Chapter 3 ( Miscellaneous)

class 12 maths chapter 3 Miscellaneous solution

Question 1: Let , $A=\begin{bmatrix} 0&1\\0&0\end{bmatrix}$ show that , $(aI+bA)^n=a^nI+na^{n-1}bA$ where I is the identity matrix of order 2 and $n\in N$ .

Solution: It is given that $A=\begin{bmatrix} 0&1\\0&0\end{bmatrix}$
$(aI+bA)=a\begin{bmatrix} 1&0\\0&1\end{bmatrix}+b\begin{bmatrix} 0&1\\0&0\end{bmatrix}$

$\Rightarrow (aI+bA)=\begin{bmatrix} a&0\\0&a\end{bmatrix}+\begin{bmatrix} 0&b\\0&0\end{bmatrix}$

$\Rightarrow (aI+bA)=\begin{bmatrix} a&b\\0&a\end{bmatrix}$
LHS.
$\Rightarrow (aI+bA)^n=\begin{bmatrix} a&b\\0&a\end{bmatrix}^n$
RHS.
$a^nI+na^{n-1}bA=a^n\begin{bmatrix} 1&0\\0&1\end{bmatrix}+na^{n-1}b\begin{bmatrix} 0&1\\0&0\end{bmatrix}$

$\Rightarrow a^nI+na^{n-1}bA=a^n\begin{bmatrix} 1&0\\0&1\end{bmatrix}+na^{n-1}b\begin{bmatrix} 0&1\\0&0\end{bmatrix}$

$\Rightarrow a^nI+na^{n-1}bA=\begin{bmatrix} a^n&0\\0&a^n\end{bmatrix}+\begin{bmatrix} 0&na^{n-1}b\\0&0\end{bmatrix}$

$\Rightarrow a^nI+na^{n-1}bA=\begin{bmatrix} a^n&na^{n-1}b\\0&a^n\end{bmatrix}$
Since,

$(aI+bA)^n=a^nI+na^{n-1}bA$
Hence,

$\begin{bmatrix} a&b\\0&a\end{bmatrix}^n=\begin{bmatrix} a^n&na^{n-1}b\\0&a^n\end{bmatrix}$

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

For n=1, we have:

P(1): $\begin{bmatrix} a&b\\0&a\end{bmatrix}^1=\begin{bmatrix} a^n&1a^{1-1}b\\0&a^n\end{bmatrix}$

$\Rightarrow\begin{bmatrix} a&b\\0&a\end{bmatrix}=\begin{bmatrix} a^n&b\\0&a^n\end{bmatrix}$

Therefore, the result is true for n=1 .

Let the result be true for $n=k$
That is :

P(k): $\begin{bmatrix} a&b\\0&a\end{bmatrix}^k=\begin{bmatrix} a^k&ka^{k-1}b\\0&a^k\end{bmatrix}---(1)$

Now, we have to prove that the result is true for $n=k+1$
Consider,

$P(k+1):\begin{bmatrix} a&b\\0&a\end{bmatrix}^{k+1}=\begin{bmatrix} a^{k+1}&(k+1)a^{k}b\\0&a^{k+1}\end{bmatrix}$

LHS.

$\begin{bmatrix} a&b\\0&a\end{bmatrix}^{k+1}=\begin{bmatrix} a&b\\0&a\end{bmatrix}^k\begin{bmatrix} a&b\\0&a\end{bmatrix}$

$\Rightarrow \begin{bmatrix} a&b\\0&a\end{bmatrix}^{k+1}=\begin{bmatrix} a^k&ka^{k-1}b\\0&a^k\end{bmatrix}\begin{bmatrix} a&b\\0&a\end{bmatrix} \space [from (1)]$

$=\begin{bmatrix} a^ka+0&ba^k+ka^{k-1}ab\\0+0&0+a^ka\end{bmatrix}$

$=\begin{bmatrix} a^{k+1}&(k+1)a^{k}b\\0&a^{k+1}\end{bmatrix}$

$=RHS$
Therefore, the result is true for $n=k+1$

Question 2: If $A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$ ,prove that $A^n=\begin{bmatrix}3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\end{bmatrix}, n\in N$

Solution: It is given that $A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$

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

Let the result be true for $n=1$ .
Now, we have:
$P(1):A^1=\begin{bmatrix}3^{1-1}&3^{1-1}&3^{1-1}\\3^{1-1}&3^{1-1}&3^{1-1}\\3^{1-1}&3^{1-1}&3^{1-1}\end{bmatrix}$

$\Rightarrow A^1=\begin{bmatrix}3^0&3^0&3^0\\3^0&3^0&3^0\\3^0&3^0&3^0\end{bmatrix}$

$\Rightarrow A^1=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
Therefore, the result is true for n=1.
Let the result be true for $n=k$ .

$P(k):A^k=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}---(1)$

Now, we have to prove that the result is true for $n=k+1$ .
Since,

$P(k+1):A^{k+1}=\begin{bmatrix}3^k&3^k&3^k\\3^k&3^k&3^k\\3^k&3^k&3^k\end{bmatrix}$

$LHS$

$A^{k+1}=A^kA$

$=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}-----[From (1)]$

$=\begin{bmatrix}3^{k-1}+3^{k-1}+3^{k-1}&3^{k-1}+3^{k-1}+3^{k-1}&3^{k-1}+3^{k-1}+3^{k-1}\\3^{k-1}+3^{k-1}+3^{k-1}&3^{k-1}+3^{k-1}+3^{k-1}&3^{k-1}+3^{k-1}+3^{k-1}\\3^{k-1}+3^{k-1}+3^{k-1}&3^{k-1}+3^{k-1}+3^{k-1}&3^{k-1}+3^{k-1}+3^{k-1}\end{bmatrix}$

$=\begin{bmatrix}3.3^{k-1}&3.3^{k-1}&3.3^{k-1}\\3.3^{k-1}&3.3^{k-1}&3.3^{k-1}\\3.3^{k-1}&3.3^{k-1}&3.3^{k-1}\end{bmatrix}$

$=\begin{bmatrix}3^k&3^k&3^k\\3^k&3^k&3^k\\3^k&3^k&3^k\end{bmatrix}$

$=RHS$

Therefore, the result is true for $n=k+1$ .

Question 3: If $A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$ ,prove that $A^n=\begin{bmatrix}1+2n&-4n\\n&1-2n\end{bmatrix}$ ,where is any positive integer.

Solution: It is given that $A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$

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

For n=1 , we have:

$P(1):A^1=\begin{bmatrix}1+2(1)&-4(1)\\(1)&1-2(1)\end{bmatrix}$

$=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$

$=A$

Therefore, the result is true for $n=1$

Let the result be true for n=k .

$A^k=\begin{bmatrix}1+2k&-4k\\k&1-2k\end{bmatrix},k\in N---(1)$

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

Since,

$A^{k+1}=\begin{bmatrix}1+2(k+1)&-4(k+1)\\(k+1)&1-2(k+1)\end{bmatrix}$

$LHS$

$A^{k+1}=A^kA$

$=\begin{bmatrix}1+2k&-4k\\k&1-2k\end{bmatrix}\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$

$=\begin{bmatrix}3+6k-4k&-4-8k+4k\\3k+1-2k&-4k-1+2k\end{bmatrix}$

$=\begin{bmatrix}3+2k&-4-4k\\k+1&-1-2k\end{bmatrix}$

$RHS$

$\begin{bmatrix}1+2(k+1)&-4(k+1)\\(k+1)&1-2(k+1)\end{bmatrix}$

$\begin{bmatrix}1+2k+2&-4k-4\\(k+1)&1-2k-2\end{bmatrix}$

$\begin{bmatrix}3+2k&-4k-4)\\(k+1)&-1-2k\end{bmatrix}$
Hence,

$LHS=RHS$

Therefore, the result is true for $n=k+1$

Question 4: If $A$ and $B$ are symmetric matrices, prove that $AB-BA$ is a skew symmetric matrix.

Solution: It is given that $A$ and $B$ are symmetric matrices.

Therefore, we have:

$A'=A$ and $B'=B ----(1)$
Now,

$(AB-BA)'=(AB)'-(BA)'$

$=B'A'-A'B'$

$=BA-AB$

$=-(AB-BA)$

Hence,

$(AB-BA)'=-(AB-BA)$

Thus, $AB-BA$ is a skew symmetric matrix.

Question 5: Show that the matrix $B'AB$ is symmetric or skew symmetric according as $A$ is symmetric orskew symmetric.

Solution: We suppose that $A$ is a symmetric matrix, then

$A'=A------(1)$

Consider,

$(B'AB)'=\{B'(AB)\}'$

$=(AB)'(B')'---[since (AB)'=B'A']$

$=B'A'B---[(B')'=B]$

$=B'(A'B)$

$=B'(AB)---[From (1)]$

Therefore,

$(B'AB)'=B'AB$

Thus, if $A$ is symmetric matrix, then $B'AB$ is a symmetric matrix.

Now, we suppose that $A$ is a skew symmetric matrix, then

$A'=-A---(2)$

Consider,

$(B'AB)'=\{B'(AB)\}'$

$=(AB)'(B')'---[since (AB)'=B'A']$

$=B'A'B---[(B')'=B]$

$=B'(A'B)$

$=B'(-AB)---[From (2)]$

$=-B'AB$

Therefore

$(B'AB)'=-B'AB$

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.

Question 6: Find the values of $x,y,z$ if the matrix $A=\begin{bmatrix}0&2y&z\\x&y&-z\\x&-y&z\end{bmatrix}$ satisfy the equation $A'A=I$

Solution: It is given that $A=\begin{bmatrix}0&2y&z\\x&y&-z\\x&-y&z\end{bmatrix}$

Therefore,

$A'=\begin{bmatrix}0&x&x\\2y&y&-y\\z&-z&z\end{bmatrix}$
Now,

$A'A=I$

Hence,

$\Rightarrow \begin{bmatrix}0&x&x\\2y&y&-y\\z&-z&z\end{bmatrix}\begin{bmatrix}0&2y&z\\x&y&-z\\x&-y&z\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

$\Rightarrow \begin{bmatrix}0+x^2+x^2&0+xy-xy&0-xz+xz\\0+xy-xy&4y^2+y^2+y^2&2yz-yz-yz\\0-xz+xz&2yz-yz-yz&z^2+z^2+z^2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

$\Rightarrow \begin{bmatrix}2x^2&0&0\\0&6y^2&0\\0&0&3z^2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

On comparing the corresponding elements, we have:

$2x^2=1$

$\Rightarrow x=\pm\frac{1}{\sqrt2}$

$6y^2=1$

$\Rightarrow y=\pm\frac{1}{\sqrt6}$

$3z^2=1$

$\Rightarrow z=\pm\frac{1}{\sqrt3}$
Hence,
$x=\pm\frac{1}{\sqrt2},y=\pm\frac{1}{\sqrt6}$ and $z=\pm\frac{1}{\sqrt3}$

Question 7:For what values of x: $\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0$ ?

Solution: We have:

$\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0$

$\Rightarrow \begin{bmatrix}1+4+1&2+0+0&0+2+2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0$

$\Rightarrow \begin{bmatrix}6&2&4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0$

$\Rightarrow \begin{bmatrix}6(0)+2(2)+4x\end{bmatrix}=0$

$\Rightarrow 4+4x=0$

$\Rightarrow 4x=-4$

$\Rightarrow x=-1$

Thus, the required value of $x=-1$ .

Question 8: If $A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ , show that $A^2-5A+7I=0$

Solution : It is given that $A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$

Therefore,

$A^2=AA$

$=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$

$=\begin{bmatrix}3(3)+1(-1)&3(1)+1(2)\\-1(3)+2(-1)&-1(1)+2(2)\end{bmatrix}$

$=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$

Now,

$LHS=A^2-5A+7I$

$=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$

$=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$

$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

$=RHS$

Thus, $A^2-5A+7I=0$

Question 9: Find , if $\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$

Solution: $\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$

Hence,

$\begin{bmatrix}x+0-2&0-10+0&2x-5-3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$

$\Rightarrow \begin{bmatrix}x-2&-10&2x-8\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$

$\Rightarrow \begin{bmatrix}x(x-2)-40+2x-8\end{bmatrix}=0$

$\Rightarrow x^2-2x+2x-48=0$

$x^2=48$

$\Rightarrow x=\pm4\sqrt3$

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

product x y z

I 10000 2000 18000
II 6000 20000 8000

(a) If unit sale prices of $x,y$ and $z$ are ₹ 2.50 ,₹ 1.50and₹ 1.00, 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 ₹2 , ₹1 and 50paise respectively.
Find the gross profit.

Solution: (a) The unit sale prices of $x,y$ and $z$ are ₹2.50 , ₹1.50 and ₹ 1.00respectively.

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

Product matrix: $\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}$
Sale price matirx: $\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}$

The total revenue $=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}$

$=\begin{bmatrix}10000(2.50)+2000(1.50)+18000(1.00)\\6000(2.50)+20000(1.50)+8000(1.00)\end{bmatrix}$

$=\begin{bmatrix}25000+3000+18000\\15000+30000+8000\end{bmatrix}$

$=\begin{bmatrix}46000\\53000\end{bmatrix}$

Total revenue in I market= ₹46000

Total revenue in II market=₹53000

(b) The unit costs of $x,y$ and $z$ are ₹2.00 , ₹1.00 and 50 paise respectively.

Consequently,

total cost price of each market:

$=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}$

$=\begin{bmatrix}10000(2.00)+2000(1.00)+18000(0.50)\\6000(2.00)+20000(1.00)+8000(0.50)\end{bmatrix}$

$=\begin{bmatrix}20000+2000+9000\\12000+20000+4000\end{bmatrix}$

$=\begin{bmatrix}31000\\36000\end{bmatrix}$

Profit: $=\begin{bmatrix}46000\\53000\end{bmatrix}-\begin{bmatrix}31000\\36000\end{bmatrix}$

$=\begin{bmatrix}15000\\17000\end{bmatrix}$

Thus, the gross profit in market I is ₹ $15000$ and in market II is ₹ $17000$ .

Question 11: Find the matrix $X$ so that $X\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$

Solution: $X\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$

The matrix given on the R.H.S. of the equation is a $2\times3$

matrix and the one given on the L.H.S.of the equation is a $2\times3$ matrix.

now, Let $X=\begin{bmatrix}a&c\\b&d\end{bmatrix}$
Therefore,

$\begin{bmatrix}a&c\\b&d\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$

$\Rightarrow\begin{bmatrix}a+4c&2a+5c&3a+6c\\b+4d&2b+5d&3b+6d\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$

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

$a+4c=7,2a+5c=-8,3a+6c=-9$

$b+4d=2,2b+5d=4,3b+6d=6$

Now,

solving $a+4c=7$ and $2a+5c=-8$
we get ,

$a=1,c=-2$

again solving

$b+4d=2$ and $2b+5d=4$

We get,

$b=2,d=0$

Hence the required matrix $X=\begin{bmatrix}1&-2\\2&0\end{bmatrix}$

Question 12: If $A$ and $B$ are square matrices of the same order such that $AB=BA$ ,then prove by induction that $AB^n=B^nA$ .Further prove that $(AB)^n= A^nB^n$ for all $n\in N$

Solution: Given $A$ and $B$ are square matrices of the same order such that $AB=BA$ .

To prove: $P(n):AB^n=B^nA,n\in N$

For $n=1$ ,we have:

$P(1):AB=BA [Given]$ .

therefore, the result is true for $n=1$ .

Let the result is true for $n=k$

$P(k):AB^k=B^kA---(1)$

Now, we prove that the result is true for $n=k+1$

$P(k+1):AB^{k+1}=B^{k+1}A$

$LHS=AB^{k+1}$

$=AB^KB$

$=(B^kA)B\space ---[By 1]$

$=B^k(AB)----[Associative \ law]$

$=B^k(BA)\ [AB=BA]$

$=(B^kB)A\ [Associative\ law]$

$=B^{k+1}A$

Therefore, the result is true for $n=k+1$ .

Now, we have to prove that $(AB)^n=A^nB^n$ For all $n\in N$

For $n=1$ , we have:

$P(1):(AB)^1=A^1B^1=AB$

Therefore, the result is true for $n=1$ .

Let the result be true for $n=k$ .

$P(k):(AB)^k=A^kB^k---(2)$

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

$P(k+1):(AB)^{k+1}=A^{k+1}B^{k+1}$

$LHS=(AB)^{k+1}$

$=(AB)^{k}(AB)$

$=(A^kB^k)(AB)---[By 2]$

$=A^k(B^kA)A \ [Associative \ law]$

$=A^k(AB^k)B \ [AB^n=B^nA]$

$=(A^kA)(B^kB) \ [Associative \ law]$

$=A^{k+1}B^{k+1}$

Therefore, the result is true for $n=k+1$ .

Question 13: If $A=\begin{bmatrix}\alpha&\beta\\ \gamma&\alpha\end{bmatrix}$ is such that $A^n=I$ then,

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

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

(D) $1-\alpha^2-\beta\gamma=0$

Solution: It is given that $A=\begin{bmatrix}\alpha&\beta\\ \gamma&\alpha\end{bmatrix}$
Therefore,

$A^2=AA$

$=\begin{bmatrix}\alpha&\beta\\ \gamma&\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\ \gamma&\alpha\end{bmatrix}$

$=\begin{bmatrix}\alpha^2+\beta\gamma&\alpha\beta-\beta\alpha\\ \alpha\gamma-\alpha\gamma&\beta\gamma+\alpha^2\end{bmatrix}$

$=\begin{bmatrix}\alpha^2+\beta\gamma&0\\ 0&\beta\gamma+\alpha^2\end{bmatrix}$
Now, $A^2=I$
Hence,

$=\begin{bmatrix}\alpha^2+\beta\gamma&0\\ 0&\beta\gamma+\alpha^2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

On comparing the corresponding elements, we have:

$\alpha^2+\beta\gamma=1$

$\Rightarrow\alpha^2+\beta\gamma-1=0$

$\Rightarrow1-\alpha^2-\beta\gamma=0$
Thus, the correct option is C.

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

(D) None of these

Solution: If the matrix $A$ is both symmetric and skew symmetric, then

$A'=A$ and $A'=A$

Hence,

$\Rightarrow A=-A$

$\Rightarrow A+A=0$

$\Rightarrow 2A=0$

$\Rightarrow A=0$

Therefore, $A$ is a zero matrix.

Thus, the correct option is B.

Question 15: If $A$ is a square matrix such that $A^2=A$ , then $(I+A)^3-7A$ is equal to

(A) $A$ (B) $I-A$ (C) $I$ (D) $3A$

Solution: It is given that $A$ is a square matrix such that $A^2=A$ .

Now,

$(I+A)^3-7A=I^3+A^3+3I^2A+3A^2I-7A$

$=I+A^2A+3A+3A^2-7A$

$=I+AA+3A+3A-7I \ [A^2=A]$

$=I+A^2+6A-7A$

$=I+A+6A-7A$

$=I$

Thus, the correct option is $C$ .

Case study problem matrix 2 — Amit, Biraj and chirag were giventhe task of creating a square matrix of order 2

Chapter 3 ( Miscellaneous)

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