CBSE Class 12 Mathematics HOTs Matrices and Determinants Set 02

Question. Write the number of all possible matrices of order \( 2 \times 2 \) with each entry 1, 2 or 3.
Answer: As matrix is of order \( 2 \times 2 \), so there are 4 entries possible. Each entry has 3 choices i.e. 1, 2 or 3. So, the number of ways to make such matrices is \( 3 \times 3 \times 3 \times 3 = 81 \).

Question. Write the element \( a_{23} \) of a \( 3 \times 3 \) matrix \( A = [a_{ij}] \) whose elements \( a_{ij} \) are given by \( a_{ij} = \frac{|i - j|}{2} \).
Answer: Here, \( a_{ij} = \frac{|i - j|}{2} \)
\( \therefore a_{23} = \frac{|2 - 3|}{2} = \frac{1}{2} \) [For \( i = 2, j = 3 \)]

Question. The elements \( a_{ij} \) of a \( 3 \times 3 \) matrix are given by \( a_{ij} = \frac{1}{2} |-3i + j| \). Write the value of element \( a_{32} \).
Answer: Here, \( a_{ij} = \frac{1}{2} |-3i + j| \)
\( \therefore a_{32} = \frac{1}{2} |-3 \cdot 3 + 2| \) [For \( i = 3, j = 2 \)]
\( = \frac{1}{2} |-9 + 2| = \frac{1}{2} |-7| = \frac{7}{2} \)

Question. For a \( 2 \times 2 \) matrix \( A = [a_{ij}] \), whose elements are given by \( a_{ij} = \frac{(i + 2j)^2}{4} \), write the value of \( a_{21} \).
Answer: Here, \( a_{ij} = \frac{(i + 2j)^2}{4} \)
\( \dots a_{21} = \frac{(2 + 2 \cdot 1)^2}{4} = 4 \) [For \( i = 2, j = 1 \)]

Question. For a \( 2 \times 2 \) matrix, \( A = (a_{ij}) \) whose elements are given by \( a_{ij} = \frac{i}{j} \), write the value of \( a_{12} \).
Answer: Here, \( a_{ij} = \frac{i}{j} \Rightarrow a_{12} = \frac{1}{2} \) [For \( i = 1, j = 2 \)]

Question. If a matrix has 5 elements, then write all possible orders it can have.
Answer: The possible orders are \( 1 \times 5 \) or \( 5 \times 1 \).

Question. If \( \begin{bmatrix} x-y & z \\ 2x-y & w \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ 0 & 5 \end{bmatrix} \), find the value of \( x + y \).
Answer: Here, \( \begin{bmatrix} x-y & z \\ 2x-y & w \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ 0 & 5 \end{bmatrix} \)
\( \Rightarrow x - y = -1, z = 4, 2x - y = 0, w = 5 \)
Solving these equations for \( x \) and \( y \), we get \( x = 1, y = 2 \)
\( \therefore x + y = 1 + 2 = 3 \).

Question. If \( \begin{bmatrix} a+4 & 3b \\ 8 & -6 \end{bmatrix} = \begin{bmatrix} 2a+2 & b+2 \\ 8 & a-8b \end{bmatrix} \), write the value of \( a - 2b \).
Answer: Given, \( \begin{bmatrix} a+4 & 3b \\ 8 & -6 \end{bmatrix} = \begin{bmatrix} 2a+2 & b+2 \\ 8 & a-8b \end{bmatrix} \)
By equality of matrices, we get \( a + 4 = 2a + 2, 3b = b + 2, -6 = a - 8b \)
On solving these equations, we get \( a = 2, b = 1 \).
So \( a - 2b = 0 \).

Question. If \( \begin{bmatrix} x \cdot y & 4 \\ z+6 & x+y \end{bmatrix} = \begin{bmatrix} 8 & w \\ 0 & 6 \end{bmatrix} \), write the value of \( (x+y+z) \).
Answer: Here, \( \begin{bmatrix} x \cdot y & 4 \\ z+6 & x+y \end{bmatrix} = \begin{bmatrix} 8 & w \\ 0 & 6 \end{bmatrix} \)
\( \Rightarrow x \cdot y = 8, w = 4, z + 6 = 0, x + y = 6 \)
\( \Rightarrow z = -6, x + y = 6 \)
\( \Rightarrow x + y + z = 6 - 6 = 0 \).

Question. Find the value of \( a \) if \( \begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \).
Answer: Given, \( \begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \)
On comparing the corresponding elements, we get \( a - b = -1, 2a - b = 0 \)
On subtraction, we get \( a = 1 \).

Question. Find the value of \( b \) if \( \begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \).
Answer: Refer to the previous question. We have \( a - b = -1 \) and \( 2a - b = 0 \), which gives \( a = 1 \). Substituting \( a = 1 \) into \( a - b = -1 \) gives \( 1 - b = -1 \Rightarrow b = 2 \).

Question. If \( \begin{bmatrix} x-y & 2y \\ 2y+z & x+y \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 9 & 5 \end{bmatrix} \), then write the value of \( (x + y + z) \).
Answer: On comparing the corresponding elements, we get \( x - y = 1, 2y = 4, 2y + z = 9, x + y = 5 \)
\( \Rightarrow y = 2, x = 3, z = 5 \)
\( \therefore x + y + z = 3 + 2 + 5 = 10 \).

Question. If \( \begin{bmatrix} 2x+1 & 2y \\ 0 & y^2+1 \end{bmatrix} = \begin{bmatrix} x+3 & 10 \\ 0 & 26 \end{bmatrix} \), write the value of \( (x + y) \).
Answer: Given, \( \begin{bmatrix} 2x+1 & 2y \\ 0 & y^2+1 \end{bmatrix} = \begin{bmatrix} x+3 & 10 \\ 0 & 26 \end{bmatrix} \)
\( \Rightarrow 2x + 1 = x + 3, 2y = 10, y^2 + 1 = 26 \)
\( \Rightarrow \) From the first two equations, we get \( x = 2, y = 5 \) which also satisfies \( y^2 + 1 = 26 \).
\( \therefore x + y = 2 + 5 = 7 \).

Question. If \( \begin{bmatrix} x & x-y \\ 2x+y & 7 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 8 & 7 \end{bmatrix} \), then find the value of \( y \).
Answer: Given, \( \begin{bmatrix} x & x-y \\ 2x+y & 7 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 8 & 7 \end{bmatrix} \)
\( \Rightarrow x = 3, x - y = 1, 2x + y = 8 \)
\( \Rightarrow x = 3, y = 2 \).
Also, \( 2x + y = 8 \) is satisfied by \( x = 3 \) and \( y = 2 \).
\( \therefore y = 2 \).

Question. If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \), then for what value of \( \alpha \) is \( A \) an identity matrix?
Answer: Given, \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \)
If \( A \) is an identity matrix, then
\( \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
By equality of two matrices, corresponding elements are equal.
\( \therefore \cos \alpha = 1 \Rightarrow \alpha = 0 \) and \( \sin \alpha = 0 \Rightarrow \alpha = 0 \)
\( \therefore \alpha = 0 \).

Question. If \( \begin{bmatrix} x+y & 1 \\ 2y & 5 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 4 & 5 \end{bmatrix} \), then find \( x \).
Answer: We are given that \( \begin{bmatrix} x+y & 1 \\ 2y & 5 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 4 & 5 \end{bmatrix} \)
By equality of two matrices, we get \( 2y = 4 \Rightarrow y = 2 \).
Also, \( x + y = 7 \Rightarrow x + 2 = 7 \Rightarrow x = 5 \).

Question. If \( \begin{bmatrix} 3y-x & -2x \\ 3 & 7 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ 3 & 7 \end{bmatrix} \), then find \( y \).
Answer: We have \( \begin{bmatrix} 3y-x & -2x \\ 3 & 7 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ 3 & 7 \end{bmatrix} \)
By equality of two matrices, we have \( 3y - x = 5 \) and \( -2x = -2 \Rightarrow x = 1 \).
Putting the value of \( x \), we get \( 3y - 1 = 5 \Rightarrow 3y = 6 \Rightarrow y = 2 \).

Question. Find the values of \( x \) and \( y \), if \( \begin{bmatrix} 3x+y & -y \\ 2y-x & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix} \).
Answer: We are given that \( \begin{bmatrix} 3x+y & -y \\ 2y-x & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix} \)
By equality of two matrices, we get \( 3x + y = 1, -y = 2, 2y - x = -5 \).
Now, \( -y = 2 \Rightarrow y = -2 \).
Also, \( 3x + y = 1 \Rightarrow 3x + (-2) = 1 \Rightarrow 3x = 3 \Rightarrow x = 1 \).
Also, \( 2y - x = -5 \) is satisfied by \( x = 1 \) and \( y = -2 \).
Therefore, \( x = 1, y = -2 \).

Question. Find the value of \( x \) from the following: \( \begin{bmatrix} 2x-y & 5 \\ 3 & y \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix} \).
Answer: We are given that \( \begin{bmatrix} 2x-y & 5 \\ 3 & y \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix} \)
By equality of two matrices, we get \( y = -2 \) and \( 2x - y = 6 \).
\( \therefore 2x - (-2) = 6 \Rightarrow 2x = 6 - 2 = 4 \Rightarrow x = 2 \).

Question. If \( \begin{bmatrix} 15 & x+y \\ 2 & y \end{bmatrix} = \begin{bmatrix} 15 & 8 \\ x-y & 3 \end{bmatrix} \), then find the value of \( x \).
Answer: We have, \( \begin{bmatrix} 15 & x+y \\ 2 & y \end{bmatrix} = \begin{bmatrix} 15 & 8 \\ x-y & 3 \end{bmatrix} \)
By equality of two matrices, we get \( x + y = 8, x - y = 2 \), and \( y = 3 \).
Now, \( y = 3 \)
\( \therefore x + y = 8 \Rightarrow x + 3 = 8 \Rightarrow x = 5 \).
Also, \( x - y = 2 \) is satisfied by \( x = 5 \) and \( y = 3 \).
\( \therefore x = 5 \).

Question. If \( \begin{bmatrix} 2x & 1 \\ 5 & x+2y \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 5 & 0 \end{bmatrix} \), then find the values of \( x \) and \( y \).
Answer: We have, \( \begin{bmatrix} 2x & 1 \\ 5 & x+2y \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 5 & 0 \end{bmatrix} \)
By equality of two matrices, we get \( 2x = 4 \Rightarrow x = 2 \) and \( x + 2y = 0 \).
Now, \( x + 2y = 0 \Rightarrow 2 + 2y = 0 \Rightarrow 2y = -2 \Rightarrow y = -1 \).
\( \therefore x = 2 \) and \( y = -1 \).

Question. If \( \begin{bmatrix} x+2y & -y \\ 3x & 4 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 6 & 4 \end{bmatrix} \), then find the values of \( x \) and \( y \).
Answer: We have, \( \begin{bmatrix} x+2y & -y \\ 3x & 4 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 6 & 4 \end{bmatrix} \)
By equality of two matrices, we get \( x + 2y = -4, -y = 3 \Rightarrow y = -3 \) and \( 3x = 6 \Rightarrow x = \frac{6}{3} = 2 \).

Question. If \( \begin{bmatrix} x+2y & 3y \\ 4x & 2 \end{bmatrix} = \begin{bmatrix} 0 & -3 \\ 8 & 2 \end{bmatrix} \), then find the values of \( x \) and \( y \).
Answer: By equality of two matrices, we get \( x + 2y = 0, 3y = -3, 4x = 8 \).
From \( 3y = -3 \), we get \( y = -1 \).
From \( 4x = 8 \), we get \( x = 2 \).
These values satisfy \( x + 2y = 0 \).
Therefore, \( x = 2 \) and \( y = -1 \).