Question 12 Marks
If $A$ is a non$-$singular symmetric matrix, write whether $A^{-1}$ is symmetric or skew$-$symmetric.
AnswerLet $A$ be an invertible symmetric matrix. Then,
$|\text{A}|\neq0\text{ and }\text{A}^\text{T}=\text{A}$
Now, $(A^{-1})^T = (A^T)^{-1}$
$\Rightarrow (A^{-1})^T = A^{-1} [\because A^T = A]$
Thus, $A^{-1}$ is symmetric matrix.
View full question & answer→Question 22 Marks
Evaluate the determinant:$ \begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}$
Answer Let $\text{A}=\begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}.$ By expanding along the first row, we have: $|\text{A}|=3\begin{vmatrix}1&-2\\3&1\end{vmatrix}+4\begin{vmatrix}1&-2\\2&1\end{vmatrix}+5\begin{vmatrix}1&1\\2&3\end{vmatrix}$$=3(1+6)+4(1+4)+5(3-2)$
$=3(7)+4(5)+5(1)$
$=21+20+5=46$
View full question & answer→Question 32 Marks
If $A$ is a square matrix of order $3$ such that $adj\ (2A) = k\ adj\ (A),$ then write the value of $k.$
AnswerFor any matrix $A$ of order $n, adj (\lambda\text{A})=\lambda^{\text{n}-1}=\lambda^{\text{n}-1} (adj\ A)$ where $\lambda$ is a constant.
Thus, for matrix $A$ of order $3,$ we have
$adj\ (2A) = 2^{3-1} (adj\ A)$
$\Rightarrow adj\ (2A) = 2^2 (adj\ A)$
$\Rightarrow adj\ (2A) = 4(adj) (A)$
$\Rightarrow k\ adj\ (A) = 4 adj\ (A) [ \because adj\ (2A) = k\ adj\ (A)]$
$\Rightarrow k = 4$
View full question & answer→Question 42 Marks
If $A$ is a non$-$singular square matrix such that $|A| = 10,$ find $|A^{-1}|.$
Answer$\big|\text{A}^{-1}\big|=\Big|\frac{1}{\text{A}}\Big|$
$=\Big|\frac{1}{\text{A}}\Big|$
$=\frac{1}{10}\ \big[\because|\text{A}|=10\text{ (Given})\big]$
Hence, $\big|\text{A}^{-1}\big|=\frac{1}{10}$
View full question & answer→Question 52 Marks
Show that points:
A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
AnswerArea of triangle ABC = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}a&b+c&1\\b&c+a&1\\c&a+b&1\end{bmatrix}\end{vmatrix}$$=\bigg|\frac{1}{2}\left[a(c+a-a-b)-(b+c)(b-c)+1\left\{b(a+b)-c(c+a)\right\}\right]\bigg|$
$=\bigg|\frac{1}{2}\left[a(c-b)-(b^2-c^2)+(ab+b^2-c^2-ac)\right]\bigg|$
$=\bigg|\frac{1}{2}(ac-ab-b^2+c^2+ab+b^2-c^2-ac)\bigg|$
$=\begin{vmatrix}\frac{1}{2}\times0\end{vmatrix}=0$
Therefore, points A, B and C are collinear.
View full question & answer→Question 62 Marks
If $A$ is an invertible matrix such that $|A^{-1}| = 2,$ find the value of $|A|.$
Answer$\big|\text{A}^{-1}\big|=2$
$\Big|\frac{1}{\text{A}}\Big|=2$
$\frac{1}{|\text{A}|}=2$
$\therefore|\text{A}|=\frac{1}{2}$
Hence, $|\text{A}|=\frac{1}{2}$
View full question & answer→Question 72 Marks
If $\text{A}=\begin{bmatrix}2&4\\4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{n}\\1\end{bmatrix},\text{B}=\begin{bmatrix}8\\11\end{bmatrix}$and AX = B, then find n.
AnswerHere,
$\begin{bmatrix}2&4\\4&3\end{bmatrix}\begin{bmatrix}\text{n}\\1\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{n}+4\\4\text{n}+3\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}$
$\Rightarrow2\text{n}+4=8$
$\Rightarrow2\text{n}=4$
$\Rightarrow\text{n}=2$
View full question & answer→Question 82 Marks
If $\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0,$ find x.
Answer$\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0$
$\Rightarrow9(2\text{x}+5)-3(5\text{x}+2)=0$
$\Rightarrow18\text{x}+45-15\text{x}-6=0$
$\Rightarrow3\text{x}+39=0$
$\Rightarrow3\text{x}=-39$
$\Rightarrow\text{x}=\frac{-39}{3}$
$\Rightarrow\text{x}=-13$
View full question & answer→Question 92 Marks
If $A$ is a square matrix of order $3$ with determinant $4,$ then write the value of $|-A|.$
Answer$|A| = 4$
Here,
Order of the matrix $(n) = 3$
Using properties of matrices, we get
$|kA| = k^n|A| [$For a square matrix of order $n$ and constant $k]$
$\Rightarrow |-A| $
$= (-1)^3 |A| $
$= (-1) \times 4 $
$= -4$
View full question & answer→Question 102 Marks
Examine the consistency of the system of equations:
2x - y = 5
x + y = 4
AnswerMatrix form of given equations is AX = B $\Rightarrow \ \begin{bmatrix}2&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\4\end{bmatrix}$
$\therefore\ \text{A}=\begin{bmatrix}2&-1\\1&1\end{bmatrix}\text{and B}=\begin{bmatrix}5\\4\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}2&-1\\1&1\end{vmatrix}=2-(-1)=3\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 112 Marks
If $A$ is a square matrix of order $3$ such that $|$adj $A| = 64,$ find $|A|.$
AnswerFor any square matrix of order $n,$
$|$adj $A| = |A|^{n-1}$
$\Rightarrow 64 = |A|^2 [\because |$adj $A| = 64]$
$\Rightarrow|\text{A}|=\pm8$
View full question & answer→Question 122 Marks
If $\text{x}\in\text{N}$ and $\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8,$ then find the value of x.
Answer$\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8$
$\Rightarrow(\text{x}+3)2\text{x}-(-2)(-3\text{x})=8$
$\Rightarrow2\text{x}^2+6\text{x}-6\text{x}=8$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2-4=0$
$\Rightarrow\text{x}^2=4$
$\Rightarrow\text{x}=2$ $[\text{x}\neq-2\ \because\text{x}\in\text{N}]$
View full question & answer→Question 132 Marks
Without expanding, show that the values of the following determinant are zero$:$
$\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}0&2&7\\12&3&5\\16&4&3 \end{vmatrix} [$Applying $C_1 \rightarrow C_1 - 4C_2]$
$\Rightarrow\triangle=0$
View full question & answer→Question 142 Marks
Let $A = [a_{ij}]$ be a square matrix of order $3 \times 3$ and $C_{ij}$ denote cofactor of $a_{ij}$ in $A.$ if $|A| = 5,$ write the value of $a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33.}$
AnswerIf $A = a_{ij}$ is a square matrix of order $n$ and $C_{ij}$ is a cofactor of $a_{ij},$ then
$\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given, $|A| = 5$ and matrix $A$ is of order $3 \times 3$
Since $a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}$ represent expansion of $A$ along third column, we get
$\Rightarrow a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} = |A| = 5$
$\Rightarrow a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} = 5$
View full question & answer→Question 152 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x + 2y = 5,
3x + 6y = 15
AnswerUsing the equations, we get
$\text{D}=\begin{vmatrix}1&2\\3&6\end{vmatrix}=6-6=0$
$\text{D}_1=\begin{vmatrix}5&2\\15&6\end{vmatrix}=30-30=0$
$\text{D}_2=\begin{vmatrix}1&5\\3&15\end{vmatrix}=15-15=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2$
Hence, the system of linear equation has infinitely many solutions.
View full question & answer→Question 162 Marks
If $\text{A}=\begin{bmatrix}1&2\\4&2\end{bmatrix}$, then show that $\left|2\text{A}\right|=4\left|\text{A}\right|$
AnswerThe given matrix is $\text{A}=\begin{bmatrix}1&2\\4&2\end{bmatrix}$
$\therefore2\text{A}=2\begin{bmatrix}1&2\\4&2\end{bmatrix}=\begin{bmatrix}2&4\\8&4\end{bmatrix}$
$\therefore\text{L.H.S.}=|2\text{A}|=\begin{vmatrix}2&4\\8&4\end{vmatrix}=2\times4-4\times8=8-32=-24$
$\text{Now},|\text{A}|=\begin{vmatrix}1&2\\4&2\end{vmatrix}=2-8=-6$
$\therefore\text{R.H.S.}=4|\text{A}|=4\times\left(-6\right)=-24$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 172 Marks
Write the adjoint of the matrix $\text{A}=\begin{bmatrix} -3 & 4 \\ 7 & -2 \end{bmatrix}.$
AnswerLet $C_{ij}$ be a cofactor of $a_{ij}$ in $A.$
Now,
$C_{11} = -2$
$C_{12} = -7$
$C_{21} = -4$
$C_{22} = -3$
$\therefore\ \text{adj A}=\begin{bmatrix} -2 & -7 \\ -4 & -3 \end{bmatrix}^\text{T}$
$=\begin{bmatrix} -2 & -4 \\ -7 & -3 \end{bmatrix}$
View full question & answer→Question 182 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
$=1\begin{vmatrix}6&10\\11&38\end{vmatrix}-3\begin{vmatrix}2&10\\31&38\end{vmatrix}+5\begin{vmatrix}2&6\\31&11\end{vmatrix}$
$=(228-110)-3(76-310)+5(22-186)$
$=1(118)-3(-234)+5(-164)$
$=118+702-820$
$=0$
View full question & answer→Question 192 Marks
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}1&bc&a(b+c)\\1&ca&b(c+a)\\1&ab&c(a+b)\end{vmatrix}=0$
Answer$\text{Given}:\ \begin{vmatrix}1&bc&a(b+c)\\1&ca&b(c+a)\\1&ab&c(a+b)\end{vmatrix}=\begin{vmatrix}1&bc&ab+ac\\1&ca&bc+ba\\1&ab&ca+cb\end{vmatrix}$
$\text{Operating}\ \text{C}_3\rightarrow\text{C}_3+\text{C}_2\ \begin{vmatrix}1&bc&ab+bc+ac\\1&ca&ab+bc+ca\\1&ab&ab+bc+ca\end{vmatrix}$
$=(ab+bc+ca)\begin{vmatrix}1&bc&1\\1&ca&1\\1&ab&1\end{vmatrix}$
$=(ab+bc+ca)(0)=0$ $\left[\because\text{two columns are identical Proved.}\right]$
View full question & answer→Question 202 Marks
If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.
AnswerSince A & B are square matrix of the same order, by the property of determinants we get
|AB| = |A| × |B|
|A| = 3, AB = I
⇒ |AB| = 1
⇒ |A| × |B| = 1
⇒ 3 × |B| = 1
$\Rightarrow|\text{B}|=\frac{1}{3}$
View full question & answer→Question 212 Marks
Write the value of $a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23}.$
AnswerWe know that in a square matrix of order $n,$ the sum of the products of elements of a row $($or a column$)$ with the cofactors of the corresponding elements of some other row $($or column$)$ is zero. Therefore,
$A = [a_{ij}]$ is a square matrix of order $n.$
$\Rightarrow\sum\limits_{\text{n}}^{\text{i}=1}\text{a}_{\text{ij}}\text{C}_\text{kj}=0$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ik}=0$
$\Rightarrow a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$
$[$Since the elements are of first row and the cofactors are of elements of second row$]$
$\Rightarrow a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$
View full question & answer→Question 222 Marks
Find the value of the determinant $\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
$\triangle=\begin{vmatrix}4200&1\\4205&1\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C_1]$
$\triangle=4200 - 4202$
$\triangle=-2$
View full question & answer→Question 232 Marks
If $A = [A_{ij}]$ is a $3 \times 3$ scalar matrix such that $a_{11} = 2,$ then write the value of $|A|.$
AnswerA scalar matrix is a digonal matrix, in which all the diagonal elements are equal to a given scalar number.
Given, $A = [a_{ij}]$ is $3 \times 3$ matrix, where $a_{11} = 2$
$\Rightarrow\text{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$\Rightarrow\text{A}=\begin{vmatrix}2&0&0\\0&2&0\\0&0&2\end{vmatrix}$
$\Rightarrow|\text{A}|=2\times\begin{vmatrix}2&0\\0&2\end{vmatrix} [$Expanding along $C_1]$
$\Rightarrow|\text{A}|=2\times2\times2$
$\Rightarrow|\text{A}|=8$
View full question & answer→Question 242 Marks
Evaluate the following determinant:
$\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
$=\text{a}\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}-\text{h}\begin{vmatrix}\text{h}&\text{f}\\\text{g}&\text{c} \end{vmatrix}+\text{g}\begin{vmatrix}\text{h}&\text{b}\\\text{g}&\text{f} \end{vmatrix}$
$=\big(\text{bc}-\text{f}^2\big)-\text{h}\big(\text{hc}-\text{fg}\big)+\text{g}\big(\text{hf}-\text{gb}\big)$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{g}^2\text{b}$
$=\text{abc}+2\text{fgh}-\text{af}^2-\text{ch}^2-\text{bg}^2$
View full question & answer→Question 252 Marks
If $A$ is a square matrix of order $3$ such that $|A| = 5,$ write the value of $|$adj $A|.$
AnswerFor any square matrix of order $n,$
$|$adj $A| = |A|^{n-1}$
$\Rightarrow |$adj $A| $
$= |A|^2$
$= 5^2$
$= 25$
View full question & answer→Question 262 Marks
If $\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix},$ find |AB|.
Answer$\Rightarrow\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$
⇒ |A| = -1 - 6 = -7
$\Rightarrow\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
⇒ |B| = -2 + 12 = 10
If A and B are square matrix of the same order, then |AB| = |A| |B|.
⇒ |AB| = |A| |B|
⇒ |AB| = -7 × 10 = -70
View full question & answer→Question 272 Marks
If $\text{A}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},\text{B}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ find adj $(AB).$
Answer$\text{A}\times\text{B}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$
$A \times B$ is a non$-$singular matrix. Therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $a_{ij}$ in $A.$
The cofactors of element $A$ are given by
$C_{11} = d$
$C_{12} = -c$
$C_{21} = -b$
$C_{22} = a$
$\therefore\ \text{adj A}=\begin{bmatrix} \text{d} & -\text{c} \\ -\text{b} & \text{a} \end{bmatrix}^\text{T}$
$=\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
View full question & answer→Question 282 Marks
Find the maximum value of $\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ we get
$\triangle=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\0&0&\cos\theta \end{vmatrix}$
$=\sin\theta\cos\theta$
$=\frac{\sin2\theta}{2}$
We know that $-1\leq\sin2\theta\leq1$
$\therefore$ Maximum value of $\triangle=\frac{1}{2}\times1=\frac{1}{2}$
View full question & answer→Question 292 Marks
Write the value of $\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
Answer$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
$=\text{a}^2-\text{iab}+\text{iab}-\text{i}^2\text{b}^2-(-\text{c}^2-\text{icd}+\text{icd}+\text{i}^2\text{d}^2)$
$=\text{a}^2-\text{i}^2\text{b}^2+\text{c}^2-\text{i}^2\text{d}^2$
Here, $\text{i}^2=-1$
$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}=\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2$
View full question & answer→Question 302 Marks
If $w$ is an imaginary cube root of unity, find the value of $\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
Answer$\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
$=\begin{vmatrix}1+\text{w}+\text{w}^2&\text{w}&\text{w}^2\\\text{w}+\text{w}^2+1&\text{w}^2&1\\\text{w}^2+1+\text{w}&1&\text{w}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
$=\begin{vmatrix}0&\text{w}&\text{w}^2\\0&\text{w}^2&1\\0&1&\text{w}\end{vmatrix} [\because 1 + w + w^2 = 0, w$ is the imaginary cube root of unity$]$
View full question & answer→Question 312 Marks
Write the value of the determinant $\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
Answer$\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
$=\begin{vmatrix}2&3&4\\5&6&8\\2&3&4\end{vmatrix} [$Taking $2x$ common from $R_3]$
$=0$
View full question & answer→Question 322 Marks
Find area of the triangle with vertices at the point given: (2, 7), (1, 1), (10, 8)
AnswerArea of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}2&7&1\\1&1&1\\10&8&1\end{bmatrix}\end{vmatrix}$$=\bigg|\frac{1}{2}\left[2(1-8)-7(1-10)+1(8-10)\right]\bigg|$
$=\bigg|\frac{1}{2}\left[2(-7)-7(-9)-2\right]\bigg|$
$=\bigg|\frac{1}{2}(-14+63-2)\bigg|=\bigg|\frac{1}{2}(63-16)\bigg|$
$=\begin{vmatrix}\frac{47}{2}\end{vmatrix}=\frac{47}{2}\text{sq.units}$
View full question & answer→Question 332 Marks
Find the value of x from the following: $\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
Answer$\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
$\Rightarrow2\text{x}^2-8=0$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2=\frac{8}{2}=4$
$\Rightarrow\text{x}=\sqrt{4}=\pm2$
View full question & answer→Question 342 Marks
If $A$ is a square matrix, then write the matrix adj $(A^T) − ($adj $A)^T.$
AnswerIn a non$-$singular matrix, adj $A^T = ($adj $A)^T.$
$\Rightarrow ($adj $A^T) - ($adj $A)^T$
$=$ Null matrix
View full question & answer→Question 352 Marks
Evaluate: $\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
Answer$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
$=\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ$
$=\cos(15^\circ+75^\circ)$ $\big[\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}=\cos(\text{A}+\text{B})\big]$
$=\cos90^\circ$
$=0$
$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}=0$
View full question & answer→Question 362 Marks
Find the value of x, if:
if $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10,$ find the value of x.
AnswerGiven, $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10$
$\Rightarrow12\text{x}-14=10$
$\Rightarrow12\text{x}=24$
$\Rightarrow\text{x}=2$
View full question & answer→Question 372 Marks
If the matrix $\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$ is a singular, find the value of x.
AnswerA matrix is said to be singular if its determinant is zero. since the given matrix is singular, we get
$\text{A}=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=0$
$\Rightarrow5\text{x}+20=0$ [Expanding]
$\Rightarrow\text{x}=-\frac{20}{5}$
$\Rightarrow\text{x}=-4$
View full question & answer→Question 382 Marks
Examine the consistency of the system of equations:
x + 3y = 5
2x + 6y = 8
AnswerMatrix from of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&3\\2&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\8\end{bmatrix}$
$\therefore\ \text{A}=\begin{bmatrix}1&3\\2&6\end{bmatrix}\text{and B}=\begin{bmatrix}5\\8\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&3\\2&6\end{vmatrix}=6-6=0$
$\text{Now},\ \text{(adj. A)B}=\begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix}=\begin{bmatrix}33-24\\-10+8\end{bmatrix}=\begin{bmatrix}6\\-2\end{bmatrix}\neq0$
Therefore, given equations are inconsistent, i.e., have no common solution.
View full question & answer→Question 392 Marks
Find area of the triangle with vertices at the point given:
(1, 0), (6, 0), (4, 3)
AnswerArea of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}1&0&1\\6&0&1\\4&3&1\end{bmatrix}\end{vmatrix}$$=\bigg|\frac{1}{2}\left[1(0-3)-0(6-4)+1(18-0)\right]\bigg|$
$=\bigg|\frac{1}{2}(-3+18)\bigg|=\bigg|\frac{15}{2}\bigg|=\frac{15}{2}\text{sq.units}$
View full question & answer→Question 402 Marks
On expanding by first row, the value of the determinant of $3 \times 3$ square matrix $A = [a_{ij}]$ is $a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13},$ where$ [C_{ij}]$ is the cofactor of $a_{ij}$ in $A.$ Write the expression for its value on expanding by second column.
AnswerIf $A = a_{ij}$ is a square matrix of order $n,$ then the sum of the products of elements of a row $($or a column$)$ with their cofactors is always equal to det $(A). $
Therefore, $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given,$ |A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} [$Expanding along $R_1]$
Now, $|A| = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}$
$[$Expanding along $R_2] [a_{12}, a_{22}$ and $a_{32}$ are elements of $C_3]$
View full question & answer→Question 412 Marks
State whether the matrix $\begin{vmatrix}2&3\\6&4\end{vmatrix}$ is singular or non-singular.
AnswerLet $\triangle=\begin{vmatrix}2&3\\6&4\end{vmatrix}$
= 2 × 4 - 6 × 3
= 18 - 18 = -10
A matrix is said to be singular if its determinant is equal to zero. Since $\triangle=-10\neq0,$ the given matrix is non-singular.
View full question & answer→Question 422 Marks
If $\text{A}=\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix},$ find $\left|\text{A}\right|.$
Answer$\text{Let A}=\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix}.$
By expanding along the first row, we have:
$\left|\text{A}\right|=1\begin{vmatrix}1&-3\\4&-9\end{vmatrix}-1\begin{vmatrix}2&-3\\5&-9\end{vmatrix}-2\begin{vmatrix}2&1\\5&4\end{vmatrix}$
$=1(-9+12)-1(-18+15)-2(8-5)$
$ =1(3)-1(-3)-2(3)$
$ =3+3-6$
$=6-6$
$=0$
View full question & answer→Question 432 Marks
Evaluate the following determinant: $\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
AnswerLet $\triangle$ be the determinant.
$\triangle=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
Applying $R_3 \rightarrow R_3 - R_2,$ we get
$\Rightarrow\triangle=\begin{vmatrix}1&4&9-4\\4&9&16-9\\9&16&25-16 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&4&5\\4&9&7\\9&16&9\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&5&5\\4&13&7\\9&25&9 \end{vmatrix} [$Applying $C_2 \rightarrow C_1 + C_2]$
$\Rightarrow\triangle=\begin{vmatrix}1&0&0\\4&-7&-13\\9&-20&-36 \end{vmatrix} [$Applying $C_2 \rightarrow 5C_1 - C_2$ and $C_3 \rightarrow 5C_1 - C_3]$
$\Rightarrow\triangle=1(7\times36-13\times20)$
$\Rightarrow\triangle=252-260=-8$
View full question & answer→Question 442 Marks
A matrix $A$ of order $3 \times 3$ has determinant $5.$ What is the value of $|3A|$?
AnswerIf $A = [a_{ij}]$ is $a$ square matrix of order $n$ and $k$ is a constant, then
$|kA| = k^n|A|$
Here,
Number of rows $= n$
$k$ is a common factor from each row of $k$
$|3A| = 3^3|A| $
$= 27 \times 5 $
$= 135 [$Given matrix is $3 \times 3$ such that $|A| = 5]$
Thus, $|3A| = 135$
View full question & answer→Question 452 Marks
Examine the consistency of the system of equations:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\4\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}1&1&1\\2&3&2\\a&a&2a\end{vmatrix}$
$\Rightarrow\ \text{|A|}=1(6a-2a)-1(4a-2a)+1(2a-3a)=4a-2a-a=a\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 462 Marks
Examine the consistency of the system of equations:
x + 2y = 2
2x + 3y = 3
AnswerMatrix form of given equation is AX = B $\Rightarrow\ \begin{bmatrix}1&2\\2&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}$
$\therefore \ \text{A}=\begin{bmatrix}1&2\\2&3\end{bmatrix}\text{and B}=\begin{bmatrix}2\\3\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&2\\2&3\end{vmatrix}\text{and B}=3-4=-1\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 472 Marks
Find the inverse of the matrix $\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}.$
Answer$\text{A}=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix}$
Cofactors of $A$ are$:$
$C_{11} = 0, C_{21} = 3$
$C_{12} = -2, C_{22} = 1$
$\text{Adj A}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$
View full question & answer→Question 482 Marks
If $|A| = 2,$ where $A$ is $2 \times 2$ matrix, find $|$adj $A|.$
AnswerFor any square matrix $A$ of order $n, |$adj $A| = |A|^{n-1}$
Given, $|A| = 2$
Here, order is $2$
$\Rightarrow |$ adj $A| = |2|^{2-1} = 2$
View full question & answer→Question 492 Marks
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}=0$
Answer$\text{Let}\ \triangle=\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}$
$=\triangle=(-1)^3\begin{vmatrix}0&-a&b\\a&0&c\\-b&-c&0\end{vmatrix}$ [Taking (-1) common from each row]
Interchanging rows and columns in the determinants on R.H.S.,
$\triangle=-\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}\ \Rightarrow\triangle=-\triangle\ \Rightarrow\triangle+\triangle=0$
$\Rightarrow 2\triangle=0\ \Rightarrow\triangle=0$ Proved.
View full question & answer→Question 502 Marks
Write the value of the determinant $\begin{vmatrix}2&3&4\\2\text{x}&3\text{x}&4\text{x}\\5&6&8\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}2&3&4\\2\text{x}&3\text{x}&4\text{x}\\5&6&8\end{vmatrix}$
$=\text{x}\begin{vmatrix}2&3&4\\2&3&4\\5&6&8\end{vmatrix} [$Taking out $x$ common from $R_2]$
$=0$
View full question & answer→