Question 12 Marks
Find matrices X and Y, if $\text{X}+\text{Y}=\begin{bmatrix}5&2\\0&9\end{bmatrix}$ and $\text{X}-\text{Y}=\begin{bmatrix}3&6\\0&-1\end{bmatrix}$
AnswerGiven: $(\text{X}+\text{Y})+(\text{X}-\text{Y})=\begin{bmatrix}5&2\\0&9\end{bmatrix}+\begin{bmatrix}3&6\\0&-1\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}5+3&2+6\\0+0&9-1\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}8&8\\0&8\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}8&8\\0&8\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}4&4\\0&4\end{bmatrix}$
Now,
$(\text{X}+\text{Y})-(\text{X}-\text{Y})=\begin{bmatrix}5&2\\0&9\end{bmatrix}-\begin{bmatrix}3&6\\0&-1\end{bmatrix}$
$\Rightarrow\text{X}+\text{Y}-\text{X}+\text{Y}=\begin{bmatrix}5-3&2-6\\0-0&9+1\end{bmatrix}$
$\Rightarrow2\text{Y}=\begin{bmatrix}2&-4\\0&10\end{bmatrix}$
$\Rightarrow\text{Y}=\frac{1}{2}\begin{bmatrix}2&-4\\0&10\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}1&-2\\0&5\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}4&4\\0&4\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}1&-2\\0&5\end{bmatrix}$
View full question & answer→Question 22 Marks
If $A = [a_{ij}]$ is a $2 \times 2$ matrix such that $a_{ij} = i + 2j$, write $A$.
AnswerHere,
$a_{ij} = i + 2j$
$\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}$
$=\begin{bmatrix}1+2(1)&1+2(2)\\2+2(1)&2+2(2)\end{bmatrix}$
$=\begin{bmatrix}3&5\\4&6\end{bmatrix}$
Hence,
$\text{A}=\begin{bmatrix}3&5\\4&6\end{bmatrix}$
View full question & answer→Question 32 Marks
f $A$ is a matrix of order $3\times 4$ and $B$ is a matrix of order $4\times 3$, find the order of the matrix of $AB.$
AnswerOrder of $A=3 \times 4$
Order of $B=4 \times 3$
Order of $A_{3 \times 4} \times B_{4 \times 3}=3 \times 3$
So,
Order of $A B=3 \times 3$
View full question & answer→Question 42 Marks
Write the number of all possible matrices of order $2\times 2$ with each entry $1, 2$ or $3.$
AnswerAs matrices is of order $2\times 2,$ so there are $4$ entries possible.
Each entry has $3$ choices that are $1, 2$ or $3$
So, number of ways to make up such matrices are $3\times 3\times 3\times 3$ i.e, $3^4 $ times or $81$ times
View full question & answer→Question 52 Marks
If $\text{A}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}-4\\3\end{bmatrix},$ write AB.
Answer$\text{AB}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}\begin{bmatrix}-4\\3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-16+9\\-4+6\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix} -7\\2\end{bmatrix}$
View full question & answer→Question 62 Marks
If $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix},$ write $A^2$.
AnswerGiven: $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2+0&0+0\\0+0&0+\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2&0\\0&\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ $(\because\ \text{i} ^2=-1)$
View full question & answer→Question 72 Marks
For any square matrix write whether $AA^T$ is symmetric or skew-symmetric.
Answer$\big(\text{AA}^\text{T}\big)^\text{T}=\big(\text{A}^\text{T}\big)^\text{T}\times\text{A}^\text{T}$ $\big\{\text{since, (AB)}^\text{T}=\text{B}^\text{T}\text{A}^\text{T}\big\}$
$\therefore\ \big(\text{AA}^\text{T}\big)^\text{T}=\big(\text{AA}^\text{T}\big)\ \dots(\text{i})$ $\big\{\text{since, }(\text{A}^\text{T})^\text{T}=\text{A}\big\}$
We know that, a square matrix A is symmetric if $A^T = A$
So, from equation (i)
$(AA^T)$ is a symmetric matric.
View full question & answer→Question 82 Marks
Let $A$ and $B$ be square matrices of the order $"3 \times 3",$ Is $(AB)^2 = A^2B^2?$ Give reasons.
AnswerYes, $(AB)^2 = A^2 B^2,$ if $AB= BA.$
If $AB = BA,$ then
$(AB)^2 = (A B)(A B)$
$= A(BA)B ($associative law$)$
$= A(AB)B$
$= A^2B^2$
View full question & answer→Question 92 Marks
Give example of matrices:
A and B such that AB = O but A ≠ 0, B ≠ 0.
AnswerLet $\text{A}=\begin{bmatrix}0&2\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\therefore\ \text{AB}=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Thus, AB = 0 while A ≠ 0 and B ≠ 0
View full question & answer→Question 102 Marks
Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&2-2&-2+2\\4+1&\text{x}+0&6-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&0&0\\5&\text{x}&5\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\text{x}-\text{y}+3=6$
$\Rightarrow\text{x}-\text{y}=6-3$
$\Rightarrow\text{x}-\text{y}=3\ \dots(1)$
Also,
$\text{x}=2\text{x}+\text{y}$
$\Rightarrow-\text{x}=\text{y}\ \dots(2)$
Putting the value of y in eq. (1), we get
$\text{x}-(-\text{x})=3$
$\Rightarrow2\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{2}$
Putting the value of x in eq. (2), we get
$-\Big(\frac{3}{2}\Big)=\text{y}$
$\Rightarrow\text{y}=-\frac{3}{2}$
View full question & answer→Question 112 Marks
Find x, y, z and t, if.
$2\begin{bmatrix}\text{x}&5\\7&\text{y}-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
Answer$2\begin{bmatrix}\text{x}&5\\7&\text{y}-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}&10\\14&2\text{y}-6\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}+3&14\\15&2\text{y}-4\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
Comparing the corresponding elements from both sides,
$2\text{x}+3=7\Rightarrow2\text{x}=4\Rightarrow\text{x}=2$
$2\text{y}-4=14\Rightarrow2\text{y}=18\Rightarrow\text{y}=9$
Hence, x = 2, y = 9
View full question & answer→Question 122 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:$\text{a}_\text{ij}=\frac{(\text{i}+\text{j})^2}{2}$
AnswerHere,
$\text{a}_\text{ij}=\frac{(\text{i}+\text{j})^2}{2}$
$\text{a}_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2,$ $\text{a}_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{13}=\frac{(1+3)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
$\text{a}_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$ and $\text{a}_{23}=\frac{(2+3)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2}$
Required matrix = $\text{A}=\begin{bmatrix}2&\frac{9}{2}&8\\\frac{9}{2}&8&\frac{25}{2}\end{bmatrix}$
View full question & answer→Question 132 Marks
If $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix},$ show that $A - A^T$ is a skew symmetric matrix.
AnswerGiven:$\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$
$\text{A}-\text{A}^{\text{T}}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}-\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}^{\text{T}}$
$=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}-\begin{bmatrix}3 &1 \\-4 & -1 \end{bmatrix}$
$=\begin{bmatrix}3-3 & -4-1 \\1+4 & -1+1 \end{bmatrix}$
$\text{A}-\text{A}^{\text{T}}=\begin{bmatrix}0 & -5 \\5 & 0 \end{bmatrix}\ \dots( \text{i})$
$-(\text{A}-\text{A}^{\text{T}})^\text{T}=-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}^{\text{T}}$
$=-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}$
$-(\text{A}-\text{A}^{\text{T}})^{\text{T}}-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}\ \dots(\text{ii})$
From equation (i) and (ii)
$(\text{A}-\text{A}^{\text{T}})^{\text{T}}=-(\text{A}-\text{A}^{\text{T}})^{\text{T}}$
We know that, x is skewsym metric matrix if $x = -x^T$
So, $(A - A^T)$ is skewsym metric matrix.
View full question & answer→Question 142 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:
$a_{ij} = i.j$
AnswerHere,
$a_{ij} = i.j. 1 \leq i \leq 2$ and $1 \leq j \leq 3$
$a_{11} = 1 \times 1 = 1, a_{12} = 1 \times 2 = 2, a_{13} = 1 \times 3 = 3$
$a_{21} = 2 \times 1 = 2, a_{22} = 2 \times 2 = 4$ and $a_{23} = 2 \times 3 = 6$
View full question & answer→Question 152 Marks
Compute the indicated products:
$\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}\begin{bmatrix}\text{a}&-\text{b}\\\text{b}&\text{a}\end{bmatrix}$
Answer$\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}\begin{bmatrix}\text{a}&-\text{b}\\\text{b}&\text{a}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}\times\text{a}+\text{b}\times\text{b}&\text{a}\times(-\text{b})+\text{b}\times\text{a}\$-\text{b})\times\text{a}+\text{a}\times\text{b}&(-\text{b})\times(-\text{b})+\text{a}\times\text{a}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}^2+\text{b}^2&-\text{ab}+\text{ab}\\-\text{ab}+\text{ab}&\text{b}^2+\text{a}^2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}^2+\text{b}^2&0\\0&\text{a}^2+\text{b}^2\end{bmatrix}$
View full question & answer→Question 162 Marks
If a matrix has 5 elements, write all possible orders it can have.
AnswerWe know that if a matrix is of order m×n,then it has mn elements.If the matrix has 5 elements, then the number of elements will be 1×5 or 5×1, i.e. there will be 2 possible orders of the matrix.
View full question & answer→Question 172 Marks
Give example of matrices:
A, B and C such that AB = AC but B ≠ C, A ≠ 0
AnswerLet $\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&0\\-1&0\end{bmatrix},\ \text{C}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
Here,
$\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\-1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$
$\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$\begin{bmatrix}0&0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
LHS = RHS
So,
for A ≠ 0, BC ≠ 0 but AB = AC
We have,
$\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&0\\-1&0\end{bmatrix},\ \text{C}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
View full question & answer→Question 182 Marks
If $\text{X}-\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}$ and $\text{X}+\text{Y}=\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix},$ find X and Y.
AnswerHere,
$\text{X}-\text{Y}+\text{X}+\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}+\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}1+3&1+5&1+1\\1-1&1+1&0+4\\1+11&0+8&0+0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$
Now,
$(\text{X}-\text{Y})-(\text{X}+\text{Y})=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}-\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow\text{X}-\text{Y}-\text{X}-\text{Y}=\begin{bmatrix}1-3&1-5&1-1\\1+1&1-1&0-4\\1-11&0-8&0-0\end{bmatrix}$
$\Rightarrow-2\text{Y}=\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=-\frac{1}{2}\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
View full question & answer→Question 192 Marks
If $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix},$ find $AA^T$.
AnswerGiven: $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos^2\text{x}+\sin^2\text{x}&\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}\\\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}&\sin^2\text{x}+\cos^2\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→Question 202 Marks
Find x, y satisfying the matrix equation.
$\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=0$
AnswerGiven: $\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}+3\text{y}-8\\\text{x}+5\text{y}-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow2\text{x}+3\text{y}-8=0$
$\Rightarrow2\text{x}+3\text{y}=8\ \dots(1)$
Also,
$\text{x}+5\text{y}-11=0$
$\Rightarrow\text{x}+5\text{y}=11$
$\Rightarrow\text{x}=11-5\text{y}\ \dots(2)$
Putting the value of x in eq. (1), we get
$2(11-5\text{y})+3\text{y}=8$
$\Rightarrow22-10\text{y}+3\text{y}=8$
$\Rightarrow-7\text{y}=8-22$
$\Rightarrow-7\text{y}=-14$
$\Rightarrow\text{y}=2$
Putting the value of y in eq. (2), we get
$\text{x}=11-5(2)$
$\Rightarrow\text{x}=11-10$
$\Rightarrow\text{x}=1$
$\therefore\ \text{x}=1$ and $\text{y}=2$
View full question & answer→Question 212 Marks
If $\text{A}=\begin{bmatrix}9&1\\7&8\end{bmatrix},\text{ B}=\begin{bmatrix}1&5\\7&12\end{bmatrix},$ find matrix C such that 5A + 3B + 2C is a null matrix.
AnswerGiven, $5\text{A}+3\text{B}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow5\begin{bmatrix}9&1\\7&8\end{bmatrix}+3\begin{bmatrix}1&5\\7&12\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45&5\\35&40\end{bmatrix}+\begin{bmatrix}3&15\\21&36\end{bmatrix}+2\text{C}\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45+3&5+15\\35+21&40+36\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}48&20\\56&76\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}48&20\\56&76\end{bmatrix}$
$\Rightarrow\text{C}=\frac{1}{2}\begin{bmatrix}-48&-20\\-56&-76\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}$
View full question & answer→Question 222 Marks
If $A$ is $2\times 3$ matrix and $B$ is a matrix such that $A^T B$ and $BA^T$ both are defined, then what is the order of $B?$
AnswerOrder of $A = 2 \times 3$
Order of $A^T= 3 \times 2$
Let Order of $B = m \times n$
Given: $A^TB$ and $BA^T$ are defined
If ${A^T}_{3\times 2}B_{m\times n}$_exists, then the number of columns in $A^T$ must be equal to number of rows in $B.$
$\Rightarrow m = 2$
If_$B_{m\times n} {A^T}_{3\times 2}$_exists, then the number of columns in $B$ must be equal to number of rows in $A^T$
$\Rightarrow n = 3$
$\therefore$ Order of $B = 2 \times 3$
View full question & answer→Question 232 Marks
A trust fund has Rs. 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
- Rs. 1800
- Rs. 2000
AnswerIf Rs. x are invested in the first type of bond and Rs. (30000 - x) are invested in the second type of bond,
Then the matrix $\text{A}=\begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}$ represents investment and the matrix $\text{B}=\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}$ represents rate of interest.
- $\begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\frac{5\text{x}}{100}+\frac{7(30000-\text{x})}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}$
$\Rightarrow\frac{5\text{x}+210000-7\text{x}}{100}=1800$
$\Rightarrow210000-2\text{x}=180000$
$\Rightarrow2\text{x}=30000$
$\Rightarrow\text{x}=15000$
Thus,
Amount invested in the first bond = Rs. 15000
Amount invested in the second bond = Rs. (30000 - 15000)
= Rs. 15000
- $ \begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\frac{5\text{x}}{100}+\frac{7(30000-\text{x})}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}$
$\Rightarrow\frac{5\text{x}+210000-7\text{x}}{100}=2000$
$\Rightarrow210000-2\text{x}=200000$
$\Rightarrow2\text{x}=10000$
$\Rightarrow\text{x}=5000$
Thus,
Amount invested in the first bond = Rs. 5000
Amount invested in the second bond = Rs. (30000 - 5000)
= Rs. 25000 View full question & answer→Question 242 Marks
Compute the products AB and BA whichever exists the following cases:
$\text{A}=\begin{bmatrix}3&2\\-1&0\\-1&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}$
AnswerHere, $\text{A}=\begin{bmatrix}3&2\\-1&0\\-1&1\end{bmatrix},\text{B}=\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}$
Order of A = 3 × 2 and order of B = 2 × 3
So, AB and BA Both exith and order of AB = 3 × 3 and order of BA = 2 × 2
$\text{AB}=\begin{bmatrix}3&2\\-1&0\\-1&0\end{bmatrix}\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}$
$=\begin{bmatrix}(3)(4)+(2)(0)&(3)(5)+(2)(1)&(3)(6)+(2)(2)\$-1)(4)+(0)(0)&(-1)(5)+(0)(0)&(-1)(6)+(0)(2)\$-1)(4)+(1)(0)&(-1)(5)+(1)(1)&(-1)(6)+(1)(2)\end{bmatrix}$
$\begin{bmatrix}12+0&15+2&18+4\\-4+0&-5+0&-6+0\\-4+0&-5+0&-6+2\end{bmatrix}$
$=\begin{bmatrix}12&17&22\\-4&-5&-6\\-4&-4&-4\end{bmatrix}$
$\text{BA}=\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}\begin{bmatrix}3&2\\-1&0\\-1&1\end{bmatrix}$
$=\begin{bmatrix}(4)(3)+(5)(-1)+(6)(-1)&(4)(2)+(5)(0)+(6)(1)\$0)(3)+(1)(-1)+(2)(-1)&(0)(2)+(1)(0)+(2)(1)\end{bmatrix}$
$=\begin{bmatrix}12-5-6&8+0+6\\0-1-2&0+0+2\end{bmatrix}$
$=\begin{bmatrix}1&14\\-3&2\end{bmatrix}$
Hence,
$\text{AB}=\begin{bmatrix}12&17&22\\-4&-5&-6\\-4&-4&-4\end{bmatrix},\text{BA}=\begin{bmatrix}1&14\\-3&2\end{bmatrix}$
View full question & answer→Question 252 Marks
If A and B are symmetric matrices, then write the condition for which AB is also symmetric.
AnswerGiven that,
A and B are symmetric matrices, so
$\Rightarrow A^T = A$ and $B^T = B$
Now,
$\big(\text{AB}\big)^\text{T}=\text{B}^\text{T}\times\text{A}^\text{T}$ $\big\{\text{since, (AB)}^\text{T}=\text{B}^\text{T}\text{A}^\text{T}\big\}$
$\big(\text{AB}\big)^\text{T}=\text{BA}\ \dots(\text{i})$ $\big\{\text{since, B}^\text{T}=\text{B},\text{A}^\text{T}=\text{A}\big\}$
For AB to be symmetric matrix
$(AB)^T = AB$
From equation (i) and (ii),
AB = BA
So,
For AB to be symmetric matrix we must have AB = BA.
View full question & answer→Question 262 Marks
Find the values of x and y, if $2\begin{bmatrix}1&3\\0&\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
AnswerGiven,
$2\begin{bmatrix}1&3\\0&\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&6\\0&2\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+\text{y}&6+0\\0+1&2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+\text{y}&6\\1&2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
$2+\text{y}=5$
$\Rightarrow\text{y}=5-2$
$\Rightarrow\text{y}=3$
And $2\text{x}+2=8$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}=3$
Hence,
$\text{x}=3,\text{y}=3$
View full question & answer→Question 272 Marks
If I is the identity matrix and A is a square matrix such that $A^2 = A$, then what is the value of $(I + A)^2 = 3A$?
AnswerGiven,
A is a square matrix such that $A^2 = A$
Now,
$(I + A)^2- 3A = (I + A)(I + A) - 3A$
$\Rightarrow (I + A)^2- 3A = I \times I + I \times A + A \times I + A \times A - 3A$ {using distributive property}
$\Rightarrow (I + A)^2- 3A = I + A + A + A^2- 3A$ {using $I \times I = I, IA = AI = A$}
$\Rightarrow (I + A)^2- 3A = I + 2A + A - 3A$ {since, $A^2= A$}
$\Rightarrow (I + A)^2- 3A = I$
View full question & answer→Question 282 Marks
For what value of $x$, is the matrix $\text{A}=\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}$ a skew-symmetric matrix?
AnswerSince, A is a skew symmetric matrix.
$\therefore a^T= -A$
$\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}^{\text{T}}=-\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}0&-1&\text{x}\\1&0&-3\\-2&3&0\\ \end{bmatrix}=\begin{bmatrix}0&-1&2\\1&0&-3\\-\text{x}&3&0 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
⇒ x = 2
Hence, the value of x is 2.
View full question & answer→Question 292 Marks
Let $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix}$ and $\text{C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}.$ Find each of the following:
$2\text{A}-3\text{B}$
AnswerGiven, $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{ B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{ C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}$
$2\text{A}-3\text{B}$
$=2\begin{bmatrix}2&4\\3&2\end{bmatrix}-3\begin{bmatrix}1&3\\-2&5 \end{bmatrix}$
$=\begin{bmatrix}4&8\\6&4\end{bmatrix}-\begin{bmatrix}3&9\\-6&15 \end{bmatrix}$
$=\begin{bmatrix}4-3&8-9\\6+6&4-15\end{bmatrix}$
$=\begin{bmatrix}1&-1\\12&-11\end{bmatrix}$
Hence,
$2\text{A}-3\text{B}=\begin{bmatrix}1&-1\\12&-11\end{bmatrix}$
View full question & answer→Question 302 Marks
If $\text{A}=\begin{bmatrix}1&2\\3&4 \end{bmatrix}$, find $A+A^T$.
AnswerGiven: $\text{A}=\begin{bmatrix}1&2\\3&4 \end{bmatrix}$
$\text{A}^{\text{T}}=\begin{bmatrix}1&3\\2&4 \end{bmatrix}$
$\text{A+A}^{\text{T}}=\begin{bmatrix}1&2\\3&4 \end{bmatrix}+\begin{bmatrix}1&3\\2&4 \end{bmatrix}$
$\Rightarrow\text{A+A}^{\text{T}}=\begin{bmatrix}1+1&2+3\\3+2&4+4 \end{bmatrix}$
$\Rightarrow\text{A+A}^{\text{T}}=\begin{bmatrix}2&5\\5&8 \end{bmatrix}$
View full question & answer→Question 312 Marks
If $\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6 \end{bmatrix}=\begin{bmatrix}2\text{a}+2&\text{b}+2\\8&\text{a}-8\text{b} \end{bmatrix},$ write the value of a - 2b.
Answer$\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6 \end{bmatrix}=\begin{bmatrix}2\text{a}+2&\text{b}+2\\8&\text{a}-8\text{b} \end{bmatrix}$
Form the above equation,
$\therefore$ a + 4 = 2a + 2
⇒ a = 2
$\therefore$ 3b = b + 2
⇒ 2b = 2
⇒ b = 1
a - 2b
= 2 - 2 × 1
= 0
View full question & answer→Question 322 Marks
If $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix},$ then find the values of $x, y, z$ and $w.$
Answer$\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix}$
The corresponding entries of the two equal matrices are equal,
$\Rightarrow xy = 8 ...(1),$
$w = 4 ...(2),$
$z + 6 = 0 ...(3),$
And $x + y = 6 ...(4)$
From equation $(2)$ and equation $(3)$ we get $z = -6$ and $w = 4.$
From equation $(4)$ we have,
$x + y = 6$
$\Rightarrow x = 6 - y,$
Subsituting value of $x$ in equation $(1)$ we get,
$\Rightarrow (6 - y)y = 8$
$\Rightarrow y^2 - 6y + 8 = 0$
$\Rightarrow (y - 2)(y - 4) = 0,$
$\Rightarrow y = 2, 4$
Subsituting the value of $y$ in equation $(1)$ we get,
$\Rightarrow x = 4, 2$
Therefore, value of $x, y, z, w$ are $2, 4, -6, 4$ or $4, 2, -6, 4.$
View full question & answer→Question 332 Marks
Let A and B be matrices of orders 3×2 and 2×4 respectively. Write the order of matrix AB.
AnswerSince, the order of matrix A is 3×2 and order of matrix B is 2×4
So, the order of AB will be the "number of rows of A × number of columns of B" = 3×4
View full question & answer→Question 342 Marks
Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+2+4&\text{z}-3+5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+6&\text{z}+2\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\therefore\ \text{x}+\text{y}=4\ \dots(1)$
Also,
$\text{y}+6=9$
$\Rightarrow\text{y}=3$
$\text{z}+2=12$
$\Rightarrow\text{z}=10$
Putting the value of y in eq. (1), we get
$\text{x}+3=4$
$\Rightarrow\text{x}=4-3$
$\Rightarrow\text{x}=1$
$\therefore\ \text{x}=1,\text{ y}=3$ and $\text{z}=10$
View full question & answer→Question 352 Marks
If $\text{A}=\begin{bmatrix}2&1&4\\4&1&5 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}3&-1\\2&2\\1&3\end{bmatrix}.$ Write the order of AB and BA.
AnswerOrder of $A=2 \times 3$
Order of $B=3 \times 2$
So,
$\mathrm{A}_{2 \times 3} \times \mathrm{B}_{3 \times 2}$ has order $=2 \times 2$
$\mathrm{B}_{3 \times 2} \times \mathrm{A}_{2 \times 3}$ has order $=3 \times 3$
Hence,
Order of $A B=2 \times 2$
Order of $B A=3 \times 3$
View full question & answer→Question 362 Marks
If A and B are square matrices of the same order, explain, why in general:
$(A + B)^2 \neq A^2 + 2AB + B^2$
Answer$\text { LHS }=(A+B)^2$
$=(A+B)(A+B)$
$=A(A+B)+B(A+B)$
$=A^2+A B+B A+B^2$
We know that a matrix does not have commutative property. So,
$A B \neq B A$
Thus,
$(A+B)^2 \neq A^2+2 A B+B^2$
View full question & answer→Question 372 Marks
If $A$ is a skew-symmetric matrix and $n$ is an even natural number, write whether $A^n$ is symmetric or skew-symmetric or neither of these two.
AnswerIf A is a skew-symmetric matrix, then $A^T = -A.$
$(A^n)^T = (A^T)^n [$For all $n \in N]$
$\Rightarrow (A^n)^T = (-A)^n [\because A^T = -A]$
$\Rightarrow (A^n)^T = (-1)^n A^n$
$\Rightarrow (A^n)^T = A^n,$ if n is even or $-A^n$, if $n$ is odd.
Hence, $A^n$ is a symmetric when n is an even natural number.
View full question & answer→Question 382 Marks
Find x, y, a and b if $\begin{bmatrix}3\text{x}+4\text{y}&2&\text{x}-2\text{y}\\\text{a}+\text{b}&2\text{a}-\text{b}&^-1\end{bmatrix}=\begin{bmatrix}2&2&4\\5&-5&-1\end{bmatrix}$
AnswerSince the corresponding elements of two equal matrices are equal,
$\begin{bmatrix}3\text{x}+4\text{y}&2&\text{x}-2\text{y}\\\text{a}+\text{b}&2\text{a}-\text{b}&^-1\end{bmatrix}=\begin{bmatrix}2&2&4\\5&-5&-1\end{bmatrix}$
⇒ 3x + 4y = 2 ...(1)
⇒ x - 2y = 4
⇒ x = 4 + 2y ...(2)
Putting the value of x in eq. (1), we get
3(4 + 2y) + 4y = 2
⇒ 12 + 6y + 4y = 2
⇒ 12 + 10y = 2
⇒ 10y = 2 - 12
⇒ 10y = -10
$\Rightarrow\text{y}=\frac{-10}{10}=-1$
Putting the value of y in eq. (2), we get
x = 4 + 2(-1)
⇒ x = 4 - 2 = 2
a + b = 5
⇒ a = 5 - b ...(3)
⇒ 2a - b = -5 ...(4)
Putting the value of a in eq. (4), we get
2(5 - b) - b = -5
⇒ 10 - 2b - b = -5
⇒ 10 - 3b = -5
⇒ -3b = -15
$\Rightarrow\text{b}=\frac{-15}{-3}$
⇒ b = 5
Putting the value of b in eq. (3), we get
a = 5 - 5
⇒ a = 0
$\therefore$ x = 2, y = -1, a = 0 and b = 5
View full question & answer→Question 392 Marks
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are given by:
$\text{a}_\text{ij}=\frac{|2\text{i}-3\text{j}|}{2}$
AnswerHere,
$\text{a}_{11}=\frac{|2(1)-3(1)|}{2}=\frac{1}{2},$ $\text{a}_{12}=\frac{|2(1)-3(2)|}{2}=2$
$\text{a}_{21}=\frac{|2(2)-3(1)|}{2}=\frac{1}{2},$ $\text{a}_{22}=\frac{|2(2)-3(2)|}{2}=1$
Using equation (i)
$\text{A}=\begin{bmatrix}\frac{1}{2}&2\\\frac{1}{2}&1\end{bmatrix}$
View full question & answer→Question 402 Marks
If $\begin{bmatrix}\text{x}+3&\text{z}+4&2\text{y}-7\\4\text{x}+6&\text{a}-1&0\\\text{b}-3&3\text{b}&\text{z}+2\text{c}\end{bmatrix}=\begin{bmatrix}0&6&3\text{y}-2\\2\text{x}&-3&2\text{c}-2\\2\text{b}+4&-21&0\end{bmatrix}$ Obtain the values of a, b, c, x, y and z.
AnswerSince all the corresponding elements of a matrix are equal, x + 3 = 0 ⇒ x = -3 Also, 2y - 7 = 3y - 2 ⇒ 2y - 3y = -2 + 7 ⇒ -y = 5 ⇒ y = -5⇒ z + 4 = 6
⇒ z = 6 - 4 ⇒ z = 2⇒ a - 1 = -3
⇒ a = -3 + 1 ⇒ a = -2 3b = -21 ⇒ b = -7⇒ z + 2c = 0
⇒ 2 = -2c ⇒ c = -1 Thus, x = -3, y = -5, a = -2, b = -7 and c = -1
View full question & answer→Question 412 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:$a_{ij} = i + j$
AnswerHere,
$a_{ij} = i + j$
$a_{11} = 1 + 1 = 2, a_{12} = 1 + 2 = 3, a_{13} = 1 + 3 = 4$
$a_{21} = 2 + 1 = 3, a_{22} = 2 + 2 = 4$ and $a_{23} = 2 + 3 = 5$
Required matrix = $\text{A}=\begin{bmatrix}2&3&4\\3&4&5\end{bmatrix}$
View full question & answer→Question 422 Marks
If $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix},$ prove that $A - A^T$ is a skew symmetric matrix.
AnswerGiven: $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix}$
$\text{A}^{\text{T}}=\begin{bmatrix}2 & 4 \\3 & 5 \end{bmatrix}$
Now,
$(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix}-\begin{bmatrix}2 & 4 \\3 & 5 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}2-2 & 3-4 \\4-3 & 5-5 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^\text{T})=\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}\ ...(\text{i})$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}^\text{T}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}0 & 1 \\-1 & 0 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=-\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})=(\text{A}-\text{A}^{\text{T}})^{\text{T}} $ [Using eq.(i)
Thus, $(A - A^T)$ is a skew-symmetric matrix.
View full question & answer→Question 432 Marks
If $\begin{bmatrix}\text{x}&3\text{x}-\text{y}\\2\text{x}+\text{z}&3\text{y}-\text{w}\end{bmatrix}=\begin{bmatrix}3&2\\4&7\end{bmatrix},$ find x, y, z, w.
AnswerSince all the corresponding elements of a matrix are equal,
$\begin{bmatrix}\text{x}&3\text{x}-\text{y}\\2\text{x}+\text{z}&3\text{y}-\text{w}\end{bmatrix}=\begin{bmatrix}3&2\\4&7\end{bmatrix}$
x = 3 ...(1)
3x - y = 2 ...(2)
Putting the value of x in eq. (2), we get
3(3) - y = 2
⇒ 9 - y = 2
⇒ -y = -7
⇒ y = 7
2x + z = 4 ...(3)
Putting the value of x in eq. (3), we get
2(3) + z = 4
⇒ 6 + z = 4
⇒ z = 4 - 6
⇒ z = -2
3y - w = 7 ...(4)
Putting the value of x in eq. (4), we get
3(7) - w = 7
⇒ 21 - w = 7
⇒ 21 - 7 = w
⇒ w = 14
$\therefore$ x = 3, y = 7, z = -2 and w = 14
View full question & answer→Question 442 Marks
If $\text{A}=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ satisfies $\text{A}^4=\lambda\text{A},$ then write the value of $\lambda.$
Answer$\text{A}^2=\text{A}\cdot\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+1&1+1\\1+1&1+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&2\\2&2\end{bmatrix}$
Now,
$\text{A}^4=\text{A}^2\cdot\text{A}^2$
$\Rightarrow\text{A}^4=\begin{bmatrix}2&2\\2&2\end{bmatrix}\begin{bmatrix}2&2\\2&2\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}4+4&4+4\\4+4&4+4\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}8&8\\8&8\end{bmatrix}$
Also,
$\text{A}^4=\lambda\text{A}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\lambda\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\begin{bmatrix}\lambda&\lambda\\\lambda&\lambda\end{bmatrix}$
$\therefore\ \lambda=8$
View full question & answer→Question 452 Marks
If $\text{A}=\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}$ is identity matrix, then write the value of a.
AnswerHere,
$\text{A}=\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\text{I}$
$\Rightarrow\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\begin{bmatrix}1&0\\0&1 \end{bmatrix}$
The corresponding elements of equal matrices are equal.
$\therefore \cos\text{a}=1$
$\Rightarrow\text{a}=0^{\circ}$
View full question & answer→Question 462 Marks
If $A$ and $B$ are square matrices of the same order, explain, why in general:
$(A + B)(A - B) \neq A^2 - B^2.$
AnswerLHS $= (A + B)(A - B)$
$= A(A - B) + B(A - B)$
$= A^2 - AB + BA - B^2$
We know that a matrix does not have commutative property. So,
$AB ≠ BA$
Thus,
$(A + B)(A - B) \neq A^2 - B^2$
View full question & answer→Question 472 Marks
Give example of matrices:
A and B such that AB ≠ BA
AnswerLet $ \text{A}=\begin{bmatrix}\text{a}&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}$
$\text{A}\text{B}=\begin{bmatrix}\text{a}&0\\0&0\end{bmatrix}\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}$
$=\begin{bmatrix}0+0&\text{a}\text{b}+0\\0+0&0+0\end{bmatrix}$
$\text{AB}=\begin{bmatrix}0&\text{a}\text{b}\\0&0\end{bmatrix}$
$\text{BA}=\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}\begin{bmatrix}0&\text{a}\\0&0\end{bmatrix}$
$ =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$ \text{BA}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
From equation (i) and (ii)
AB ≠ BA
When $ \text{A}=\begin{bmatrix}\text{a}&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}$
View full question & answer→Question 482 Marks
Compute the products AB and BA whichever exists the following cases:
$\text{A}=\begin{bmatrix}1&-1&2&3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}$
Answer$\text{AB}=\begin{bmatrix}1&-1&2&3\end{bmatrix}\begin{bmatrix}0\\1\\3\\2\end{bmatrix}$
$\Rightarrow\text{AB}=[0+(1)+6+6]$
$\Rightarrow\text{AB}=[11]$
Also,
$\text{BA}=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}\begin{bmatrix}1&-1&2&3\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&0&0&0\\1&-1&2&3\\3&-3&6&9\\2&-2&4&6\end{bmatrix}$
View full question & answer→Question 492 Marks
If $A$ is a square matrix such that $A^2 = A$, then write the value of $7A − (I + A)^3$, where I is the identity matrix.
Answer$A^2=A$
$ A^3=A^2=A$
$ 7 A-(I+A)^3$
$ =7 A-\left(B^3+A^3+3 A^2 I+3 A I^2\right)$
$ =7 A-(I+A+3 A+3 A)$
$ =7 A-(I+7 A)$
$ =-1$
View full question & answer→Question 502 Marks
If $B$ is a symmetric matrix, write whether the matrix $AB\ A^T$ is symmetric or skew-symmetric.
AnswerIf B is a skew-symmetric matrix, then $B^T - B$.
$(\text{ABA}^\text{T})^\text{T}=(\text{A}^\text{T})^\text{T}\text{B}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{ABC})^\text{T}=\text{C}^\text{T}\text{B}^\text{T}\text{A}^\text{T}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=\text{AB}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{A}^\text{T})^\text{T}=\text{A}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=-\text{ABA}^\text{T}$ $\big[\because\ \text{B}^\text{T}=\text{B}\big]$
Hence, $ABA^T$ is a skew-symmetric matrix.
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