Question 13 Marks
Express the matrix $\begin{bmatrix} 0 & \frac{9}{2} & \frac{9}{2} \\ -\frac{9}{2} & 0 & -\frac{3}{2} \\ -\frac{9}{2} & \frac{3}{2} & 0 \end{bmatrix} $ as the sum of a symmetric and skew symmetric matrix.
Answer$\text{Writing A'} =\begin{bmatrix} 1 & -6 & -4 \\ 3 & 8 & 6 \\ 5 & 3 & 5 \end{bmatrix} $
$\therefore \frac{1}{2} (\text{A + A')} = \frac{1}{2} \begin{bmatrix} 2 & -3 & 1 \\ -3 & 16 & 9 \\ 1 & 9 & 10 \end{bmatrix} = \begin{matrix} 1 & -\frac{3}{2} & \frac{1}{2} \\ -\frac{3}{2} & 8 & \frac{9}{2} \\ \frac{1}{2} & \frac{9}{2} & 5 \end{matrix}$
$\frac{1}{2} \text{(A - A)} = \frac{1}{2} \begin{bmatrix} 0 & 9 & 9 \\ -9 & 0 & -3 \\ -9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \frac{9}{2} & \frac{9}{2} \\ -\frac{9}{2} & 0 & -\frac{3}{2} \\ -\frac{9}{2} & \frac{3}{2} & 0 \end{bmatrix} $
$\therefore \text{A} = \begin{bmatrix} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{2} & \frac{1}{2} \\ -\frac{3}{2} & 8 & \frac{9}{2} \\ \frac{1}{2} & \frac{9}{2} & 5 \end{bmatrix} + \begin{bmatrix} 0 & \frac{9}{2} & \frac{9}{2} \\ -\frac{9}{2} & 0 & -\frac{3}{2} \\ -\frac{9}{2} & \frac{3}{2} & 0 \end{bmatrix} $
The first matrix is symmetric and second is skew symmetric.
View full question & answer→Question 23 Marks
Express the following matrix as the sum of a symmetric and a skew symmetric matrix: $ \begin{bmatrix} 1 & 3 & 5 \\ - 6 & 8 & 3 \\ - 4 & 6 & 5 \end{bmatrix} $
Answer${A = \begin{bmatrix} 1 & 3 & 5 \\ -6 & 8 & 3 \\- 4 & 6 & 5 \end{bmatrix}\Rightarrow A^{'} = \begin{bmatrix}1 & - 6 & -4 \\- 3 & 8 & 6\\5 & 3 & 5\end{bmatrix}}$ $\therefore \frac{A + A'}{2} = \begin{bmatrix} 1 & -\frac{3}{2} & \frac{1}{2} \\ -\frac{ 3}{2} & 8 & \frac{9}{2} \\ \frac{1}{2} & \frac{9}{2} & 5 \end{bmatrix}$
$ \frac{A + A'}{2} = \begin{bmatrix} 0 & \frac{9}{2} & \frac{9}{2} \\ -\frac{ 9}{2} & 0 & -\frac{3}{2} \\ -\frac{9}{2} & \frac{3}{2} & 0 \end{bmatrix}$
$ \therefore \text{A} = \begin{bmatrix} 1 & -\frac{3}{2} & \frac{1}{2} \\ -\frac{ 3}{2} & 8 & \frac{9}{2} \\ \frac{1}{2} & \frac{9}{2} & 5\end{bmatrix} + \begin{bmatrix} 0 & \frac{9}{2} & \frac{9}{2} \\ -\frac{ 9}{2} & 0 & -\frac{3}{2} \\ -\frac{9}{2} & \frac{3}{2} & 0 \end{bmatrix}$ First is symmetric and the other skew-symmetric.
View full question & answer→Question 33 Marks
Find the matrix A such that
$\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\text{A}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}$
AnswerLet $\text{A}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}-\text{a}&2\text{y}-\text{b}&2\text{z}-\text{c}\\\text{x}&\text{y}&\text{z}\\-3\text{x}+4\text{a}&-3\text{y}+4\text{b}&-3\text{z}+4\text{c}\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}$
By comparing the elements of second row, we get
x = 1, y = -2 z = -5
By comparing the elements of first row, we get
2x - a = -1
⇒ 2 - a = -1
⇒ a = 3
Also,
2y - b = -8
⇒ -4 - b = -8
⇒ b = 4
And
2z - c = -10
⇒ -10 - c = -10
⇒ c = 10
$\therefore\ \text{A}=\begin{bmatrix}1&-2&-5\\3&4&0\end{bmatrix}$
View full question & answer→Question 43 Marks
If $\text{A}=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix},$ find x satisfying $0<\text{x}<\frac{\pi}{2}$ when A + AT = I
AnswerGiven,
$\text{A}=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\text{A}+\text{A}^\text{T}=\text{I}$
$\Rightarrow\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}+\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\cos\text{x}+\cos\text{x}&\sin\text{x}-\sin\text{x}\\-\sin\text{x}+\sin\text{x}&\cos\text{x}+\cos\text{x}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\cos\text{x}&0\\0&2\cos\text{x}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
$2\cos\text{x}=1$
$\cos\text{x}=\frac{1}{2}$ since $0<\text{x}<\frac{\pi}{2}$
So,
$\text{x}=\frac{\pi}{3}.$
View full question & answer→Question 53 Marks
Given: $3\begin{bmatrix}x & y \\z & w \end{bmatrix} = \begin{bmatrix}x & 6 \\-1 & 2w \end{bmatrix} + \begin{bmatrix}4 & x + y \\z + w & 3 \end{bmatrix},$ find the values of x, y, z and w.
AnswerGiven:$3\begin{bmatrix}x&y\\ z&w\end{bmatrix}=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4&x+y\\ z+w&3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3x&3y\\3z&3w\end{bmatrix}=\begin{bmatrix}x+4&6+x+y\\-1+z+w&2w+3\end{bmatrix}$
Equating corresponding entries, we have
3x = x + 4 ⇒ 2x = 4 ⇒ x = 2
And 3y = 6 + x + y ⇒ 2y = 6 + 2 ⇒ 2y = 8 ⇒ y = 4
And 3z = -1 + z + w ⇒ 2z - w = -1 ...(i)
And 3w = 2w + 3 ⇒ w = 3
Putting w = 3 in eq. (i), 2z - 3 = -1 ⇒ 2z = 2 ⇒ z = 1
$\therefore$ x = 2, y = 4, z = 1, w = 3
View full question & answer→Question 63 Marks
If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?
AnswerWe know that if a matrix is of order m × n, then it has mn elements.
The possible orders of a matrix with 8 elements are given below:
1 × 8, 2 × 4, 4 × 2, 8 × 1
Thus, there are 4 possible orders of the matrix.
The possible orders of a matrix with 5 elements are given below:
1 × 5, 5 × 1 Thus, there are 2 possible orders of the matrix.
View full question & answer→Question 73 Marks
If A = [aij] is a skew-symmetric matrix, then write the value of $\sum_\text{i}\text{a}_\text{ij}.$
AnswerGiven: A = [aij] is a skew-symmetric matrix.
⇒ aij = -aij [From all values of i, j]
⇒ aij = -aij [From all values of i]
⇒ aij + aij = 0
⇒ 2aij = 0
⇒ aij = 0 [From all values of i]
$\sum_\text{i}\text{a}_\text{ij}=0+0+...+0$ [i times]
Thus,
$\sum_\text{i}\text{a}_\text{ij}=0$
View full question & answer→Question 83 Marks
Find $\text{X if Y}=\begin{bmatrix}3&2\\1&4\end{bmatrix}$ and $2\text{X}+\text{Y}=\begin{bmatrix}1&0\\-3&2\end{bmatrix}$
Answer$2\text{X}+\text{Y}=\begin{bmatrix}1&0\\-3&2\end{bmatrix}$
$\Rightarrow2\text{X}+\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}1&0\\-3&2\end{bmatrix}$
$\Rightarrow2\text{X}+\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}1&0\\-3&2\end{bmatrix}$
$\Rightarrow2\text{X}=\frac{1}{2}\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}1-3&0-2\\-3-1&2-4\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}$
$\therefore\ \text{X}=\frac{1}{2}=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}$
View full question & answer→Question 93 Marks
Given, $\text{A}=\begin{bmatrix}2&4&0\\3&9&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&4\\2&8\\1&3\end{bmatrix}$ is $(\text{AB})'=\text{B}'\text{A}'\ ?$
AnswerWe have, $\text{A}=\begin{bmatrix}2&4&0\\3&9&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&4\\2&8\\1&3\end{bmatrix}$
$\therefore\ \text{AB}=\begin{bmatrix}2+8+0&8+32+0\\3+18+6&12+72+18\end{bmatrix}$
$\Rightarrow\ \text{AB}=\begin{bmatrix}10&40\\27&102\end{bmatrix}$
$\Rightarrow\ (\text{AB})'=\begin{bmatrix}10&27\\40&102\end{bmatrix}$
Also, $\text{B}'=\begin{bmatrix}1&2&1\\4&8&3\end{bmatrix}$ and $\text{A}'=\begin{bmatrix}2&3\\4&9\\0&6\end{bmatrix}$
$\therefore\ \text{B}'\text{A}'=\begin{bmatrix}2+8+0&3+18+6\\8+32+0&12+72+18\end{bmatrix}$
$=\begin{bmatrix}10&27\\40&102\end{bmatrix}$
$\therefore\ (\text{AB})'=\text{B}'\text{A}'$
View full question & answer→Question 103 Marks
Let $\text{A}=\begin{bmatrix}-1&0&2\\3&1&4 \end{bmatrix},\text{ B}=\begin{bmatrix}0&-2&5\\1&-3&1 \end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&-5&2\\6&0&-4 \end{bmatrix}.$ Compute 2A - 3B + 4C.
AnswerHere,
$2\text{A}-3\text{B}+4\text{C}=2\begin{bmatrix}-1&0&2\\3&1&4\end{bmatrix}-3\begin{bmatrix}0&-2&5\\1&-3&1\end{bmatrix}+4\begin{bmatrix}1&-5&2\\6&0&-4\end{bmatrix}$
$\Rightarrow2\text{A}-3\text{B}+4\text{C}=\begin{bmatrix}-2&0&4\\6&2&8\end{bmatrix}-\begin{bmatrix}0&-6&15\\3&-9&3\end{bmatrix}+\begin{bmatrix}4&-20&8\\24&0&-16\end{bmatrix}$
$\Rightarrow2\text{A}-3\text{B}+4\text{C}=\begin{bmatrix}-2-0+4&0+6-20&4-15+8\\6-3+24&2+9+0&8-3-16\end{bmatrix}$
$\Rightarrow2\text{A}-3\text{B}+4\text{C}=\begin{bmatrix}2&-14&-3\\27&11&-11\end{bmatrix}$
View full question & answer→Question 113 Marks
Define a symmetric matrix. Prove that for $\text{A}=\begin{bmatrix}2&4 \\5&6 \end{bmatrix},$ A + AT is a symmetric matrix where AT is the transpose of A.
AnswerA squrae matrix A is called a syammetric matrix, if AT = A. Given: $\text{A}=\begin{bmatrix}2&4 \\5&6 \end{bmatrix}$ $\text{A}^{\text{T}}=\begin{bmatrix}2&5 \\4&6 \end{bmatrix}$ Now, $\text{A}+\text{A}^{\text{T}}=\begin{bmatrix}2&4 \\5&6 \end{bmatrix}+\begin{bmatrix}2&5 \\4&6 \end{bmatrix}$ $\Rightarrow\text{A}+\text{A}^{\text{T}}=\begin{bmatrix}4 & 9 \\9&12 \end{bmatrix}\ \dots(\text{i})$ $(\text{A}+\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}4 & 9 \\9&12 \end{bmatrix}^{\text{T}}$ $=\begin{bmatrix}4 & 9 \\9&12 \end{bmatrix}$ $\Rightarrow\text{A}+\text{A}^{\text{T}} $ [From eq. (1)] $\therefore\ (\text{A}+\text{A}^{\text{T}})^{\text{T}}=(\text{A}+\text{A}^{\text{T}})$ Thus, (A + AT)
is a symmetric matrix. View full question & answer→Question 123 Marks
Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Answer(B’AB)’ = [B’(AB]’ = (AB)’ (B’)’ [$\because$ (CD)’ = D’C’]
⇒ (B’AB)’ = B’A’B ……….(i)
Case I: A is a symmetric matrix, then ⇒ A’ = A
$\therefore$ From eq. (i) (B’AB)’ = B’AB
$\therefore$ B’AB is a symmetric matrix.
Case II: A is a skew symmetric matrix. ⇒ A’ = –A
Putting A’ = –A in eq. (i), (B’AB)’ = B’(–A)B = –B’AB
$\therefore$ B’AB is a skew symmetric matrix.
View full question & answer→Question 133 Marks
If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
AnswerA and B are symmetric matrices. ⇒ A’ = A and B’ = B ...(i)
Now, (AB – BA)’ = (AB)’ – (BA)’
⇒ (AB – BA)’ = B’A’ – A’B’ [Reversal law]
⇒ (AB – BA)’ = BA – AB [Using eq. (i)]
⇒ (AB – BA)’ = –(AB – BA)
Therefore, (AB – BA) is a skew symmetric.
View full question & answer→Question 143 Marks
In the matrix $\text{A}=\begin{bmatrix}\text{a}&1&\text{x}\\2&\sqrt{3}&\text{x}^2-\text{y}\\0&5&\frac{-2}{5}\end{bmatrix},$ write:
- The order of the matrix A.
- The number of elements.
- Write elements a23, a31, a12.
AnswerWe have, $\text{A}=\begin{bmatrix}\text{a}&1&\text{x}\\2&\sqrt{3}&\text{x}^2-\text{y}\\0&5&\frac{-2}{5}\end{bmatrix}$ - If a matrix has M rows and N columns, the order of matrix is M × N. Therefore, the order of the matrix A is 3 × 3.
- If a matrix has M rows and N columns, the number of elements is MN. Therefore, the number of elements is 3 × 3 = 9.
- We know that aij, is a representation of element lying in the ith row and jth column
$\therefore$ a23 = x2 - y, a31 = 0, a12 = 1
View full question & answer→Question 153 Marks
If $\text{A}=\begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix},\text{B}=\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix},$ find (AB)T
AnswerHere,
$\text{AB}=\begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix}\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}6-4-2&8+8-1\\-3-0+4&-4+0+2\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0&15\\1&-2\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}0&1\\15&-2\end{bmatrix}$
View full question & answer→Question 163 Marks
Find matrix $\text{A},\text{if}\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}+\text{A}=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}$
AnswerGiven,
$\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}+\text{A}=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}-\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}9-1&-1-2&4+1\\-2-0&1-4&3-9\end{bmatrix}$
Hence,
$\text{A}=\begin{bmatrix}8&-3&5\\-2&-3&-6\end{bmatrix}$
View full question & answer→Question 173 Marks
Find X if $\text {Y} = \begin{bmatrix}3&2\\1&4\end{bmatrix} \text{and}\ \text{2X + Y} = \begin{bmatrix}1&0\\-3&2\end{bmatrix}.$
Answer$\text{2X + Y}=\begin{bmatrix}1&0\\-3&2\end{bmatrix}\Rightarrow\text{2X}=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\text{Y} $
$\Rightarrow\text{2X}=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}$
$\Rightarrow\text{2X}=\begin{bmatrix}1-3&0-2\\-3-1&2-4\end{bmatrix}=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}$
View full question & answer→Question 183 Marks
Find matrices X and Y, if $2\text{X}-\text{Y}=\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}$ and $\text{X}+2\text{Y}=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
AnswerGiven: $(2\text{X}-\text{Y})=\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}\ \dots(1)$
$(\text{X}+2\text{Y})=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}\ \dots(2)$
Multiplying eq. (1) by eq. (2), we get
$2(2\text{X}-\text{Y})=2\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}$
$\Rightarrow4\text{X}-2\text{Y}=\begin{bmatrix}12&-12&0\\-8&4&2\end{bmatrix}\ \dots(3)$
From eq. (3) and eq. (4), we get
$(4\text{X}-2\text{Y})+(\text{X}+2\text{Y})=\begin{bmatrix}12&-12&0\\-8&4&2\end{bmatrix}+\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
$\Rightarrow5\text{X}=\begin{bmatrix}12+3&-12+2&0+5\\-8-2&4+1&2-7\end{bmatrix}$
$\Rightarrow5\text{X}=\begin{bmatrix}15&-10&5\\-10&5&-5\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}15&-10&5\\-10&5&-5\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}$
Putting the value of X in eq. (2), we get
$(\text{X}+2\text{Y})=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}+2\text{Y}=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
$\Rightarrow2\text{Y}=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}-\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}$
$\Rightarrow2\text{Y}=\begin{bmatrix}3-3&2+2&5-1\\-2+2&1-1&-7+1\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}0&2&2\\0&0&-3\end{bmatrix}$
View full question & answer→Question 193 Marks
If $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix},\text{ B}=\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix},\text{C}=\begin{bmatrix}-1&2&3\\2&1&0\end{bmatrix},$ find 2B + 3A and 3C - 4B.
Answer$2\text{B}+3\text{A}=2\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix}+3\begin{bmatrix}2&3\\5&7\end{bmatrix}$
It is not possible to add these matrices because the number of elements in B are not equal to the number of elements in A. So, 2B + 3A does not exist.
$\Rightarrow3\text{C}-4\text{B}=3\begin{bmatrix}-1&2&3\\2&1&0\end{bmatrix}-4\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix}$
$\Rightarrow3\text{C}-4\text{B}=\begin{bmatrix}-3&6&9\\6&3&0\end{bmatrix}-\begin{bmatrix}-4&0&8\\12&16&4\end{bmatrix}$
$\Rightarrow3\text{C}-4\text{B}=\begin{bmatrix}-3+4&6-0&9-8\\6-12&3-16&0-4\end{bmatrix}$
$\Rightarrow3\text{C}-4\text{B}=\begin{bmatrix}1&6&1\\-6&-13&-4\end{bmatrix}$
View full question & answer→Question 203 Marks
For two matrices A and B, $\text{A}=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix},\text{B}=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$ verify that (AB)T = BTAT.
AnswerGiven,
$\text{A}=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix},\text{B}=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$
$(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}^\text{T}\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}2+0+15&-2+20\\4+0+0&-4+2+0\end{bmatrix}^\text{T}$ $=\begin{bmatrix}1&0&5\\-1&2&0\end{bmatrix}\begin{bmatrix}2&4\\1&1\\3&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}17&0\\4&-2\end{bmatrix}^\text{T}=\begin{bmatrix}2+0+15&4+0+0\\-2+2+0&-4+2+0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}17&4\\0&-2\end{bmatrix}=\begin{bmatrix}17&4\\0&-2\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$
View full question & answer→Question 213 Marks
For the following matrices verify the distributivity of matrix, multiplication over matrix addtion i.e., A(B + C) = AB + AC.
$\text{A}=\begin{bmatrix}1&-1\\0&2\end{bmatrix},\text{B}=\begin{bmatrix}-1&0\\2&1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}0&1\\1&-1\end{bmatrix}$
Answer$\text{A}(\text{B}+\text{C})=\text{AB}+\text{AC}$
$\Rightarrow\begin{bmatrix}1&-1\\0&2\end{bmatrix}\begin{pmatrix}\begin{bmatrix}-1&0\\2&1\end{bmatrix}+\begin{bmatrix}0&1\\1&-1\end{bmatrix}\end{pmatrix}$ $=\begin{bmatrix}1&-1\\0&2 \end{bmatrix}\begin{bmatrix}-1&0\\2&1\end{bmatrix}+\begin{bmatrix}1&-1\\0&2\end{bmatrix}\begin{bmatrix}0&1\\1&-1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-1\\0&2 \end{bmatrix}\begin{bmatrix}-1+0&0+1\\2+1&1-1\end{bmatrix}$ $=\begin{bmatrix}-1-2&0-1\\0+4&0+2\end{bmatrix}+\begin{bmatrix}0-1&1+1\\0+2&0-2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-1\\0&2 \end{bmatrix}\begin{bmatrix}-1&1\\0&3\end{bmatrix}=\begin{bmatrix}-3&-1\\4&2\end{bmatrix}+\begin{bmatrix}-1&2\\2&-2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-1-3&1-0\\0+6&0+0\end{bmatrix}=\begin{bmatrix}-3-1&-1+2\\4+2&2-2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-4&1\\6&0\end{bmatrix}=\begin{bmatrix}-4&1\\6&0\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
View full question & answer→Question 223 Marks
Find the matrix A such that
$ \begin{bmatrix}4\\1\\3\end{bmatrix}\text{A}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}$
AnswerLet $\text{A}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}4\\1\\3\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}&\text{z}\end{bmatrix}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}4\text{x}&4\text{y}&4\text{z}\\\text{x}&\text{y}&\text{z}\\3\text{x}&3\text{y}&3\text{z}\end{bmatrix}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
⇒ 4x = -4 ...(1)
4y = 8 ...(2)
4z = 4 ...(3)
⇒ x = -1, y = 2 and z = 1
$\therefore\ \text{A}=\begin{bmatrix}-1&2&1\end{bmatrix}$
View full question & answer→Question 233 Marks
$\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$, then verify that A'A = I
Answer$\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$
$\therefore\ \text{A}'=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$
$\text{A}'\text{A}=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$
$\begin{bmatrix}(\cos\alpha)(\cos\alpha)+(-\sin\alpha)(-\sin\alpha)&(\cos\alpha)(\sin\alpha)+(-\sin\alpha)(\cos\alpha)\$\sin\alpha)(\cos\alpha)+(\cos\alpha)(-\sin\alpha)&(\sin\alpha)(\sin\alpha)+(\cos\alpha)(\cos\alpha)\end{bmatrix}$
$=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\sin^2\alpha+\cos^2\alpha\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
Hence, we have verified that A'A = I
View full question & answer→Question 243 Marks
construct a 2×2 matrix A = [aij] whose elemants aij are given by aij $=\begin{cases}\frac{|-3\text{i}+\text{j}|}{2},&\text{if i }\neq\text{j}\$\text{i}+\text{j})^2,&\text{if i }=\text{j}\end{cases}$
AnswerLet us the matrix be $\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22} \end{bmatrix}$
For the entries a12 and a21 we have i $\neq$ j, so by the
Condition we have $\text{a}_{12}=\frac{|-3\text{i}+\text{j}|}{2},\text{ a}_{21}=\frac{|-3\text{i}+\text{j}|}{2}$
$\Rightarrow\text{a}_{12}=\frac{|-3+2|}{2},\text{ a}_{21}=\frac{|-3.2+1|}{2}$
$\Rightarrow\text{a}_{12}=\frac{|-1|}{2},\text{ a}_{21}=\frac{|-5|}{2}$
$\Rightarrow\text{a}_{12}=\frac{1}{2},\text{ a}_{21}=\frac{5}{2}$
For the entries a11 and a22 we have i = j, so by the given condition we have
$\Rightarrow\text{a}_{11}=(\text{i + j})^2, \text{ a}_{22}=(\text{i + j})^2$
$\Rightarrow\text{a}_{11}=(1+1)^2,\text{ a}_{22}=(2+2)^2$
$\Rightarrow\text{a}_{11}=4,\text{ a}_{22}=16$
So, $\text{A}=\begin{bmatrix}4&\frac{1}{2}\\\frac{5}{2}&16 \end{bmatrix}$
View full question & answer→Question 253 Marks
If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?
AnswerWe know that, if a matrix is of order m × n, it has mn elements, where m and n are natural numbers.
We have, m × n = 28
⇒ (m, n) = {(1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1)}
So, the possible orders are 1 × 28, 2 × 14, 4 × 7, 7 × 4, 14 × 2, 28 × 1
Also, if it has 13 elements, then m × n = 13
(m, n) = {(1, 13), (13, 1)}
Hence, the possible orders are 1 × 13, 13 × 1.
View full question & answer→Question 263 Marks
Find values of a and b if A = B, where:
$\text{A}=\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2\text{a}+2&\text{b}^2+2\\8&\text{b}^2-5\text{b}\end{bmatrix}$
AnswerConsider, A = B $\Rightarrow\ \begin{bmatrix}\text{a}+4&3\text{b}\\8&-6\end{bmatrix}=\begin{bmatrix}2\text{a}+2&\text{b}^2+2\\8&\text{b}^2-5\text{b}\end{bmatrix}$ $\Bigg\{\because\ \text{A}=\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6\end{bmatrix}\text{ and B}=\begin{bmatrix}2\text{a}+2&\text{b}^2+2\\8&\text{b}^2-5\text{b}\end{bmatrix}\Bigg\}$
By equality of matrices, we have a + 4 = 2a + 2 ⇒ a = 2 Also, -6 = b2 - 5b ⇒ -6 = 3b - 2 - 5b $[\because$ b2 = 3b - 2$]$ ⇒ 2b = 4 ⇒ b = 2 $\therefore$ a = 2 and b = 2 View full question & answer→Question 273 Marks
Given the matrices
$\text{A}=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix},\text{B}=\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}$ and $\text{C}=\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$ Verify that (A + B) + C = A + (B + C).
AnswerHere,
$\text{LHS}=(\text{A}+\text{B})+\text{C}$
$=\begin{pmatrix}\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}\end{pmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$
$=\begin{pmatrix}\begin{bmatrix}2+9&1+7&1-1\\3+3&-1+5&0+4\\0+2&2+1&4+6\end{bmatrix}\end{pmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$
$=\begin{bmatrix}11&8&0\\6&4&4\\2&3&10\end{bmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$
$=\begin{bmatrix}11+2&8-4&0+3\\6+1&4-1&4+0\\2+9&3+4&10+5\end{bmatrix}$
$=\begin{bmatrix}13&4&3\\7&3&4\\11&7&15\end{bmatrix}$
$\text{RHS}=\text{A}+(\text{B}+\text{C})$
$=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{pmatrix}\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{pmatrix}\begin{bmatrix}9+2&7-4&-1+3\\3+1&5-1&4+0\\2+9&1+4&6+5\end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4 \end{bmatrix}+\begin{bmatrix}11&3&2\\4&4&4\\11&5&11\\\end{bmatrix}$
$=\begin{bmatrix}2+11&1+3&1+2\\3+4&-1+4&0+4\\0+11&2+5&4+11\\\end{bmatrix}$
$=\begin{bmatrix}13&4&3\\7&3&4\\11&7&15\\\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.
View full question & answer→Question 283 Marks
If $\text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix},$ verify that ATA = I2.
AnswerGiven: $\text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$
Now,
$\text{A}^\text{T}\text{A}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}(\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha)&(\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\$\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha)&(\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha)\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
View full question & answer→Question 293 Marks
Show that $\text{AB}\neq\text{BA}$ in the following cases:
$\text{A}=\begin{bmatrix}5&-1\\6&7\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&1\\3&4\end{bmatrix}$
Answer$\text{AB}=\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}10-3&5-4\\12+21&6+28\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}7&1\\33&34\end{bmatrix}\ \dots(1)$
Also,
$\text{BA}=\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}10+6&-2+7\\15+24&-3+28\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}16&5\\39&25\end{bmatrix}\ \dots(2)$
$\therefore\ \text{AB}\neq\text{BA}$ From eqs. (1) and (2)
View full question & answer→Question 303 Marks
If A, B are square matrices of same order and B is a skew-symmetric matrix, show that A′BA is skew symmetric.
AnswerSince, A and B are square matrices of same order and B is a skew symmetric matrix i.e. B’ = -B. Now, we have to prove that A’BA is a skew-symmetric matrix. $\therefore$ A'BA' = A'BA' = BA'A' $[\because$ AB' = B'A'$]$ = A'BA' = A'(-B)A = -A'BA
Hence, A’BA is a skew-symmetric matrix.
View full question & answer→Question 313 Marks
Let $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix},$ verify that
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$
AnswerGiven: $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
Given,
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$
$\begin{pmatrix} \begin{bmatrix} 2&-3\\-7&5\end{bmatrix}+\begin{bmatrix} 1&0\\2&-4\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}^\text{T}+\begin{bmatrix}1&0\\2&-4\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}2+1&-3+0\\-7+2&5-4\end{bmatrix}^\text{T}$ $=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}+\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-3\\-5&1\end{bmatrix}^\text{T}=\begin{bmatrix}2+1&-7+2\\-3+0&5-4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-5\\-3&1\end{bmatrix}=\begin{bmatrix}3&-5\\-3&1\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$
View full question & answer→Question 323 Marks
If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then find $\lambda$ so that $\text{A}^2 = 5\text{A} + \lambda\text{I}.$
AnswerGiven, $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
And
$\text{A}^2=5\text{A}+\lambda\text{I}$
$\Rightarrow\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}9-1&3+2\\-3+2&-1+4\end{bmatrix}=\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}$
$\Rightarrow\begin{bmatrix}8&5\\-5&3\end{bmatrix}=\begin{bmatrix}15+\lambda&5\\-5&10+\lambda\end{bmatrix}$
Since, Corresponding entries of equal matrices are equal, So
$8=15+\lambda$
$\lambda=8-15$
$\lambda=-7$
View full question & answer→Question 333 Marks
Let $\text{A} = \begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix},$show that (aI + bA)n = anI + nan-1 bA where I is the identity matrix of order 2 and $ \text{n} \in \text{N}.$
AnswerUsing Mathematical Induction, we see the result is true for n = 1, for
(aI + bA)n = anI + nan - 1 bA
Given: p(k) is true, i.e. (aI + bA)k = akI + kak - 1bA
To prove: (aI + bA)k + 1 = ak + 1I + (k + 1)ak bA
Proof: L.H.S. = (aI + bA)k + 1 = (aI + bA)k (aI + bA)
= (akI + kak - 1 bA)(aI + bA)
= ak + 1 I × I + kakbAI + akbAI + kak - 1b2A.A
= ak + 1 I × I + kakbAI + akbAI + kak - 1b2.0
= ak +1I + (k +1) akbA = R.H.S.
Thus, p(k + 1) is true, therefore, p(n) is true.
View full question & answer→Question 343 Marks
Find the matrix A such that
$\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=\text{A}$
AnswerLet $\text{A}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=\text{A}$
$ \Rightarrow\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}-2-1+0&0+1+3&-2+0+3\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}-3&4&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}-3+0-1\end{bmatrix}=[\text{x}]$
$\Rightarrow\begin{bmatrix}-4\end{bmatrix}=[\text{x}]$
The corresponding elements of two equal matrices are equal.
$\therefore\ \text{x}=-4$
$\therefore\ \text{A}=[-4]$
View full question & answer→Question 353 Marks
Find the matrix A such that
$\text{A}=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$
AnswerLet $\text{A}=\begin{bmatrix}\text{x}&\text{a}\\\text{y}&\text{b}\\\text{z}&\text{c}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}&\text{a}\\\text{y}&\text{b}\\\text{z}&\text{c}\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+4\text{a}&2\text{x}+5\text{a}&3\text{x}+6\text{a}\\\text{y}+4\text{b}&2\text{y}+5\text{b}&3\text{y}+6\text{b}\\\text{z}+4\text{c}&2\text{z}+5\text{c}&3\text{z}+6\text{c}\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$
By comparing the corresponding elements, we get
x + 4a = -7 and 2x + 5a = -8
⇒ a = -2 and x = 1
Also,
y + 4b = 2 and 2y + 5b = 4
⇒ b = 0 and y = 2
And
z + 4c = 11 and 2z + 5b = 10
⇒ c = 4 and z = -5
$\therefore\ \text{A}=\begin{bmatrix}1&-2\\2&0\\-5&4\end{bmatrix}$
View full question & answer→Question 363 Marks
Using elementary transformation, find the inverse of each of the matrices, $\begin{bmatrix}3 & 10 \\2 & 7 \end{bmatrix}$
AnswerLet $\text{A}=\begin{bmatrix}3&10\\2&7\end{bmatrix}$
since $\text{A = IA}\ \Rightarrow \begin{bmatrix}3&10\\2&7\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\text{A}$
$\Rightarrow\ \begin{bmatrix}1&3\\2&7\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \left[ \text R_1\rightarrow \text R_1- \text R_2\right]$
$\Rightarrow\ \begin{bmatrix}1&3\\0&1\end{bmatrix}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}\text{A} \ \ \ \ \ \ \ \left[\text R_2\rightarrow \text R_2-2 \text R_1\right]$
$\Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}\text{A} \ \ \ \ \ \ \left[\text R_1\rightarrow \text R_1-3\text R_2\right]$
$\therefore\ \ \ \ \ \ \text{A}^{-1}=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}$
View full question & answer→Question 373 Marks
Find A2 - 5A + 6I if A = $\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}.$
Answer$\text{A}^2-5\text{A}+6\text{I}=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}-5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}+6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}4 + 0 + 1&0 + 0 - 1&2 + 0 + 0\\4 + 2 + 3&0 + 1 - 3&2 + 3 + 0\\2 - 2 + 0&0 - 1 - 0&1 - 3 + 0\end{bmatrix} - \begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix} + \begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$
$=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix} + \begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix} $
$= \begin{bmatrix}5 - 10 + 6& -1-0 + 0&2 - 5 + 0\\9 - 10 + 0&-2 - 5 + 6&5-15 + 0\\0-5 + 0&-1 + 5+0&-2-0 + 6\end{bmatrix}$
$=\begin{bmatrix}1&-1&-3\\-1&-1&-10\\-5&4&4\end{bmatrix}$
View full question & answer→Question 383 Marks
If $\text A = \begin{bmatrix}3&-2\\4&-2\end{bmatrix} \text {and I} = \begin{bmatrix}1&0\\0&1\end{bmatrix},$ find k so that A2 = kA - 2I.
AnswerGiven: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix} \text{and}\ \text{I}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\text{A}^2=k\text{A}-2\text{I}\Rightarrow\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=k\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}3k&-2k\\4k&-2k\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3k-2&-2k-0\\4k-0&-2k-2\end{bmatrix}$
Equating corresponding entries, we have
3k - 2 = 1 ⇒ 3k = 3
k = 1
And 4k = 4 ⇒ k = 1 and -4 = -2k - 2
⇒ 2k = 2 ⇒ k = 1
$\therefore$ k = 1
View full question & answer→Question 393 Marks
Give an example of matrices A, B and C such that AB = AC, where A is nonzero matrix, but $\text{B}\neq\text{C}.$
AnswerLet $\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}2&3\\4&0\end{bmatrix}$ and $\text{C}=\begin{bmatrix}2&3\\4&4\end{bmatrix}\ [\because\ \text{B}\neq\text{C}]$
$\therefore\ \text{AB}=\begin{bmatrix}1&0\\0&0\end{bmatrix}.\begin{bmatrix}2&3\\4&0\end{bmatrix}=\begin{bmatrix}2&3\\0&0\end{bmatrix}\ ....(\text{i})$
And $\text{AC}=\begin{bmatrix}1&0\\0&0\end{bmatrix}.\begin{bmatrix}2&3\\4&0\end{bmatrix}=\begin{bmatrix}2&3\\0&0\end{bmatrix}\ ....(\text{ii})$
Thus, we see that AB = AC [using Eq. (i) and (ii)]
where, A is non-zero matrix but $\text{B}\neq\text{C}.$
View full question & answer→Question 403 Marks
If A and B are square matrices of the same order such that AB = BA, then prove by induction that AB’’ = B’’A. Further prove that (AB)’’ = A’’B’’ for all n ⇒ N.
AnswerGiven: AB = BA ...(i)
Let p(n): ABn = BnA ...(ii)
For n = 1, p(n): becomes AB = BA
$\therefore$ p(1) is true for n = 1.
For n = k, p(k): ABk = BkA
Multiplying both sides by B, ABkB = BkAB ⇒ ABk+1 = BkAB
⇒ ABk+1 = Bk+1A [From eq. (i)]
$\therefore$ p(k +1) is also true.
Therefore, p(n) is true for all $\text{n}\in\text{N}$ by P.M.I.
View full question & answer→Question 413 Marks
If $\text{A}=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix},$ show that A2 = I3.
AnswerGiven, $\text{A}=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}$ $\text{A}^2=\text{A.A}$ $=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}$ $=\begin{bmatrix}16-3-12&-4+0+4&-16+4+12\\12+0-12&-3+0+4&-12+0+12\\12-3-9&-3+0+3&-12+4+9\end{bmatrix}$ $=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ $=\text{I}_3$ Hence, $\text{A}^2=\text{I}_3$
View full question & answer→Question 423 Marks
Find the values of a, b, c and d if: $3\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}=\begin{bmatrix}\text{a}&6\\-1&2\text{d}\end{bmatrix}+\begin{bmatrix}4&\text{a}+\text{b}\\\text{c}+\text{d}&3\end{bmatrix}.$
AnswerConsider, $3\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}=\begin{bmatrix}\text{a}&6\\-1&2\text{d}\end{bmatrix}+\begin{bmatrix}4&\text{a}+\text{b}\\\text{c}+\text{d}&3\end{bmatrix}$ $\Rightarrow\ \begin{bmatrix}3\text{a}&3\text{b}\\3\text{c}&3\text{d}\end{bmatrix}=\begin{bmatrix}\text{a}+4&6+\text{a}+\text{b}\\\text{c}+\text{d}-1&3+2\text{d}\end{bmatrix}$
By equality of matrices, we get
3a = a + 4, 3b = 6 + a + b and 3d = 3 + 2d
⇒ a = 2, b = 4 and d = 3.
View full question & answer→Question 433 Marks
If $\text{A} = \begin{bmatrix}3 & 1 \\-1 & 2\end{bmatrix},$ show that A2 - 5A + 7I = 0.
AnswerGiven: $\text{A} = \begin{bmatrix}3 & 1 \\-1 & 2\end{bmatrix},$ $\therefore \ \text{A}^{2}-5\text{A}+7\text{I}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$= \begin{bmatrix}9-1&3 + 2\\-3-2&-1 + 4\end{bmatrix}-\begin{bmatrix}15 & 5 \\-5 & 10\end{bmatrix} + \begin{bmatrix}7 & 0 \\0 & 7 \end{bmatrix} $$= \begin{bmatrix}8 & 5 \\-5 & 3 \end{bmatrix} - \begin{bmatrix}15 & 5\\-5& 10 \end{bmatrix}+ \begin{bmatrix}7 & 0 \\0 & 7 \end{bmatrix}$
$=\begin{bmatrix}8-15&5-5\\-5+5&3-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$$=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}-7+7&0+7\\0+7&-7+7\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0 = \text{R.H.S.}$
View full question & answer→Question 443 Marks
If $\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix},$ then verify that ATA = I2.
Answer$\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$ $\therefore\ \text{A}'=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$ $\text{A}'\text{A}=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$ $\begin{bmatrix}(\cos\alpha)(\cos\alpha)+(-\sin\alpha)(-\sin\alpha)&(\cos\alpha)(\sin\alpha)+(-\sin\alpha)(\cos\alpha)\$\sin\alpha)(\cos\alpha)+(\cos\alpha)(-\sin\alpha)&(\sin\alpha)(\sin\alpha)+(\cos\alpha)(\cos\alpha)\end{bmatrix}$ $\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\sin^2\alpha+\cos^2 \alpha\end{bmatrix}$ $=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$ Hence, we have verified that A'A = I
View full question & answer→Question 453 Marks
Solve for x and y:
$\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=0.$
AnswerWe have, $\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=0$ $\Rightarrow\ \begin{bmatrix}2\text{x}\\\text{x}\end{bmatrix}+\begin{bmatrix}3\text{y}\\5\text{y}\end{bmatrix}=\begin{bmatrix}-8\\-11\end{bmatrix}=0$
$\therefore\ \begin{bmatrix}2\text{x}+3\text{y}-8\\\text{x}+5\text{y}-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow\ 2\text{x}+3\text{y}-8=0\ ...(\text{i})$
and $\text{x}+5\text{y}-11=0\ ....(\text{ii})$
Solving equation (i) and (ii), we get
$\text{x}=1$ and $\text{y}=2$
View full question & answer→Question 463 Marks
If $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{ B}=\text{diag}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-6&3&4\end{pmatrix},$ find.
$2\text{A}+3\text{B}-5\text{C}$
Answer Given, $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{B}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-\text{b}&3&4\end{pmatrix}$ $2\text{A}+3\text{B}-5\text{C}$
$\Rightarrow2\text{A}+3\text{B}-5\text{C}=2\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}-5\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}$
$\Rightarrow2\text{A}+3\text{B}-5\text{C}=\begin{bmatrix}4&0&0\\0&-10&0\\0&0&18\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&-12\end{bmatrix}-\begin{bmatrix}-30&0&0\\0&15&0\\0&0&20\end{bmatrix}$
$\Rightarrow2\text{A}+3\text{B}-5\text{C}=\begin{bmatrix}4+3+30&0+0-0&0+0-0\\0+0-0&-10+3-15&0+0-0\\0+0-0&0+0-0&18-12-20\end{bmatrix}$
$\Rightarrow2\text{A}+3\text{B}-5\text{C}=\begin{bmatrix}37&0&0\\0&-22&0\\0&0&-14\end{bmatrix}=\text{diag}\begin{pmatrix}37&-22&-14\end{pmatrix}$
View full question & answer→Question 473 Marks
Find X and Y, if:
- $\text {X + Y} = \begin{bmatrix}7&0\\2&5\end{bmatrix} \text {and}\ \text{X} - \text{Y} = \begin{bmatrix}3&0\\0&3\end{bmatrix}$
- $2\text {X }+ 3\text {Y} = \begin{bmatrix}2&3\\4&0\end{bmatrix} \text {and}\ 3\text{X }+\text{ 2Y} = \begin{bmatrix}-2&-2\\-1&5\end{bmatrix}$
Answer - $\text{Given: X }+\text{ Y}=\begin{bmatrix}7&0\\2&5\end{bmatrix}...\text{(i)}\ \text{and X} -\text{ Y}=\begin{bmatrix}3&0\\0&3\end{bmatrix}...\text{(ii)}$
Adding eq. (i) and (ii), we get
$2\text {X}=\begin{bmatrix}7&0\\2&5\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}7+3&0+0\\2+0&5+3\end{bmatrix}=\begin{bmatrix}10&0\\2&8\end{bmatrix} $
$\Rightarrow \text{X}=\frac{1}{2}\cdot\begin{bmatrix}10&0\\2&8\end{bmatrix}=\begin{bmatrix}5&0\\1&4\end{bmatrix}$
Subtracting eq. (i) and (ii), we get
$\text{2Y}=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}7-3&0-0\\2-0&5-3\end{bmatrix}=\begin{bmatrix}4&0\\2&2\end{bmatrix}$
$\text{Y}=\frac{1}{2}\cdot\begin{bmatrix}4&0\\2&2\end{bmatrix}=\begin{bmatrix}2&0\\1&1\end{bmatrix}$
- $\text{Given} :2\text{ X} +3\text{Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix}...\text{(iii)}$
$\text{and}\ 3\text{ X} + 2\text{Y}=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}...\text{(iv)}$
Multiplying eq. (iii) by 2, $4\text{X} +6\text{Y}=2\begin{bmatrix}2&3\\4&0\end{bmatrix}=\begin{bmatrix}4&6\\8&0\end{bmatrix}...\text{(v)}$
Multipiying eq. (iv) by 3, $\text{9X + 6Y}=3\begin{bmatrix}2&-2\\-1&5\end{bmatrix}=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}...\text{(vi)}$
Subtract (vi) - Eq. (v) $=\text{5X}=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}-\begin{bmatrix}4&6\\8&0\end{bmatrix}$
$=\begin{bmatrix}6-4&-6-6\\-3-8&15-0\end{bmatrix}=\begin{bmatrix}2&-12\\-11&15\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}2&-12\\-11&15\end{bmatrix}=\begin{bmatrix}\frac{2}{5}&-\frac{12}{5}\\-\frac{11}{5}&{3}\end{bmatrix}$
Now, From eq. (iii), $\text{3Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\text{X}=\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\begin{bmatrix}\frac{2}{5}&-\frac{12}{5}\\-\frac{11}{5}&3\end{bmatrix}$
$\Rightarrow\text{3Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix}-\begin{bmatrix}\frac{4}{5}&-\frac{24}{5}\\-\frac{22}{5}&6\end{bmatrix}$
$=\begin{bmatrix}2-\frac{4}{5}&3+\frac{24}{5}\\4+\frac{22}{5}&0-6\end{bmatrix}=\begin{bmatrix}\frac{6}{5}&\frac{39}{5}\\\frac{42}{5}&-6\end{bmatrix}$
$\Rightarrow\text{Y}=\frac{1}{3}\begin{bmatrix}\frac{6}{5}&\frac{39}{5}\\\frac{42}{5}&-6\end{bmatrix}=\begin{bmatrix}\frac{2}{5}&\frac{13}{5}\\\frac{14}{5}&-2\end{bmatrix}$
View full question & answer→Question 483 Marks
If the matric $\text{A}=\begin{bmatrix}5 & 2&\text{x} \\\text{y} & \text{z}&-3\\4&\text{t}&-7\end{bmatrix}$ is a symmetric matrix, find x, y, z and t.
AnswerGiven: $\text{A}=\begin{bmatrix}5 & 2&\text{x} \\\text{y} & \text{z}&-3\\4&\text{t}&-7\end{bmatrix}$ $\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}5&\text{y} & 4 \\2&\text{z}&\text{t}\\\text{x}&-3&-7 \end{bmatrix}$
Since A is a symmetricmatric matrix, AT = A.
$\Rightarrow\begin{bmatrix}5&\text{y} & 4 \\2&\text{z}&\text{t}\\\text{x}&-3&-7 \end{bmatrix}=\begin{bmatrix}5&2 &\text{x}\\\text{y} & \text{z}&-3\\4&\text{t}&-7 \end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ x = 4
y = 2
z = z
t = -3
Hence, x = 2, y = 2, t = -3, and z can have any value.
View full question & answer→Question 493 Marks
$\text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$, then verify that A'A = I
Answer$ \text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\therefore\ \text{A}'=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$
$\text{A}'\text{A}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$=\begin{bmatrix}(\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha)&(\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\$\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha)&(\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha)\end{bmatrix}$
$=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
Hence, we have verified that A'A = I
View full question & answer→Question 503 Marks
If $\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=\text{A},$ then find the value of A.
AnswerWe have, $\begin{bmatrix}2&1&3\end{bmatrix}_{1\times3}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}_{3\times3}\begin{bmatrix}1\\0\\-1\end{bmatrix}_{3\times1}=\text{A}$ $\therefore\ \begin{bmatrix}2&1&3\end{bmatrix}_{1\times3}\begin{bmatrix}-1+0+1\\-1+0+0\\0+0-1\end{bmatrix}_{3\times1}=\text{A}$ $\Rightarrow\ \begin{bmatrix}2&1&3\end{bmatrix}_{1\times3}\begin{bmatrix}0\\-1\\-1\end{bmatrix}_{3\times1}=\text{A}$
$\Rightarrow\ [0-1-3]=\text{A}$
$\Rightarrow\ \text{A}=[-4]$
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