 3}}{\left[ {\begin{array}{*{20}{c}}{2.5}\\{1.5}\\{1.0}\end{array}} \right]_{3 \times 1}}\] \[ = \left[ {\begin{array}{*{20}{c}}{25,000 + 3000 + 18,000}\\{15,000 + 30,000 + 8000}\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}{46,000}\\{53,000}\end{array}} \right]\] Revenue form market I is Rs 46,000
Revenue form market II is Rs 53,000
Profit = S.P. - C.P
\[\therefore text{profit matrix} = \left[ {\begin{array}{*{20}{c}}{0.50}\\{0.50}\\{0.50}\end{array}} \right]\] \[text{profit} = \left[ {\begin{array}{*{20}{c}}{10000}&{2000}&{18000}\\{6000}&{20,000}&{8000}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{0.5}\\{0.5}\\{0.5}\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}{5000 + 1000 + 9000}\\{3000 + 10000 + 4000}\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}{15000}\\{17000}\end{array}} \right]\]

Question (11)

Find the matrix X so that $X\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]$

Solution

\[\left[ x \right]\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]\] x must be 2×2 matrix \[\text{Let} x = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] \[x\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}{a + 4b}&{2a + 5b}&{3a + 6b}\\{c + 4d}&{2c + 5d}&{3c + 6d}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]\] \[ \Rightarrow a + 4b = - 7 - - - (1)\] \[2a + 5b = - 8 - - - (2)\] \[2 \times 1\] \[\begin{array}{l}2a + 8b = - 14\\\underline { \pm 2a \pm 5b = \mp 8} \\0\;\;\,\;\; + 3b = - 6\end{array}\] \[b = - 2\] substituting b = -2 in (1)
\[a - 8 = - 7\] \[ \Rightarrow a = - 7 + 8 = 1\] \[c + 4d = 2 - - - (3)\] \[c = 2 - 4d\] \[2c + 5d = 4\] \[2\left( {2 - 4d} \right) + 5d = 4\] \[4 - 8d + 5d = 4\] \[ - 3d = 0\] \[d = 0\] \[c = 2 - 4d = 2 - 0 = 2\] \[x = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\2&0\end{array}} \right]\]

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

\[A{B^n} = {B^n}A\] Let us prove for n=1
LHS = AB
LHS = BA ( ∴ AB = BA)
LHS = RHS
∴ It is true for n=1
Let us assume it is true for n = k
∴ AB^k = B^kA
Let us prove for n = k+1
LHS = AB^k+1
LHS = AB^k. B
LHS = B^kAB
LHS= B^kBA ( AB = BA)
LHS = B^k+1 A = LHS
∴ It is true for n = k +1
∴ By PMI it is true for all n ∈ N

(ii) (AB)ⁿ = AⁿBⁿ
Let us prove for n = 1
LHS = (AB)¹ = AB
RHS = A¹B¹ = AB = LHS
It is true for n = 1
Let us assume it is true for n = k
∴ (AB)^k = A^k B^k
Let us prove for n = k +1 , i.e. we will prove (AB)^k+1 = A^k+1 B^k+1
LHS = (AB)^k+1
LHS = A^kB^k AB
LHS = A^kB^k BA
LHS= A^kB^k+1 A
LHS= A^k+1B^k+1 = LHS
so PMI is true ∀ n ∈ N

Question (13)

If $A = \left[ {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right]$ 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

\[A = \left[ {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right]\] \[{A^2} = I\] \[\left[ {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}{{\alpha ^2} + \beta \gamma }&{\alpha \beta - \beta \alpha }\\{\gamma \alpha - \alpha \gamma }&{\gamma \beta + {\alpha ^2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}{{\alpha ^2} + \beta \gamma }&0\\0&{{\alpha ^2} + \gamma \beta }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[ \Rightarrow {\alpha ^2} + \beta \gamma = 1\] \[1 - {\alpha ^2} - \beta \gamma = 0\] 'C' is correct answer

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 oof these

Solution

A is symmetric ⇒ A = A' ---(1)
A is skew symmetric ∴ A' = -A ----(2)
From (1) and (2)
A = -A
2A = 0
A = 0
∴ so B is correct answer

Question (15)

If A is square matrix such that A² = A, the (I+A)³ -7A is equal to
(A) A    (B) I - A    (C) = I    (D) 3A

Solution

A² = A
(I + A)³ - 7A = I³ + 3IA(I+A) + A³ - 7A
= I + 3A(I + A) + A.A² - 7A
= I + 3AI + 3A² + A.A - 7A
=I + 3A +3A +A² - 7A
=I - A + A = I
so 'c' is correct answer