Question 13 Marks
If $\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix},$ then verify that $A^TA = I_2$.
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 23 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 33 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.
$\text{B}+\text{C}-2\text{A}$
AnswerGiven, $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{B}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-\text{b}&3&4\end{pmatrix}$ $\text{B}+\text{C}-2\text{A}$$\Rightarrow\text{B}+\text{C}-2\text{A}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}+\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}-2\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix}$
$\Rightarrow\text{B}+\text{C}-2\text{A}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}+\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}-\begin{bmatrix}4&0&0\\0&-10&0\\0&0&18\end{bmatrix}$
$\Rightarrow\text{B}+\text{C}-2\text{A}=\begin{bmatrix}1-6-4&0+0-0&0+0-0\\0+0-0&1+3+10&0+0-0\\0+0-0&0+0-0&-4+4-18\end{bmatrix}$
$\Rightarrow\text{B}+\text{C}-2\text{A}=\begin{bmatrix}-9&0&0\\0&14&0\\0&0&-18\end{bmatrix}=\text{diag}\begin{pmatrix}-9&14&-18\end{pmatrix}$
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 + A^T = 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
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}$
AnswerGiven, $\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 63 Marks
Let $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix},$ Find $A^T, B^T$ and verify that.$(2\text{A})^\text{T}=2\text{A}^\text{T}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ and $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$
$(2\text{A})^\text{T}=2\text{A}^\text{T}$
$\Rightarrow\begin{pmatrix}2\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\end{pmatrix}^\text{T}=2\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}2&-2&0\\4&2&6\\2&4&2\end{bmatrix}^\text{T}=2\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}=\begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(2\text{A})^\text{T}=2\times\text{A}^\text{T}$
View full question & answer→Question 73 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 83 Marks
If $A = [a_{ij}]$ is a skew-symmetric matrix, then write the value of $\sum_\text{i}\text{a}_\text{ij}.$
AnswerGiven: $A = [a_{ij}]$ is a skew-symmetric matrix.
$\Rightarrow a_{ij} = -a_{ij}$ [From all values of i, j]
$\Rightarrow a_{ij} = -a_{ij}$ [From all values of i]
$\Rightarrow a_{ij} + a_{ij} = 0$
$\Rightarrow 2a_{ij} = 0$
$\Rightarrow a_{ij} = 0$ [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 93 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 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 $\text{A}=\begin{bmatrix}2&4 \\5&6 \end{bmatrix}, A + A^T$ is a symmetric matrix where $A^T$ is the transpose of A.
AnswerA squrae matrix A is called a syammetric matrix, if $A^T = A$. 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, $(A + A^T)$ is a symmetric matrix.
View full question & answer→Question 123 Marks
If $\text{A}=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix},$ show that $A^2 = I_3$.
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 133 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, $A^T = A$.
$\Rightarrow\begin{bmatrix}5&\text{y} & 4 \\2&\text{z}&\text{t}\\\text{x}&-3&-7 \end{bmatrix}=\begin{bmatrix}5&2 &\text{x}\\\text{y} & \text{z}&-3\\4&\text{t}&-7 \end{bmatrix}$
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 143 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 153 Marks
If $\text{x}\begin{bmatrix}2\\3 \end{bmatrix}+\text{y}\begin{bmatrix}-1\\1 \end{bmatrix}=\begin{bmatrix}10\\5 \end{bmatrix},$ find the value of x.
Answer$\text{x}\begin{bmatrix}2\\3 \end{bmatrix}+\text{y}\begin{bmatrix}-1\\1 \end{bmatrix}=\begin{bmatrix}10\\5 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}-\text{y}\\3\text{x + y} \end{bmatrix}=\begin{bmatrix}10\\5 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
⇒ 2x - y = 10 and 3x + y = 5
⇒ y = 2x - 10 and 3x + (2x - 10) = 5
⇒ y = 2x - 10 and 5x = 15
⇒ y = 2x - 10 and x = 3
⇒ y = 2(3) - 10 and x = 3
⇒ y = -4 and x = 3
$\therefore$ x = 3 and y = -4
Hence, the value of x is 3.
View full question & answer→Question 163 Marks
If $\text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix},$ verify that $A^TA = I_2$.
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 173 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 183 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
$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$
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,
$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$
$\Rightarrow\begin{pmatrix} \begin{bmatrix} 2&-3\\-7&5\end{bmatrix}-\begin{bmatrix} 1&0\\2&-4\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}1&2\\0&-4\end{bmatrix}\begin{bmatrix}2&-7\\-3&5\end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}2-6&0+12\\-7+10&0-20\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}2-6&-7+10\\0+12&0-20\end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}-4&12\\3&-20\end{bmatrix}\end{pmatrix}^\text{T}=\begin{bmatrix}-4&3\\12&-20\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-4&3\\12&-20\end{bmatrix}=\begin{bmatrix}-4&3\\12&-20\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
View full question & answer→Question 193 Marks
Let $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix},$ Find $A^T, B^T$ and verify that.$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ and $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$
$\text{A}+\text{B}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}+\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$
$\Rightarrow\text{A}+\text{B}=\begin{bmatrix}1+1&-1+2&0+3\\2+2&1+1&3+3\\1+0&2+1&1+1\end{bmatrix}$
$\Rightarrow\text{A}+\text{B}=\begin{bmatrix}2&1&3\\4&2&6\\1&3&2\end{bmatrix}$
$\Rightarrow(\text{A}+\text{B})^\text{T}=\begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}\ \dots(1)$
Now,
$\text{A}^\text{T}+\text{B}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}+\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}+\text{B}^\text{T}=\begin{bmatrix}1+1&2+2&1+0\\-1+2&1+1&2+1\\0+3&3+3&1+1\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}+\text{B}^\text{T}=\begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}\ \dots(2)$
$\Rightarrow(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$ [From eqs. (1) and (2)]
View full question & answer→Question 203 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
$(2\text{A})^\text{T}=2\text{A}^\text{T}$
AnswerGiven: $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
$(2\text{A})^\text{T}=2\text{A}^\text{T}$
$\Rightarrow\begin{pmatrix} 2\begin{bmatrix}2&-3\\-7&5\end{bmatrix}\end{pmatrix}^\text{T}=2\begin{bmatrix} 2&-7\\-3&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix} 4&-6\\-14&10\end{bmatrix}^\text{T}=\begin{bmatrix}4&-14\\-6&10 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}4&-14\\-6&10\end{bmatrix}=\begin{bmatrix}4&-14\\-6&10\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
View full question & answer→Question 213 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,
$(\text{A}-\text{B})^\text{T}=\text{A}^\text{T}-\text{B}^\text{T}$
$\Rightarrow\begin{pmatrix} \begin{bmatrix} 2&-3\\-7&5\end{bmatrix}-\begin{bmatrix} 1&0\\2&-4\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}-\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}2-1&-3-0\\-7-2&5+4\end{bmatrix}^\text{T}\end{pmatrix}$ $=\begin{bmatrix}2-1&-7-2\\-3-0&5+4\end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}1&-3\\-9&9\end{bmatrix}\end{pmatrix}^\text{T}=\begin{bmatrix}1&-9\\-3&9\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-9\\-3&9\end{bmatrix}=\begin{bmatrix}1&-9\\-3&9\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
View full question & answer→Question 223 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 233 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 243 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 253 Marks
If $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix},$ show that $A^2 = A$.
AnswerHere,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4+3-5&-6-12+15&-10-15+20\\-2-4+5&3+16-15&5+20-20\\2+3-4&-3-12+12&-5-15+16\end{bmatrix}$
$\Rightarrow\text{A}^2-\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$\therefore\ \text{A}^2=\text{A}$
View full question & answer→Question 263 Marks
If $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix},\text{ B}=\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix},\text{C}=\begin{bmatrix}-1&2&3\\2&1&0\end{bmatrix},$ find2B + 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 273 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 $(AB)^T = B^TA^T$.
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 283 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 293 Marks
construct a $2×2$ matrix $A = [a_{ij}]$ whose elemants $a_{ij}$ are given by $a_{ij}$ $=\begin{cases}\frac{|-3\text{i}+\text{j}|}{2},&\text{if i }\neq\text{j}(\text{i}+\text{j})^2,&\text{if i }=\text{j}\end{cases}$
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 $a_{12}$ and $a_{21}$ 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 $a_{11}$ and $a_{22}$ 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 303 Marks
Show that $\text{AB}\neq\text{BA}$ in the following cases:
$\text{A}=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}$
AnswerGiven, $\text{A}=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix},\text{B}=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}$
$\text{BA}=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}$
$=\begin{bmatrix}-1+0+0&-2+1+0&-3+0+0\\0+0+1&0-1+1&0+0+0\\2+0+4&4+3+4&6+0+0\end{bmatrix}$
$\text{AB}=\begin{bmatrix}-1&-1&3\\0&1&3\\1&1&0\end{bmatrix}\ \dots(1)$
$\text{BA}=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}1&1&0\\0&-1&1\\2&3&4\end{bmatrix}$
$=\begin{bmatrix}-1+0+6&1-2+9&0+2+12\\0+0+0&0-1+0&0+1+0\\-1+0+0&1-1+0&0+1+0\end{bmatrix}$
$\text{BA}=\begin{bmatrix}5&8&14\\0&-1&1\\-1&0&1\end{bmatrix}\ \dots(2)$
$\therefore\ \text{AB}\neq\text{BA}$ From (1) and (2)
View full question & answer→Question 313 Marks
Show that $\text{AB}\neq\text{BA}$ in the following cases:
$\text{A}=\begin{bmatrix}1&3&0\\1&1&0\\4&1&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&1&0\\1&0&0\\0&5&1\end{bmatrix}$
Answer$\text{AB}=\begin{bmatrix}1&3&0\\1&1&0\\4&1&0\end{bmatrix}\begin{bmatrix}0&1&0\\1&0&0\\0&5&1\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0+3+0&1+0+0&0+0+0\\0+1+0&1+0+0&0+0+0\\0+1+0&4+0+0&0+0+0\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}3&1&0\\1&1&0\\1&4&0\end{bmatrix}\ \dots(1)$
Also,
$\text{BA}=\begin{bmatrix}0&1&0\\1&0&0\\0&5&1\end{bmatrix}\begin{bmatrix}1&3&0\\1&1&0\\4&1&0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0+1+0&0+1+0&0+0+0\\1+0+0&3+0+0&0+0+0\\0+5+4&0+5+1&0+0+0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}1&1&0\\1&3&0\\9&6&0\end{bmatrix}\ \dots(2)$
$\therefore\ \text{AB}\neq\text{BA}$ From (1) and (2)
View full question & answer→Question 323 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 333 Marks
If $\text{A}=\begin{bmatrix}1&2\\0&3 \end{bmatrix}$ is written as B + C, where B is a symmetric matrix and C is a skew- symmetric matrix, then B is equal to.
AnswerGiven: $\text{A}=\begin{bmatrix}1&2\\0&3 \end{bmatrix}$
$\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}1&0\\2&3 \end{bmatrix}$
Let $\text{B}=\frac{1}{2}(\text{A+A}^{\text{T}})=\frac{1}{2}\bigg(\begin{bmatrix}1&2\\0&3 \end{bmatrix}+\begin{bmatrix}1&0\\2&3 \end{bmatrix}\bigg)$
$=\frac{1}{2}\begin{bmatrix}1+1&2+0\\0+2&3+3 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix} 2&2\\2&6\end{bmatrix}$
$=\begin{bmatrix} 1&1\\1&3\end{bmatrix}$
Now,
$\text{B}^{\text{T}}=\begin{bmatrix}1&1\\1&3 \end{bmatrix}=\text{B}$
Therefore, B is symmetric matrix.
Let $\text{C}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})=\frac{1}{2}\bigg(\begin{bmatrix}1&2\\0&3 \end{bmatrix}-\begin{bmatrix}1&0\\2&3 \end{bmatrix}\bigg)$
$=\frac{1}{2}\begin{bmatrix}1-1&2-0\\0-2&3-3 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}0&2\\-2&2 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}0&2\\-1&0 \end{bmatrix}$
$\therefore\text{C}^{\text{T}}=\begin{bmatrix}0&1\\-1&0 \end{bmatrix}^{\text{T}}=\begin{bmatrix}0&-1\\1&0 \end{bmatrix}=-\begin{bmatrix}0&1\\-1&0 \end{bmatrix}=\text{C}$
So, C is a skew-symmetric matrix.
Now,
$\text{B + C}=\begin{bmatrix}1&1\\1&3 \end{bmatrix}+\begin{bmatrix}0&1\\-1&0 \end{bmatrix}=\begin{bmatrix}1+0&1+1\\1-1&3+0 \end{bmatrix}=\begin{bmatrix}1&2\\0&3 \end{bmatrix}=\text{A}$
$\therefore\text{B}=\begin{bmatrix}1&1\\1&3 \end{bmatrix}$
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
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 363 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 373 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
$8=15+\lambda$
$\lambda=8-15$
$\lambda=-7$
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