Question 11 Mark
Use elementary column operation $\text{C}_{2}\rightarrow\text{C}_{2} + 2\text{C}_{1}$ in the following matrix equation:
$ \begin{bmatrix} 2 & 1 \\ 2 & 0 \\ \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 2 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ \end{bmatrix} $
Answer$ \begin{bmatrix} 2 & 5 \\ 2 & 4 \\ \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 2 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -1 \\ \end{bmatrix} $
View full question & answer→Question 21 Mark
Write the number of all possible matrices of order $2\times2$ with each entry 1, 2 or 3.
AnswerNo. of possible matrices $ =3^{4}$
$\text{or 81}$
View full question & answer→Question 31 Mark
If for any $2 \times 2$ square matrix A, A(adj A) $= \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix},$ then write the value of |A|.
Answerit is given that
$\Rightarrow \text{A (adj A}) = 8 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$ \Rightarrow \text{A(adj A) } 8\text{I}_{2} \text{ }\text{ }\text{ } \dots\dots \text{(1)}$
We know that for any square matrix A of order 2, we have
$\text{A(adj A)} = |\text{A}| \text{I}_{2} \text{ }\text{ }\text{ } \dots\dots\dots \text{(2)}$
From (1) and (2), we have
|A| = 8
View full question & answer→Question 41 Mark
If A is a square matrix such that $A^2 = A$, then write the value of $7A – (I + A)^3$, where I is an identity matrix.
View full question & answer→Question 51 Mark
If $\text{A} = \begin{bmatrix} \\cos\theta & \sin\theta & \\ -\sin\theta & \cos\theta & \\ \end{bmatrix}, $ then for any natural number n, find the value of Det $(A^{n}).$
Answergetting $|\text{A}| = 1$
$|\text{A}^{\text{n}}| = 1$
View full question & answer→Question 61 Mark
If matrix A = $\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}$ and $A^2 = kA$, then write the value of k.
AnswerGiven $A^2 = kA$
$\Rightarrow\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}.\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix} = \text{k}\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 2 & -2 \\ -2 & 2 \\ \end{bmatrix} =\text{k}\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix} \Rightarrow 2\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix} = \text{k}\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}$
$\Rightarrow\text{k} = 2.$
View full question & answer→Question 71 Mark
For what value of x, is the matrix A =$\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ \text{x} & -3 & 0 \end{bmatrix}$ a skew-symmetric matrix?
AnswerA will be skew symmetric matrix if
A = – A'
$\Rightarrow\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ \text{x} & -3 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -1 & \text{x} \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -\text{x} \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix}$
Equating, we get x = 2.
View full question & answer→Question 81 Mark
If $A_{ij}$ is the cofactor of the element $a_{ij}$ of the determinant$\begin{bmatrix} 2 & -3 & 5 \$0.3em] 6 & 0 & 4 \$0.3em] 1 & 5 & -7 \end{bmatrix},$ then write the value of $a_{32}. A_{32}$.
Answer$a_{32} .A_{32} = 5\times(-1)^{3+2}\begin{vmatrix} 2 & 5 \$0.3em] 6 & 4 \end{vmatrix}$
= – 5 (8 – 30) = -5 X -22 = 110.
View full question & answer→Question 91 Mark
Find the value of x + y from the following equation:
$2\begin{bmatrix} \text{x} & 5 \\ 7 & \text{y - 3} \end{bmatrix} +\begin{bmatrix} 3 & -4 \\ 1 & 2 \\ \end{bmatrix}=\begin{bmatrix} 7 & 6 \\ 15 & 14 \\ \end{bmatrix}\dot{}$
Answer$2\text{x}+3=7$
$2\text{x}=4$
$\text{x}=2$
$2\text{y}-6+2=14$
$2\text{y}-4=14$
$2\text{y}=18$
$\text{y}=9$
$\therefore\ \text{x}+\text{y}=2+9=11$
View full question & answer→Question 101 Mark
If $A^T$= $\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \text{and B}=\begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \\ \end{bmatrix}$, then find $A^T – B^T$.
Answer$\begin{pmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{pmatrix} $.
View full question & answer→Question 111 Mark
Let A be a square matrix of order 3×3. Write the value of |2A|, where |A| = 4.
Answer|kA|=k^n|A|
K is a constant, for example, 2 in your question and n is the order of the matrix, which will be 3 according to your question.
|2A|=2^3|A|= 8|A|=8×4=32
View full question & answer→Question 121 Mark
If a matrix has 5 elements, write all possible orders it can have.
Question 131 Mark
If A = $ \begin{vmatrix} \text{2}& \text{3} \\ \text{5}& \text{-2} \\ \end{vmatrix}$, write $A^{–1}$ in terms of A.
Answer$\text{A}^{-1}=\frac{1}{19}\text{A}$.
View full question & answer→Question 141 Mark
Evaluate:
$ \begin{vmatrix} \text{cos 15}^{0}& \text{sin 15}^{0} \\ \text{sin 75}^{0} & \text{cos 75}^{0} \\ \end{vmatrix}$.
View full question & answer→Question 151 Mark
Write the adjoint of the following matrix: $\left(\begin{array}{c}2&-1\\ 4&3\end{array}\right)$
Answer$\left(\begin{array}{c}3&1\\- 4&2\end{array}\right)$
View full question & answer→Question 161 Mark
A is a square matrix of order 3 and | A | = 7. Write the value of | adj. A |.
Question 171 Mark
Find the value of x, if
$ \begin{pmatrix} \text{3x + y}&\text{-y} \\ \text{2y - x}&\text{3} \\ \end{pmatrix}= \begin{pmatrix} \text{1}&\text{2} \\ \text{-5}&\text{3} \\ \end{pmatrix}.$
View full question & answer→Question 181 Mark
For what value of x, is the following matrix singular?
$\begin{vmatrix} \text{3-2x} & \text{x + 1} \\ 2 & 4 \\ \end{vmatrix}$.
Answer$\begin{vmatrix} \text{3-2x} & \text{x + 1} \\ 2 & 4 \\ \end{vmatrix}=0$
$\Rightarrow$ 12 −8x − 2x − 2 = 0
$\Rightarrow$ x = 1.
View full question & answer→Question 191 Mark
A matrix A, of order $3 × 3$, has determinant $4$. Find the value of $|3A|$.
Answer$|3A| =3^3 \times 4 = 108$.
View full question & answer→Question 201 Mark
If $ \text{A} = \begin{pmatrix} 3 & 5 \\ 7 & 9 \\ \end{pmatrix} $ is written as A = P + Q, where P is a symmetric matrix and Q is skew symmetric matrix, then write the matrix P.
Answer$ \text{A} = \begin{bmatrix} 3 & 5 \\ 7 & 9 \\ \end{bmatrix} $
P is symmetric matrix. So, $\text{P} = \frac{1}{2} \text{(A + A}^{\text{T}})$
Q is skew symmetric matrix. So, $\text{Q} = \frac{1}{2} \text{(A + A}^{\text{T}})$
$\text{A}^{\text{T}} = \begin{bmatrix} 3 & 7 \\ 5 & 9 \\ \end{bmatrix} $
$\text{P} = \frac{1}{2}\begin{bmatrix} 6 & 12 \\ 12 & 18 \\ \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 6 & 9 \\ \end{bmatrix} $
View full question & answer→Question 211 Mark
If $\text{|A}| = 3 \text{ and } \text{A}^{-1} = \begin{bmatrix} 3 & -1 \\ -\frac{5}{3} & \frac{2}{3} \\ \end{bmatrix},$ then write the adj A.
Answer$ \text{adj A = 3} \begin{bmatrix} 3 & -1 \\ -\frac{5}{3} & \frac{2}{3} \\ \end{bmatrix} = \begin{bmatrix} 9 & -3 \\ -5 & 2 \\ \end{bmatrix} $
View full question & answer→Question 221 Mark
If $(2\ \ 1\ \ 3) \begin{pmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \text{A},$ then write the order of matrix A.
AnswerConsider, $(2\ \ 1\ \ 3) \begin{pmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \text{A},$
Order of matrix (2 1 3) is $1 \times 3.$
Order of matrix $\begin{pmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$ is $3 \times 3$
Order of matrix $\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$ is $3 \times 1$
Therefore, order of $(2\ \ 1\ \ 3) \begin{pmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \text{A},$ is $1 \times 1$
View full question & answer→Question 231 Mark
If A is a $3 \times 3$ invertible matrix, then what will be the value of k if $\text{det(A}^{-1}) = (\text{det(A)}^{\text{k}}.$
Answer$|\text{A}^{-1}| = \frac{1}{|\text{A}|} \Rightarrow \text{k} = -1$
View full question & answer→Question 241 Mark
$\text{Matrix A} = \begin{bmatrix} 0 & 2\text{b} & -2 \\ 3 & 1 & 3 \\ 3\text{a} & 3 & -1 \end{bmatrix} $ is given to be symmetric, find values of a and b.
Answer$\text{2b = 3 and 3a = - 2}$
$\text{b} = \frac{3}{2} \text{and a} = -\frac{2}{3}$
View full question & answer→Question 251 Mark
If A is a square matrix such that $\text{A}^{2} = \text{I,} $then find the simplified value of
$\text{(A - I)}^{3}+\text{(A + I)}^{3} - \text{7 A}.$
Answer$\text{(A - I)}^{3}+\text{(A + I)}^{3} - \text{7A},\ \ \ \ \ \text{A}^{2} =\text{I} \Rightarrow \text{A}^{3} = \text {A}$
$= \text{2A - A = A}$
View full question & answer→Question 261 Mark
Solve the following matrix equation for $\text{x};[\text{x } { 1 }] \ \begin{bmatrix}1 & 0\\-2 & 0\\\end{bmatrix} \ = \text{O}.$
View full question & answer→Question 271 Mark
If 2 $\ \begin{bmatrix}3 & 4 \\5 & x \\ \end{bmatrix}\ + \ \begin{bmatrix}1 & y \\0 & 1 \\ \end{bmatrix}\ = \ \begin{bmatrix}7 & 0 \\10 & 5 \\ \end{bmatrix}\ $ find $(\text{x} - \text{y}).$
View full question & answer→Question 281 Mark
Write the element $\text{a}_{23} \text{ of a 3} \times \text{3 matrix A} = \text({a}_{\text{ij})}$ whose elements $\text{a}_{\text{ij}}$ are given by $\text{a}_{\text{ij}} = \frac{|\text{i - j|}}{2}.$
Answer$\text{a}_{23} = \frac{|2 - 3|}{2} = \frac{1}{2}$
View full question & answer→Question 291 Mark
If $\begin{bmatrix} 9 &-1& 4\\ -2 & 1 & 3 \\ \end{bmatrix} =\text{A} + \begin{bmatrix} 1 & 2&-1 \\ 0 & 4 & 9 \\ \end{bmatrix},$then find the matrix A.
AnswerGiven $\begin{bmatrix} 9 &-1& 4 \\ -2 & 1 & 3 \\ \end{bmatrix} =\text{A} + \begin{bmatrix} 1 & 2&-1 \\ 0 & 4 & 9 \\ \end{bmatrix},$
$\Rightarrow\text{A} = \begin{bmatrix} 9 &-1& 4 \\ -2 & 1 & 3 \\ \end{bmatrix} - \begin{bmatrix} 1 & 2&-1 \\ 0 & 4 & 9 \\ \end{bmatrix},$
$ = \begin{bmatrix} 8 &-3& 5 \\ -2 & -3 & -6 \\ \end{bmatrix}.$
View full question & answer→Question 301 Mark
Write $A^{–1}$ for A = $\begin{bmatrix} 2 & 5 \\ 1 & 3 \\ \end{bmatrix}$.
Answer$\text{A}^{-1}=\begin{pmatrix} 3 & - 5 \\ -1 & 2 \\ \end{pmatrix}$.
View full question & answer→Question 311 Mark
For what value of x, the matrix $\begin{bmatrix} 5-\text{x} & \text{x + 1} & \\ 2 & 4 & \\ \end{bmatrix}$ is singular?
View full question & answer→Question 321 Mark
For a $2 \times 2$ matrix, $A=\left[a_{i j}\right]$, whose elements are given by $a_{i j}=\frac{i}{j}$, write the value of $a_{12}$.
Answer$\text{a}_{12}=\frac{1}{2}$.
View full question & answer→Question 331 Mark
$\text{If}\begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 2 & 5 \\ \end{pmatrix}=\begin{pmatrix} 7 & 11 \\ \text{k} & 23 \\ \end{pmatrix}, $then write the value of k.
View full question & answer→Question 341 Mark
If $\text{A}=\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \\ \end{pmatrix}$, then for what value $\alpha$ of is A an identity matrix?
View full question & answer→Question 351 Mark
If matrix A = (1, 2, 3), write AA' , where A' is the transpose of matrix A.
AnswerGiven A = (1, 2, 3),$A' = \begin{bmatrix} 1& \\ 2 &\\ 3 & \end{bmatrix} $
$AA' = ( 1\times 1 + 2\times 2 + 3\times 3) = (14)$
View full question & answer→Question 361 Mark
If A is an invertible matrix of order 3 and |A| = 5,Then find |adj. A|.
AnswerGiven $|A| = 5$We know $|adj.A| = |A|^{2}$
$\therefore |adj. A| = 5^{2} = 25$
View full question & answer→Question 371 Mark
Find the value of x and y if: $2 \begin{bmatrix} 1 & 3 \\ 0 & X \\ \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \\ \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \\ \end{bmatrix}$
Answer$2 \begin{bmatrix} 1 & 3 \\ 0 & X \\ \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \\ \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \\ \end{bmatrix}$$ \begin{bmatrix} 2 & 6 \\ 0 & 2X \\ \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \\ \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \\ \end{bmatrix}$
$ \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \\ \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \\ \end{bmatrix}$
Comparing both matrices
$2 + y =\text{ 5 and 2}x + 2 = 8$
$\Rightarrow \text{3 and 2}x = 6$
$\Rightarrow x = 3, y = 3.$
$$
View full question & answer→Question 381 Mark
If the matrix $\text{A}=\begin{bmatrix}0 & \text{a} & -3 \\2 & 0&-1\\\text{b}&1&0 \end{bmatrix}$ is skew symmetric, find the values of ‘a’ and ‘b’.
Answer$\text{A}=\begin{bmatrix}0 & \text{a} & -3 \\2 & 0&-1\\\text{b}&1&0 \end{bmatrix}$
For skew symmetric matrix
$\text{A}^\text{T} = -\text{A}$
$\begin{bmatrix}0 & \text{a} & -3 \\2 & 0&-1\\\text{b}&1&0 \end{bmatrix}^\text{T}=\begin{bmatrix}0 & -\text{a} & 3 \\-2 & 0&1\\-\text{b}&-1&0 \end{bmatrix}$
$\begin{bmatrix}0 & 2&\text{b} \\\text{a} &0&1\\-3&-1&0 \end{bmatrix}=\begin{bmatrix}0 & -\text{a} & 3 \\-2 & 0&1\\-\text{b}&-1&0 \end{bmatrix}$
$\text{b}=3$ (On comparing L.H.S. & R.H.S.)
$\text{a}=-2.$
View full question & answer→Question 391 Mark
If $3\text{A}-\text{B}=\begin{bmatrix}5 & 0 \\1 & 1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}4 & 3 \\2 & 5 \end{bmatrix},$ then find the matrix A.
Answer$3\text{A}-\text{B}=\begin{bmatrix}5 & 0 \\1 & 1 \end{bmatrix}$
We need to calculate A.
$3\text{A}=\begin{bmatrix}5 & 0 \\1 & 1 \end{bmatrix}+\text{B}$
$\therefore\ \text{B}=\begin{bmatrix}4 & 3 \\2 & 5 \end{bmatrix}\ \ (\text{given})$
$3\text{A}=\begin{bmatrix}9 & 3 \\3 & 6 \end{bmatrix}$
$\text{A}=\frac{1}{3}\begin{bmatrix}9 & 3 \\3 & 6 \end{bmatrix}$
$\text{A}=\begin{bmatrix}3 & 1 \\1 & 2 \end{bmatrix}$
View full question & answer→Question 401 Mark
Find the value of x - y, if $2\begin{bmatrix}1 & 3 \\0 & \text{x} \end{bmatrix}+\begin{bmatrix}\text{y} & 0 \\1 & 2 \end{bmatrix}=\begin{bmatrix}5 & 6 \\1 & 8 \end{bmatrix}.$
Answer$2\begin{bmatrix}1 & 3 \\0 & \text{x} \end{bmatrix}+\begin{bmatrix}\text{y} & 0 \\1 & 2 \end{bmatrix}=\begin{bmatrix}5 & 6 \\1 & 8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2 & 6 \\0 & 2\text{x} \end{bmatrix}+\begin{bmatrix}\text{y} & 0 \\1 & 2 \end{bmatrix}=\begin{bmatrix}5 & 6 \\1 & 8 \end{bmatrix}.$
$\Rightarrow\begin{bmatrix}2+\text{y} & 6+0 \\0+1 & 2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5 & 6 \\1 & 8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+\text{y} & 6 \\1 & 2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5 & 6 \\1 & 8 \end{bmatrix}$
Two matrices are equal and their corresponding entries are equal.
⇒ 2 + y = 5, 2x + 2 = 8
⇒ y = 3, x = 3
⇒ x - y = 3 - 3 = 0
View full question & answer→Question 411 Mark
If A and B are square matrices of the same order 3, such that |A| = 2 and AB = 21, write the value of |B|.
AnswerWe have, AB = 21
|AB| = |21|
⇒ |A| × |B| = 8
⇒ 2|B| = 8 (Given |A| = 2)
⇒ |B| = 4
View full question & answer→Question 421 Mark
If $\text{A}[\text{a}_{\text{ij}}]=\begin{bmatrix}2&3&-5\\1&4&9\\0&7&-2\end{bmatrix}$ and $\text{B}=[\text{b}_\text{ij}]=\begin{bmatrix}2&-1\\-3&4\\1&-2\end{bmatrix}$
Then find $a_{22} + b_{21}$
Answer$a_{22} + b_{22} = 4 + (-3) = 1$
Hence, $a_{22} + b_{21} = 1$
View full question & answer→Question 431 Mark
In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind in all the colleges.
AnswerNumber of different types of post in any college is given by,
$\text{X}=\begin{bmatrix}15\\6\\1\\1\end{bmatrix}$
Total number of posts of each kind in all the colleges = 30X
$=\begin{bmatrix}15\\6\\1\\1\end{bmatrix}$
$=\begin{bmatrix}450\\180\\30\\30\end{bmatrix}$
View full question & answer→Question 441 Mark
Compute the following:
$\begin{bmatrix}-1 & 4&-6 \\8 & 5&16\\ 2&8&5\end{bmatrix}+\begin{bmatrix}12 & 7&6 \\8 & 0&5\\3&2&4 \end{bmatrix}$
Answer$\begin{bmatrix}-1&4&-6\\8&5&16\\2&8&5\end{bmatrix}+\begin{bmatrix}12&7&6\\8&0&5\\3&2&4\end{bmatrix}$
$=\begin{bmatrix}-1+12&4+7&-6+6\\8+8&5+0&16+5\\2+3&8+2&5+4\end{bmatrix}=\begin{bmatrix}11&11&0&\\16&5&21\\5&10&9\end{bmatrix} $
View full question & answer→Question 451 Mark
Construct a $4 \times 3$ matrix whose element are:
$a_{ij} = i$
AnswerHere,
$a_{11} = 1, a_{12} = 1, a_{13}= 2,$
$a_{21} = 2, a22 = 2, a_{23} = 2$
$a_{31} = 3, a_{32} = 3, a_{33} = 3$
$a_{41} = 4, a_{42} = 4, a_{43} = 4$
Using Equation (i),
$\text{A}=\begin{bmatrix}1&1&1\\2&2&2\\3&3&3\\4&4&4\end{bmatrix}$
View full question & answer→Question 461 Mark
Compute the following:
$\begin{bmatrix}\text{a}^2+\text{b}^2 & \text{b}^2+\text{c}^2 \\ \text{a}^2+\text{c}^2 & \text{a}^2+\text{b}^2 \end{bmatrix}+\begin{bmatrix}2\text{ab} & 2\text{bc} \\-2\text{ac} &-2\text{ab}\end{bmatrix}$
Answer$\begin{bmatrix}\text{a}^2+\text{b}^2 &\text{b}^2+\text{c}^2\\ \text{a}^2+\text{c}^2&\text{a}^{2}+\text{b}^2\end{bmatrix}+\begin{bmatrix}\text{2ab } &\text{2bc}\\ \text{-2ac}&\text{-2ab}\end{bmatrix}$
$=\begin{bmatrix}\text{a}^2+\text{b}^2+2\text{ab} &\text{b}^2+\text{c}^2+2\text{bc}\\ \text{a}^2+\text{c}^2-2\text{ac}&\text{a}^{2}+\text{b}^2-\text{2ab}\end{bmatrix}$
$=\begin{bmatrix}(\text{a + b})^2 &(\text{b + c})^2\\ (\text{a - c})^2&(\text{a - b})^2\end{bmatrix}$
View full question & answer→Question 471 Mark
Construct a 4 × 3 matrix whose element are:
$\text{a}_\text{ij}=\frac{\text{i}-\text{j}}{\text{i}+\text{j}}$
AnswerHere,
$\text{a}_{11}=\frac{1-1}{1+1}=\frac{0}{2},\text{ a}_{12}=\frac{1-2}{1+2}=\frac{-1}{3},$ $\text{a}_{13}=\frac{1-3}{1+3}=\frac{-2}{4}=\frac{-1}{2}$
$\text{a}_{21}=\frac{2-1}{2+1}=\frac{1}{3},\text{ a}_{22}=\frac{2-2}{2-2}=\frac{0}{0}=0,$ $\text{a}_{23}=\frac{2-3}{2+3}=\frac{-1}{5}$
$\text{a}_{31}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2},\text{ a}_{32}=\frac{3-2}{3+2}=\frac{1}{5},$ $\text{a}_{33}=\frac{3-3}{3+3}=\frac{0}{6}=0$
$\text{a}_{41}=\frac{4-1}{4+1}=\frac{3}{5},\text{a}_{42}=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}$ and $\text{a}_{43}=\frac{4-3}{4+3}=\frac{1}{7}$
So, the required matrix is $\begin{bmatrix}0&\frac{-1}{3}&\frac{-1}{2}\\\frac {1}{3}&0&\frac{-1}{5}\\\frac{1}{2}&\frac{1}{5}&0\\ \frac{3}{5}&\frac{1}{3}&\frac{1}{7}\end{bmatrix}.$
View full question & answer→Question 481 Mark
Let $\text{A} = \begin{bmatrix}2&4\\3&2\end{bmatrix}, \text{B} = \begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{C} = \begin{bmatrix}-2&5\\3&4\end{bmatrix}.$Find each of the following:$\text{3A - C}$
Answer$\text {3A - C}=3\begin{bmatrix}2&4\\3&2 \end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}=\begin{bmatrix}6& 12\\ 9&6 \end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}$
$=\begin{bmatrix}6+2&12-5\\9-3&6-4\end{bmatrix}=\begin{bmatrix}8&7\\6&2 \end{bmatrix}$
View full question & answer→Question 491 Mark
Find the transpose of each of the following matrices:$\begin{bmatrix}1&-1\\2&3\end{bmatrix}$
Answer$\text{Let}\ \text{A}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}, \text{then}\ \text{A}^ \text{T}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}$
View full question & answer→Question 501 Mark
A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2800 as interest. However, if trust had interchanged money in bonds, they would have got ₹ 100 less as interest. Using matrix method, find the amount invested by the trust.
AnswerLet Rs. x be invested in the first bond and Rs. y be invested in the second bond.
Let A be the investment matrix and B be the interest per rupee matrix. Then,
$\text{A}=\begin{bmatrix}\text{x}&\text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}\frac{10}{100}\\\frac{12}{100}\end{bmatrix}$
Total annual interest $=\text{AB}=\begin{bmatrix}\text{x}&\text{y}\end{bmatrix}\begin{bmatrix}\frac{10}{100}\\\frac{12}{100}\end{bmatrix}=\frac{10\text{x}}{100}+\frac{12\text{y}}{100}$
$\therefore\ \frac{10\text{x}}{100}+\frac{12\text{y}}{100}=2800$
$\Rightarrow10\text{x}+12\text{y}=280000\ ...(1)$
If the rates of interest had been interchanged, then the total interest earned is Rs. 100 less than the previous interest.
$\therefore\ \frac{12\text{x}}{100}+\frac{10\text{y}}{100}=2700$
$\Rightarrow12\text{x}+10\text{y}=270000\ ...(2)$
The system of equations (1) and (2) can be expressed as
PX = Q, where $\text{P}=\begin{bmatrix}10&12\\12&10\end{bmatrix},\ \text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix},\ \text{ Q}=\begin{bmatrix}280000\\270000\end{bmatrix}$
$\big|\text{P}\big|=\begin{vmatrix}10&12\\12&10\end{vmatrix}=100-144=-44\neq0$
Thus, P is invertible.
$\therefore\ \text{X}=\text{P}^{-1}\text{Q}$
$\Rightarrow\text{X}=\frac{\text{adj }\text{P}}{\big|\text{P}\big|}\text{Q}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{(-44)}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}^\text{T}\begin{bmatrix}280000\\270000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{(-44)}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}\begin{bmatrix}280000\\270000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}\frac{2800000-3240000}{-44}\\\frac{-3360000+2700000}{-44}\end{bmatrix}=\begin{bmatrix}10000\\15000\end{bmatrix}$
$\Rightarrow\text{x}=10000$ and $\text{y}=15000$
Therefore, Rs. 10,000 be invested in the first bond and Rs. 15,000 be invested in the second bond.
View full question & answer→Question 511 Mark
In the matrix $\text{A}=\begin{bmatrix}2&5 &19 &-7\\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt{3} & 1 &-5 &17\\\end{bmatrix} $, write:
- The order of the matrix.
- The number of elements.
- write the elements $a_{13,}a_{21,}a_{24,}a_{23.}$
Answer
- There are 3 horizontal lines (rows) and 4 vertical lines (columns)in the given matrix A.
Therefore, Order of the matrix is 3 × 4.
- The number of elements in the matrix A is 3 × 4 = 12.
- $\text a_{13} \rightarrow$Element in first row and third column = 19
$\text a_{21}\rightarrow$Element in second row and first coiumn = 35
$\text{a}_{33}\rightarrow$Element in third row and third column = -5
$\text{a}_{24}\rightarrow $Element in second row and fourth column = 12
$\text{ a}_{23}\rightarrow$Element in second row and third column = $\frac{5}{2}$ View full question & answer→Question 521 Mark
Let $\text{A} = \begin{bmatrix}2&4\\3&2\end{bmatrix}, \text{B} = \begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{C} = \begin{bmatrix}-2&5\\3&4\end{bmatrix}.$Find each of the following:
$\text{BA}$
Answer$\text {BA}=\begin{bmatrix}1&3\\-2&5 \end{bmatrix}\begin{bmatrix}2&4\\3&2\end{bmatrix}=\begin{bmatrix}1(2)+3(3)& 1(4)+3(2)\\ (-2)2+5(3)&(-2)4+5(2) \end{bmatrix}=\begin{bmatrix}11&10\\11&2\end{bmatrix} $
View full question & answer→Question 531 Mark
Given an example of:
A triangular matrix.
Answer$\begin{bmatrix}1&2&3\\0&5&4\\0&0&6\end{bmatrix}$
Here, all elements below the main diagonal in upper triangular matrix are zero.
$\begin{bmatrix}1&0&0\\2&6&0\\3&4&5\end{bmatrix}$
Here, all elements above the main diagonal in lower triangular matrix are zero.
View full question & answer→Question 541 Mark
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}$$=\begin{bmatrix}\text{a(a)}+\text{b(b)}&\text{a(-b)}+\text{b(a)}\\ -\text{b(a)}+\text{a(b)}&\text{(-b)(-b)}+\text{a(a)}\end{bmatrix}$
$=\begin{bmatrix}\text{a}^2+\text{b}^2&0\\0&\text{b}^2+\text{a} ^2\end{bmatrix}$
View full question & answer→Question 551 Mark
Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$a_{ij} = j$
AnswerHere, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$a_{11} = 1, a_{12} = 2, a_{13} = 3, a_{14} = 4$
$a_{21} = 1, a_{22} = 2, a_{23} = 3, a_{24} = 4$
$a_{31} = 1, a_{32} = 2, a_{33} = 3, a_{34} = 4$
Using Equation (i),
$\text{A}=\begin{bmatrix}1&2&3&4\\1&2&3&4\\1&2&3&4\end{bmatrix}$
View full question & answer→Question 561 Mark
Given an example of:
A diagonal matrix which is not scalar.
Answer$\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}$
For a diagonal matrix which is not scalar, all elements except those in the leading diagonal should be zero and the elements in the diagonal should not be equal.
View full question & answer→Question 571 Mark
Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$a_{ij} = 2i$
AnswerHere, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$a_{11} = 2(1) = 2, a_{12} = 2(1) = 2, a_{13} = 2(1) = 2, a_{14} = 2(1) = 2$
$a_{21} = 2(2) = 4, a_{22} = 2(2) = 4, a_{23} = 2(2) = 4, a_{24} = 2(2) = 4$
$a_{31} = 2(3) = 6, a_{32} = 2(3) = 6, a_{33} = 2(3) = 6$ and $a_{34} = 2(3) = 6$
So, the required matrix is $\begin{bmatrix}2&2&2&2\\4&4&4&4\\6&6&6&6\end{bmatrix}.$
View full question & answer→Question 581 Mark
If $\text{A}[\text{a}_{\text{ij}}]=\begin{bmatrix}2&3&-5\\1&4&9\\0&7&-2\end{bmatrix}$ and $\text{B}=[\text{b}_\text{ij}]=\begin{bmatrix}2&-1\\-3&4\\1&-2\end{bmatrix}$
Then find $a_{11} + b_{11} + a_{22}b_{22}$
Answer$a_{11} b_{11} + a_{22}b_{22} = (2)(2) + (4)(4) = 4 + 16 = 20$
Hence, $a_{11}b_{11} + a_{22}b_{22} = 20$
View full question & answer→Question 591 Mark
Find the transpose of each of the following matrices:$\begin{bmatrix}-1&5&6\\ \sqrt{3}&5&6\\2&3&-1\end{bmatrix}$
Answer$\text{Let}\ \text{A}=\begin{bmatrix}-1&5&6\\ \sqrt{3}&5&6\\2&3&-1\end{bmatrix},\text{then}\ \text{A}^\text{T}=\begin{bmatrix}-1&\sqrt{3}&2\\5&5&3\\6&6&-1\end{bmatrix} $
View full question & answer→Question 601 Mark
Find the transpose of each of the following matrices:$\begin{bmatrix}5\\ \frac{1}{2}\\-1\end{bmatrix}$
Answer$\text{Let}\text{A}=\begin{bmatrix}5\\ \frac{1}{2}\\-1\end{bmatrix},\text{then}\ \text{A}^\text{T}=\begin{bmatrix}5&\frac{1}{2}&-1\end{bmatrix} $
View full question & answer→Question 611 Mark
If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.
$2\text{X}+3\text{Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix},\ 3\text{X}+2\text{Y}\begin{bmatrix}-2&2\\1&-5\end{bmatrix}$
AnswerWe have,
$3(2\text{X}+3\text{Y})-2(3\text{X}+2\text{Y})=3\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\begin{bmatrix}-2&2\\1&-5\end{bmatrix}$
$\Rightarrow6\text{X}+9\text{Y}-6\text{X}-4\text{Y}=\begin{bmatrix}6&9\\12&0\end{bmatrix}+\begin{bmatrix}4&-4\\-2&10\end{bmatrix}$
$\Rightarrow5\text{Y}=\begin{bmatrix}6+4&9-4\\12-2&0+10\end{bmatrix}$
$\Rightarrow\text{Y}=\frac{1}{5}\begin{bmatrix}10&5\\10&10\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}2&1\\2&2\end{bmatrix}\ \dots(1)$
Also,
$2(2\text{X}+3\text{Y})-3(3\text{X}+2\text{Y})=2\begin{bmatrix}2&3\\4&0\end{bmatrix}-3\begin{bmatrix}-2&2\\1&-5\end{bmatrix}$
$\Rightarrow4\text{X}+6\text{Y}-9\text{X}-6\text{Y}=\begin{bmatrix}4&6\\8&0\end{bmatrix}+\begin{bmatrix}6&-6\\-3&15\end{bmatrix}$
$\Rightarrow-5\text{X}=\begin{bmatrix}6+4&6-6\\8-3&0+15\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{-5}\begin{bmatrix}10&0\\5&15\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix}\ \dots(2)$
From (1) and (2), we get
$\text{X}=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}2&1\\2&2\end{bmatrix}$
View full question & answer→Question 621 Mark
Construct a 4 × 3 matrix whose element are:
$\text{a}_\text{ij}=2\text{i}+\frac{\text{i}}{\text{j}}$
AnswerHere,
$\text{a}_{11}=2(1)+\frac{1}{1}=\frac{2+1}{1}=\frac{3}{1}=3,$ $\text{a}_{12}=2(1)+\frac{1}{3}=\frac{4+1}{2}=\frac{5}{2},$ $\text{a}_{13}=2(1)+\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}$
$\text{a}_{21}=2(2)+\frac{2}{1}=\frac{4+2}{1}=\frac{6}{1}=6,$ $\text{a}_{22}=2(2)+\frac{2}{2}=\frac{8+2}{2}=\frac{10}{2}=5,$ $\text{a}_{23}=2(2)+\frac{2}{3}=\frac{12+2}{3}=\frac{14}{3}$
$\text{a}_{31}=2(3)+\frac{3}{1}=\frac{6+3}{1}=\frac{9}{1}=9,$ $\text{a}_{32}=2(3)+\frac{3}{2}=\frac{12+3}{2}=\frac{15}{2},$ $\text{a}_{33}=2(3)+\frac{3}{5}=\frac{18+3}{3}=\frac{21}{3}=7$
$\text{a}_{41}=2(4)+\frac{4}{1}=\frac{8+4}{1}=\frac{12}{1}=12,$ $\text{a}_{42}=2(4)+\frac{4}{2}=\frac{16+4}{2}=\frac{20}{2}=10$ and $\text{a}_{43}=2(4)+\frac{4}{3}=\frac{24+4}{3}=\frac{28}{3}$
So, the required matrix is $\begin{bmatrix}3&\frac{5}{2}&\frac{7}{3}\\6&5&\frac{14}{3}\\9&\frac{15}{2}&7\\12&10&\frac{28}{3}\end{bmatrix}.$
View full question & answer→Question 631 Mark
Compute the following:
$\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}=\begin{bmatrix}\text{a+a}& \text{b+b}\\ \text{-b+b}& \text{a+a}\end{bmatrix}=\begin{bmatrix}2\text{a}&2\text{b}\\0& 2\text{a}\end{bmatrix}$
View full question & answer→Question 641 Mark
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
AnswerLet the number of books as a 1 x 3 matrix B = $\begin{bmatrix}10\text { dozen}& 8 \text{ dozen}&10\text{ dozen}\\10 \times 12 = 120&8\times 12 = 96& 10 \times12= 120 \end{bmatrix}$
Let the selling prices of each book as a 3 x 1 matrix S = $\begin{bmatrix}80\\60\\40\end{bmatrix}$
$\therefore$ Total amount received by selling all books = BS = $\begin{bmatrix}120&96&120\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}$
= [120(80) + 96(60) + 120(40)] = [9600 + 5760 + 4800] = [20160]
Therefore, Total amount received by selling all the books = ₹ 20160
View full question & answer→Question 651 Mark
Construct a 2 × 2 matrix, A = $[\text{a}_{\text {ij}}]$, whose elements are given by:$\text a_{\text {ij}}=\frac{\text i}{\text j} $
AnswerA = $[\text a_{\text {ij}}]$ is 2 × 2 matrix where $\text a_{\text {ij}}$ = $\frac{\text {i}}{\text{j}}$$\therefore\ \text a_{\text {11}}=\frac{\text 1}{\text 1}=1 $, $\text a_{\text {12}}=\frac{\text 1}{\text 2} $
$\text a_{\text {21}}=\frac{\text 2}{\text 1}=2 $, $\text a_{22}=\frac{2}{2}=1 $
$\therefore\ \text A=\begin{bmatrix} 1& \frac{1}{2} \\2& 1 \end{bmatrix}$
View full question & answer→Question 661 Mark
Given an example of:
A row matrix which is also a column matrix,
Answer[6]
This is a matrix that contains only one element.
View full question & answer→Question 671 Mark
Compute the following:
$\begin{bmatrix}\cos^2\text{x} & \sin^2\text{x} \\ \sin^2 \text{x} & \cos^2 \text{x} \end{bmatrix}+\begin{bmatrix}\sin^2\text{x}&\cos^2\text{x}\\ \cos^2\text{x}&\sin^2\text{x}\end{bmatrix} $
Answer$\begin{bmatrix}\cos^2\text{x} & \sin^2\text{x} \\ \sin^2 \text{x} & \cos^2 \text{x} \end{bmatrix}+\begin{bmatrix}\sin^2\text{x}&\cos^2\text{x}\\ \cos^2\text{x}&\sin^2\text{x}\end{bmatrix} $
$=\begin{bmatrix}\text{cos}^2\text{x}+\sin^2\text{x}&\sin^2\text{x}+\cos^2\text{x}\\ \sin^2\text{x}+\cos^2\text{x}&\cos^2\text{x}+\sin^2\text{x}\end{bmatrix}=\begin{bmatrix}1&1\\1&1\end{bmatrix} $
View full question & answer→Question 681 Mark
Compute the following sums:
$\begin{bmatrix}2&1&3\\0&3&5\\-1&2&5\end{bmatrix}+\begin{bmatrix}1&-2&3\\2&6&1\\0&-3&1\end{bmatrix}$
Answer$\begin{bmatrix}2&1&3\\0&3&5\\-1&2&5\end{bmatrix}+\begin{bmatrix}1&-2&3\\2&6&1\\0&-3&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+1&1-2&3+3\\0+2&3+6&5+1\\-1+0&2-3&5+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-1&6\\2&9&6\\-1&-1&6\end{bmatrix}$
View full question & answer→Question 691 Mark
Let $\text{A} = \begin{bmatrix}2&4\\3&2\end{bmatrix}, \text{B} = \begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{C} = \begin{bmatrix}-2&5\\3&4\end{bmatrix}.$ Find each of the following:
$\text{AB}$
Answer$\text {AB}=\begin{bmatrix}2&4\\3&2 \end{bmatrix}\begin{bmatrix}1&3\\-2&5\end{bmatrix}=\begin{bmatrix}2(1)+4(-2)& 2(3)+4(5)\\ 3(1)+2(-2)&3(3)+2(5) \end{bmatrix}=\begin{bmatrix}-6&26\\-1&19\end{bmatrix} $
View full question & answer→Question 701 Mark
Show that the matrix $\text{A}=\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}$is a symmentic matrix.
AnswerWe have:
$\text{A}'=\begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix}=-\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}=-\text{A}$
$\therefore\ \text{ A}'=-\text{A} $
Hence, A is a skew-symmetric matrix.
View full question & answer→Question 711 Mark
Let A be a matrix of order $3 \times 4$. If $R_1$ denotes the first row of A and $C_2$ denotes its second column, then determine the orders of matrices $R_1$ and $C_2$.
AnswerThe order of $R_1$ is $1 \times 4$ and the order of $C_2$ is $3 \times 1$.
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