Miscellaneous Exercise Matrices Solution - NCERT Class 12


Chapter 3 Ex.3.ME Question 1

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

 

Solution

Video Solution

 

It is given that \(A = \left( {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right)\)

We shall prove the result by using the principle of mathematical induction.

For \(n = 1\), we have:

\(P\left( 1 \right):\left( {aI + bA} \right) = aI + b{a^0}A = aI + bA\)

Therefore, the result is true for \(n = 1\).

Let the result be true for \(n = k\)

That is, \(P\left( k \right):{\left( {aI + bA} \right)^k} = {a^k}I + k{a^{k - 1}}bA\)

Now, we have to prove that the result is true for \(n = k + 1\).

Consider,

\[\begin{align}{\left( {aI + bA} \right)^{k + 1}} &= {\left( {aI + bA} \right)^k}\left( {aI + bA} \right)\\ &= \left( {{a^k}I + k{a^{k - 1}}bA} \right)\left( {aI + bA} \right)\\ &= {a^{k + 1}}I + k{a^k}bAI + {a^k}bIA + k{a^{k - 1}}{b^2}{A^2}\\ &= {a^{k + 1}}I + \left( {k + 1} \right){a^k}bA + k{a^{k - 1}}{b^2}{A^2} \qquad \ldots \left( 1 \right)\end{align}\]

Now,

\({A^2} = \left( {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right) = 0\)

From \(\left( 1 \right)\), we have

\[\begin{align}{\left( {aI + bA} \right)^{k + 1}} &= {a^{k + 1}}I + \left( {k + 1} \right){a^k}bA + 0\\& = {a^{k + 1}}I + \left( {k + 1} \right){a^k}bA\end{align}\]

Therefore, the result is true for \(n = k + 1\).

Thus, by the principle of mathematical induction, we have:

\({\left( {aI + bA} \right)^n} = {a^n}I + n{a^{n - 1}}bA\) where \(A = \left( {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right)\),\(n \in N\)

Chapter 3 Ex.3.ME 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

Video Solution

 

It is given that \(A = \left( {\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}} \right)\)

We shall prove the result by using the principle of mathematical induction.

For \(n = 1\), we have:

\(P\left( 1 \right):\left( {\begin{array}{*{20}{c}}{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{3^0}}&{{3^0}}&{{3^0}}\\{{3^0}}&{{3^0}}&{{3^0}}\\{{3^0}}&{{3^0}}&{{3^0}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}} \right) = A\)

Therefore, the result is true for \(n = 1\) .

Let the result be true for \(n = k\).

\(P\left( k \right):{A^k} = \left( {\begin{array}{*{20}{c}} {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}} \\ {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}} \\ {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}} \end{array}} \right)\)

Now, we have to prove that the result is true for \(n = k + 1\).

Since,

\[\begin{align}{A^{k + 1}} &= A.{A^k}\\& = \left( {\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}\\{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}\\{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}&{3 \cdot {3^{k - 1}}}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{{3^{\left( {k + 1} \right) - 1}}}&{{3^{\left( {k + 1} \right) - 1}}}&{{3^{\left( {k + 1} \right) - 1}}}\\{{3^{\left( {k + 1} \right) - 1}}}&{{3^{\left( {k + 1} \right) - 1}}}&{{3^{\left( {k + 1} \right) - 1}}}\\{{3^{\left( {k + 1} \right) - 1}}}&{{3^{\left( {k + 1} \right) - 1}}}&{{3^{\left( {k + 1} \right) - 1}}}\end{array}} \right)\end{align}\]

Therefore, the result is true for \(n = k + 1\).

Thus, by the principle of mathematical induction, we have:

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

Chapter 3 Ex.3.ME Question 3

If \(A = \left( {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right)\) , 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

Video Solution

 

It is given that \(A = \left( {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right)\)

We shall prove the result by using the principle of mathematical induction.

For \(n = 1\), we have:

\[\begin{align}P\left( 1 \right):{A^1} &= \left( {\begin{array}{*{20}{c}}{1 + 2n}&{ - 4n}\\n&{1 - 2n}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right)\\& = A\end{align}\]

Therefore, the result is true for \(n = 1\).

Let the result be true for \(n = k\).

\(P\left( k \right):{A^k} = \left( {\begin{array}{*{20}{c}}{1 + 2k}&{ - 4k}\\k&{1 - 2k}\end{array}} \right),n \in N\)

Now, we have to prove that the result is true for \(n = k + 1\).

Since,

\[\begin{align}{A^{k + 1}} &= A.{A^k}\\ &= \left( {\begin{array}{*{20}{c}}{1 + 2k}&{ - 4k}\\k&{1 - 2k}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{3\left( {1 + 2k} \right) - 4k}&{ - 4\left( {1 + 2k} \right) + 4k}\\{3k + 1 - 2k}&{ - 4k - 1\left( {1 - 2k} \right)}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{3 + 6k - 4k}&{ - 4 - 8k + 4k}\\{3k + 1 - 2k}&{ - 4k - 1 + 2k}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{3 + 2k}&{ - 4 - 4k}\\{1 + k}&{ - 1 - 2k}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{1 + 2\left( {k + 1} \right)}&{ - 4\left( {k + 1} \right)}\\{1 + k}&{1 - 2\left( {k + 1} \right)}\end{array}} \right)\end{align}\]

Therefore, the result is true for \(n = k + 1\).

Thus, by the principle of mathematical induction, we have:

\({A^n} = \left( {\begin{array}{*{20}{c}}{1 + 2n}&{ - 4n}\\n&{1 - 2n}\end{array}} \right);\;n \in N\)

Chapter 3 Ex.3.ME Question 4

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

 

Solution

Video Solution

 

It is given that \(A\) and \(B\) are symmetric matrices.

Therefore, we have:

\(A' = A\) and \(B' = B\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\)

Now,

\[\begin{align}{\left( {AB - BA} \right)^\prime } &= {\left( {AB} \right)^\prime } - {\left( {BA} \right)^\prime } \qquad \left[ {{{\left( {A - B} \right)}^\prime } = A' - B'} \right]\\& = B'A' - A'B'\qquad\left[ {{{\left( {AB} \right)}^\prime } = B'A'} \right]\\ &= BA - AB\qquad\;\left[ {{\rm{Using }}\left( 1 \right)} \right]\\& = - \left( {AB - BA} \right)\end{align}\]

Hence,

\[{\left( {AB - BA} \right)^\prime } = - \left( {AB - BA} \right)\]

Thus, \(AB - BA\) is a skew symmetric matrix.

Chapter 3 Ex.3.ME Question 5

Show that the matrix \(B'AB\) is symmetric or skew symmetric according as \(A\) is symmetric or skew symmetric.

 

Solution

Video Solution

 

We suppose that \(A\) is a symmetric matrix, then

\(A' = A\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\)

Consider,

\[\begin{align} {\left( {B'AB} \right)^\prime } &= {\left\{ {B'\left( {AB} \right)} \right\}^\prime } \\ &= {\left( {AB} \right)^\prime }{\left( {B'} \right)^\prime }\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because {{\left( {AB} \right)}^\prime } = B'A'} \right] \\ &= B'A'\left( B \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because {{\left( {B'} \right)}^\prime } = B} \right] \\ &= B'\left( {A'B} \right) \\ &= B'\left( {AB} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{Using }}\left( 1 \right)} \right] \\ \end{align} \]

Therefore,

\({\left( {B'AB} \right)^\prime } = B'AB\)

Thus, if \(A\) is symmetric matrix, then \(B'AB\) is a symmetric matrix.

Now, we suppose that \(A\)is a skew symmetric matrix, then

\(A' = - A\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\)

Consider,

\[\begin{align}{\left( {B'AB} \right)^\prime } &= {\left\{ {B'\left( {AB} \right)} \right\}^\prime }\\ &= {\left( {AB} \right)^\prime }{\left( {B'} \right)^\prime }\\& = \left( {B'A'} \right)B\\ &= B'\left( { - A} \right)B \qquad \left[ {{\rm{Using }}\left( 2 \right)} \right]\\ &= - B'AB\end{align}\]

Therefore,

\({\left( {B'AB} \right)^\prime } = - B'AB\)

Thus, if \(A\) is a skew symmetric matrix, then \(B'AB\) is a skew symmetric matrix.

Hence, if \(A\) is symmetric or skew symmetric matrix, then \(B'AB\) is symmetric or skew symmetric accordingly.

Chapter 3 Ex.3.ME 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

Video Solution

 

It is given that \(A = \left( {\begin{array}{*{20}{c}}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right)\)

Therefore,

\(A' = \left( {\begin{array}{*{20}{c}}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right)\)

Now, \(A'A = I\)

Hence,

\[\begin{align}& \Rightarrow \;\left( {\begin{array}{*{20}{c}}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}{0 + {x^2} + {x^2}}&{0 + xy - xy}&{0 - xz + xz}\\{0 + xy - xy}&{4{y^2} + {y^2} + {y^2}}&{2yz - yz - yz}\\{0 - xz + zx}&{2yz - yz - yz}&{{z^2} + {z^2} + {z^2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\& \Rightarrow\; \left( {\begin{array}{*{20}{c}}{2{x^2}}&0&0\\0&{6{y^2}}&0\\0&0&{3{z^2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\end{align}\]

On comparing the corresponding elements, we have:

\[\begin{align}&2{x^2} = 1\\& \Rightarrow \quad x = \pm \frac{1}{{\sqrt 2 }}\\\\&6{y^2} = 1\\ &\Rightarrow \quad y = \pm \frac{1}{{\sqrt 6 }}\\\\&3{z^2} = 1\\ &\Rightarrow \quad z = \pm \frac{1}{{\sqrt 3 }}\end{align}\]

Thus, \(x = \pm \frac{1}{{\sqrt 2 }},\;y = \pm \frac{1}{{\sqrt 6 }}\) and \(z = \pm \frac{1}{{\sqrt 3 }}\)

Chapter 3 Ex.3.ME 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

Video Solution

 

We have:

\(\left[ {\begin{array}{*{20}{c}}1&2&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&2&0\\2&0&1\\1&0&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\2\\x\end{array}} \right] = 0\)

Hence,

\[\begin{align}& \Rightarrow \;\;\left[ {\begin{array}{*{20}{c}}{1 + 4 + 1}&{2 + 0 + 0}&{0 + 2 + 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\2\\x\end{array}} \right] = 0\\& \Rightarrow\;\; \left[ {\begin{array}{*{20}{c}}6&2&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\2\\x\end{array}} \right] = 0\\& \Rightarrow\;\; \left[ {6\left( 0 \right) + 2\left( 2 \right) + 4\left( x \right)} \right] = 0\\& \Rightarrow \;\;\left[ {4 + 4x} \right] = 0\\ &\Rightarrow\;\; 4x = - 4\\& \Rightarrow\;\; x = - 1\end{align}\]

Thus, the required value of \(x = - 1\).

Chapter 3 Ex.3.ME Question 8

If \(A = \left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)\), show that \({A^2} - 5A + 7I = 0\)

 

Solution

Video Solution

 

It is given that \(A = \left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)\)

Therefore,

\[\begin{align}{A^2} &= A.A\\ &= \left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{3\left( 3 \right) + 1\left( { - 1} \right)}&{3\left( 1 \right) + 1\left( 2 \right)}\\{ - 1\left( 3 \right) + 2\left( { - 1} \right)}&{ - 1\left( 1 \right) + 2\left( 2 \right)}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right)\end{align}\]

Now,

\[\begin{align}LHS &= {A^2} - 5A + 7I\\ &= \left( {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right) + 7\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{15}&5\\{ - 5}&{10}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}7&0\\0&7\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{ - 7}&0\\0&{ - 7}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}7&0\\0&7\end{array}} \right)\\ &= 0\\& = RHS\end{align}\]

Thus, \({A^2} - 5A + 7I = 0\)

Chapter 3 Ex.3.ME Question 9

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

 

Solution

Video Solution

 

We have

\(\left[ {\begin{array}{*{20}{c}}x&{ - 5}&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\1\end{array}} \right] = 0\)

Hence,

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

Thus, \(x = \pm 4\sqrt 3 \).

Chapter 3 Ex.3.ME Question 10

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

Market

Products

I

\(10000\)

\(2000\)

\(18000\)

II

\(6000\)

\(20000\)

\(8000\)

If unit sale prices of \(x,y\) and \(z\)are ₹\({\rm{2}}.{\rmabc}0\), ₹\({\rm{1}}.{\rm{5}}0\)and ₹\({\rm{1}}.00\), respectively, find the total revenue in each market with the help of matrix algebra.

If the unit costs of the above three commodities are ₹\({\rm{2}}.00\), ₹\({\rm{1}}.00\)and \({\rm{5}}0\) paise respectively. Find the gross profit.

 

Solution

Video Solution

 

The unit sale prices of \(x,y\) and \(z\)are ₹\({\rm{2}}.{\rm{5}}0\), ₹\({\rm{1}}.{\rm{5}}0\)and ₹\({\rm{1}}.00\) respectively.

Consequently, the total revenue in market I can be represented in the form of a matrix as:

\[\begin{align}\left[ {\begin{array}{*{20}{c}}{10000}&{2000}&{18000}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{2.50}\\{1.50}\\{1.00}\end{array}} \right] &= 10000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1.00\\ &= 25000 + 3000 + 18000\\ &= 46000\end{align}\]

The total revenue in market II can be represented in the form of a matrix as:

\[\begin{align}\left[ {\begin{array}{*{20}{c}}{6000}&{20000}&{8000}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{2.50}\\{1.50}\\{1.00}\end{array}} \right]& = 6000 \times 2.50 + 20000 \times 1.50 + 8000 \times 1.00\\& = 15000 + 30000 + 8000\\ &= 53000\end{align}\]

Thus, the total revenue in market I is ₹\(46000\) and the total revenue in market II is ₹\(53000\).

The unit costs of \(x,y\) and \(z\)are ₹\({\rm{2}}.00\), ₹\({\rm{1}}.00\)and \({\rm{5}}0\) paise respectively.

Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:

\[\begin{align}\left[ {\begin{array}{*{20}{c}}{10000}&{2000}&{18000}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{2.00}\\{1.00}\\{0.50}\end{array}} \right] &= 10000 \times 2.00 + 2000 \times 1.00 + 18000 \times 0.50\\ &= 20000 + 2000 + 9000\\ &= 31000\end{align}\]

Since the total revenue in market I is ₹\(46000\), the gross profit in this market in ₹ is

\(46000 - 31000 = 15000\)

The total cost prices of all the products in market II can be represented in the form of a matrix as:

\[\begin{align}\left[ {\begin{array}{*{20}{c}}{6000}&{20000}&{8000}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{2.00}\\{1.00}\\{0.50}\end{array}} \right] &= 6000 \times 2.00 + 20000 \times 1.00 + 8000 \times 0.50\\ &= 12000 + 20000 + 4000\\ &= 36000\end{align}\]

Since the total revenue in market I is ₹\(53000\), the gross profit in this market in ₹ is

\(53000 - 36000 = 17000\)

Thus, the gross profit in market I is ₹\(15000\) and in market II is ₹\(17000\).

Chapter 3 Ex.3.ME 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

Video Solution

 

It is given that \(X\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]\)

The matrix given on the R.H.S. of the equation is a \(2 \times 3\) matrix and the one given on the L.H.S. of the equation is a \(2 \times 3\) matrix.

Therefore, X has to be a \(2 \times 2\) matrix.

Now, let \(X = \left[ {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right]\)

Therefore,

\[\begin{align}& \Rightarrow \left[ {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]\\\\& \Rightarrow \left[ {\begin{array}{*{20}{c}}{a + 4c}&{2a + 5c}&{3a + 6c}\\{b + 4d}&{2b + 5d}&{3b + 6d}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]\end{align}\]

Equating the corresponding elements of the two matrices, we have:

\[\begin{align}a + 4c& = 7\,\,\,\,\,\,\,2a + 5c = - 8\,\,\,\,\,\,\,3a + 6c = - 9a\\b + 4d &= 2\,\,\,\,\,\,\,2b + 5d = 4\,\,\,\,\,\,\,\,\,3b + 6d = 6\end{align}\]

Now,

\[\begin{align}a + 4c &= - 7\\ \Rightarrow a &= - 7 - 4c\end{align}\]

Therefore,

\[\begin{align}&2a + 5c = - 8\\ &\Rightarrow\;\; 2\left( { - 7 - 4c} \right) + 5c = - 8\\ &\Rightarrow \;\;- 14 - 8c + 5c = - 8\\& \Rightarrow \;\;- 3c = 6\\& \Rightarrow\;\; c = - 2\end{align}\]

Hence,

\[\begin{align}& \Rightarrow \;a = - 7 - 4\left( { - 2} \right)\\ &\Rightarrow \;a = - 7 + 8\\& \Rightarrow\; a = 1\end{align}\]

Now,

\[\begin{align}b + 4d = 2\\ \Rightarrow b = 2 - 4d\end{align}\]

Therefore,

\[\begin{align}&2b + 5d = 4\\& \Rightarrow \;2\left( {2 - 4d} \right) + 5d = 4\\& \Rightarrow \;4 - 8d + 5d = 4\\ &\Rightarrow \;- 3d = 0\\& \Rightarrow \;d = 0\end{align}\]

Hence,

\[\begin{align}b &= 2 - 4d\\ \Rightarrow b &= 2\end{align}\]

Thus, \(a = 1,b = 2,c = - 2\) and \(d = 0\)

Hence, the required matrix \(X = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\2&0\end{array}} \right]\)

Chapter 3 Ex.3.ME Question 12

If \(A\) and \(B\) are square matrices of the same order such that \(AB = BA\), then prove by induction that \(A{B^n} = {B^n}A\). Further, prove that \({\left( {AB} \right)^n} = {A^n}{B^n}\) for all \(n \in N\).

 

Solution

Video Solution

 

Given: \(A\) and \(B\) are square matrices of the same order such that \(AB = BA\).

To prove: \(P\left( n \right):A{B^n} = {B^n}A,n \in N\)

For \(n = 1\), we have:

\[\begin{align}&P\left( 1 \right):AB = BA\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Given}}} \right]\\& \Rightarrow A{B^1} = {B^1}A\end{align}\]

Therefore, the result is true for \(n = 1\).

Let the result be true for \(n = k\).

\(P\left( k \right) = A{B^k} = {B^k}A\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\)

Now, we prove that the result is true for \(n = k + 1\).

\[\begin{align}A{B^{k + 1}} &= A{B^k}.B\\ &= \left( {{B^k}A} \right)B \qquad \left[ {{\rm{ By}}\;\left( 1 \right)} \right]\\ &= {B^k}\left( {AB} \right) \qquad \left[ {{\text{Associative law}}} \right]\\ &= {B^k}\left( {BA} \right) \qquad \left[ {AB = BA\;{\rm{ }}\left( {{\text{Given}}} \right)} \right]\\ &= \left( {{B^k}B} \right)A \qquad \left[ {{\text{Associative law}}} \right]\\ &= {B^{k + 1}}A\end{align}\]

Therefore, the result is true for .

Thus, by the principle of mathematical induction, we have \(A{B^n} = {B^n}A,n \in N\)

Now, we have to prove that \({\left( {AB} \right)^n} = {A^n}{B^n}\) for all \(n \in N\)

For \(n = 1\), we have:

\({\left( {AB} \right)^1} = {A^1}{B^1} = AB\)

Therefore, the result is true for \(n = 1\).

Let the result be true for \(n = k\).

\({\left( {AB} \right)^k} = {A^k}{B^k}\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\)

Now, we prove that the result is true for \(n = k + 1\).

\[\begin{align}A{B^{k + 1}} &= {\left( {AB} \right)^k}.\left( {AB} \right)\\ &= \left( {{A^k}{B^k}} \right).\left( {AB} \right) \qquad \left[ {{\rm{By }}\left( {\rm{2}} \right)} \right]\\ &= {A^k}\left( {{B^k}A} \right)B\qquad\;\;\left[ {{\text{Associative law}}} \right]\\ &= {A^k}\left( {A{B^k}} \right)B\qquad\;\;\;\left[ {A{B^n} = {B^n}A,n \in N} \right]\\ &= \left( {{A^k}A} \right).\left( {{B^k}B} \right)\qquad\;\;\left[ {{\text{Associative law}}} \right]\\ &= {A^{k + 1}}{B^{k + 1}}\end{align}\]

Therefore, the result is true for \(n = k + 1\).

Thus, by the principle of mathematical induction, we have \({\left( {AB} \right)^n} = {A^n}{B^n},n \in N\)

Chapter 3 Ex.3.ME Question 13

If \(A = \left[ {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right]\) is such that \({A^2} = I\) then,

(A) \(1 + {\alpha ^2} + \beta \gamma = 0\)

(B) \(1 - {\alpha ^2} + \beta \gamma = 0\)

C) \(1 - {\alpha ^2} - \beta \gamma = 0\)

(D) \(1 + {\alpha ^2} - \beta \gamma = 0\)

 

Solution

Video Solution

 

It is given that \(A = \left[ {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right]\)

Therefore,

\[\begin{align}{A^2} &= A.A\\ &= \left( {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right)\left( {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{{\alpha ^2} + \beta \gamma }&{\alpha \beta - \alpha \beta }\\{\alpha \gamma - \alpha \gamma }&{\beta \gamma + {\alpha ^2}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{{\alpha ^2} + \beta \gamma }&0\\0&{\beta \gamma + {\alpha ^2}}\end{array}} \right)\end{align}\]

Now, \({A^2} = I\)

Hence,

\(\left( {\begin{array}{*{20}{c}}{{\alpha ^2} + \beta \gamma }&0\\0&{\beta \gamma + {\alpha ^2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\)

On comparing the corresponding elements, we have:

\[\begin{align}&{\alpha ^2} + \beta \gamma = 1\\& \Rightarrow \; {\alpha ^2} + \beta \gamma - 1 = 0\\& \Rightarrow \; 1 - {\alpha ^2} - \beta \gamma = 0\end{align}\]

Thus, the correct option is \(C\).

Chapter 3 Ex.3.ME 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

Video Solution

 

If the matrix \(A\) is both symmetric and skew symmetric, then

\(A' = A\) and \(A' = - A\)

Hence,

\[\begin{align}& \Rightarrow \; A = - A\\& \Rightarrow \; A + A = 0\\ &\Rightarrow \; 2A = 0\\ &\Rightarrow\; A = 0\end{align}\]

Therefore, \(A\) is a zero matrix.

Thus, the correct option is \(B\).

Chapter 3 Ex.3.ME Question 15

If \(A\) is a square matrix such that \({A^2} = A\), then \({\left( {I + A} \right)^3} - 7A\) is equal to

(A) \(A\)

(B) \(I - A\)

(C) \(I\)

(D) \(3A\)

 

Solution

Video Solution

 

It is given that \(A\) is a square matrix such that \({A^2} = A\).

Now,

\[\begin{align} {\left( {I + A} \right)^3} - 7A &= {I^3} + {A^3} + 3{I^2}A + 3{A^2}I - 7A \\ &= I + {A^2}.A + 3A + 3{A^2} - 7A \\& = I + A.A + 3A + 3A - 7A \qquad \left[ {\because {A^2} = A} \right] \\& = I + {A^2} - A \\ &= I + A - A \qquad \left[ {\because {A^2} = A} \right] \\ &= I \\ \end{align} \]

Hence,

\[{\left( {I + A} \right)^3} - 7A = I\]

Thus, the correct option is \(C\).

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