Question 12 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 22 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 $C_2 → C_2 - C_1$]
$\triangle=4200 - 4202$
$\triangle=-2$
View full question & answer→Question 32 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 42 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 52 Marks
If $\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix},$ find the value of $|\text{A}|+|\text{B}|.$
Answer$\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$$\Rightarrow|\text{A}|= 0-\text{i}^2$
$\Rightarrow|\text{A}|=-(-1)=1$
Also,
$\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\Rightarrow|\text{B}|=0-1=-1$
So,
$\Rightarrow|\text{A}|+|\text{B}|=1-1=0$
View full question & answer→Question 62 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 72 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 82 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 → C_1 - 4C_2$]
$\Rightarrow\triangle=0$
View full question & answer→Question 92 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 × 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 102 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 112 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 122 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 132 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 142 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 152 Marks
If A is a square matrix such that $|A| = 2$, write the value of $\big|\text{A}\text{A}^{\text{T}}\big|.$
AnswerIn a square matrix, $A = A^T$. Since they are of same order, $AA^T = AA^T$.
Given, A = 2
$\Rightarrow AA^T= 2^2 = 4$
View full question & answer→Question 162 Marks
A is a skew-symmetric of order 3, write the value of |A|.
AnswerWe know that if a skew symmetric matrix A is of odd order, then |A| = 0
Since the order of the given matrix is 3, |A| = 0.
View full question & answer→Question 172 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 182 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 192 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 202 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 212 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 222 Marks
Write rthe value of the determinant $\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}$
Answer$\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}=\text{p}^2-(\text{p}+1)(\text{p}-1)$
$=\text{p}^2-(\text{p}^2-1)$
$=\text{p}^2-\text{p}^2+1$
$=1$
View full question & answer→Question 232 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 → R_2 - R_1$ and $R_3 → 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 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
Prove that the determinant $\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$ is independent of θ.
Answer$\triangle=\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$
$= x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)$
$=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta$
$=-x^3-x+x(\sin^2\theta+\cos^2\theta)$
$=-x^3-x+x$
$=-x^3$ (Independent of $\theta$)
Hence, $\triangle$ is independent of $\theta.$
View full question & answer→Question 282 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 × 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 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 → 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
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 322 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 332 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 342 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 352 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 362 Marks
If A is a singular matrix, then write the value of |A|.
AnswerSince A is a singular matrix
Thus, |A| = 0
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
If $\text{A}=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix},$ write $A^{-1}$ in terms of A.
Answer$|\text{A}|=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}=-19\neq0$
A 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}= -2$
$C_{12} = -5$
$C_{21} = -3$
$C_{22} = 2$
$\text{adj A}=\begin{bmatrix} -2 & -5 \\ -3 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix},$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19} \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{19}\text{A}$
View full question & answer→Question 422 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 432 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 442 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 452 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 → 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 → C_1 + C_2$]
$\Rightarrow\triangle=\begin{vmatrix}1&0&0\\4&-7&-13\\9&-20&-36 \end{vmatrix}$ [Applying $C_2 → 5C_1 - C_2$ and $C_3 → 5C_1 - C_3$]
$\Rightarrow\triangle=1(7\times36-13\times20)$
$\Rightarrow\triangle=252-260=-8$
View full question & answer→Question 462 Marks
Evaluate the determinant:$ \begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}$
AnswerLet $\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 472 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 482 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 492 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
$ \operatorname{adj}(2 A)=2^{3-1}(\operatorname{adj} A) $
$ \Rightarrow \operatorname{adj}(2 A)=2^2(\operatorname{adj} A) $
$ \Rightarrow \operatorname{adj}(2 A)=4(\operatorname{adj})(A) $
$ \Rightarrow k \operatorname{adj}(A)=4 \operatorname{adj}(A)[\because \operatorname{adj}(2 A)=k \operatorname{adj}(A)] $
$ \Rightarrow k=4$
View full question & answer→Question 502 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 512 Marks
If $|A|=2$, where $A$ is $2 \times 2$ matrix, find $|\operatorname{adj} A|$.
AnswerFor any square matrix $A$ of order $n,|\operatorname{adj} A|=|A|^{n-1}$
Given, $|A|=2$
Here, order is 2
$\Rightarrow|\operatorname{adj} A|=|2|^{2-1}=2$
View full question & answer→Question 522 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 532 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 542 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→Question 552 Marks
write the value of the determinant $\begin{vmatrix}2&-3&5\\4&-6&10\\6&-9&15\end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}2&-3&5\\4&-6&10\\6&-9&15\end{vmatrix}$
$=\begin{vmatrix}2&-3&5\\4-4&-6+6&10-10\\6&-9&15\end{vmatrix}$ [Applying $R_2 → R_2 - 2R_1$]
$=\begin{vmatrix}2&-3&5\\0&0&0\\6&-9&15\end{vmatrix}$
$=0$
View full question & answer→Question 562 Marks
Evaluate the determinants.
- $\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}$
- $\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}$
View full question & answer→Question 572 Marks
Find the value of x, if:
$\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2\text{x}&4\\6&\text{x}\end{vmatrix}$
AnswerGiven, $\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2\text{x}&4\\6&\text{x}\end{vmatrix}$
$\Rightarrow2-20=2\text{x}^2-24$
$\Rightarrow-18=2\text{x}^2-24$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}^2=3$
$\Rightarrow\text{x}=\pm\sqrt{3}$
View full question & answer→Question 582 Marks
Evaluate $\begin{vmatrix}1&\text{x}&\text{y}\\1&\text{x}+\text{y}&\text{y}\\1&\text{x}&\text{x+y}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{x}&\text{y}\\1&\text{x}+\text{y}&\text{y}\\1&\text{x}&\text{x+y}\end{vmatrix}$
Applying $R_2 → R_2 - R_1$ and $R_3 → R_3 - R_1$, we have:
$\triangle=\begin{vmatrix}1&\text{x}&\text{y}\\0&\text{y}&0\\0&0&\text{x}\end{vmatrix}$
Expanding along $C_1$, we have:
$\triangle$ = 1(xy - 0) = xy
View full question & answer→Question 592 Marks
If $\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix},$ then write the value of x.
Answer$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
$\Rightarrow (x + 1)(x + 2) - (x - 1)(x - 3) = 12 + 1$
$\Rightarrow x^2 + 3x + 2 - x^2 + 4x - 3 = 13$
$\Rightarrow 7x - 1 = 13$
$\Rightarrow 7x = 14$
$\Rightarrow x = 2$
Hence, the value of x is $2$
View full question & answer→Question 602 Marks
Write the value of the determinant $\begin{vmatrix}\text{a}&1&\text{b}+\text{c}\\\text{b}&1&\text{c}+\text{a}\\\text{c}&1&\text{a}+\text{b}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}&1&\text{b}+\text{c}\\\text{b}&1&\text{c}+\text{a}\\\text{c}&1&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}\text{a}+\text{b}+\text{c}&1&\text{b}+\text{c}\\\text{b}+\text{c}+\text{a}&1&\text{c}+\text{a}\\\text{c}+\text{a}+\text{b}&1&\text{a}+\text{b}\end{vmatrix}$ [Applying $C_1 → C_1 + C_3$]
$=\text{a}+\text{b}+\text{c}\begin{vmatrix}1&1&\text{b}+\text{c}\\1&1&\text{c}+\text{a}\\1&1&\text{a}+\text{b}\end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})\times0$
$=0$
View full question & answer→Question 612 Marks
If $\begin{bmatrix}1&0&0\\0&\text{y}&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\-1\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\0\\1\end{bmatrix}$, find x, y and z.
AnswerHere,
$\begin{bmatrix}1&0&0\\0&\text{y}&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\-1\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\0\\1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\-\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\0\\1\end{bmatrix}$
$\therefore\ \text{x}=1,\text{y}=0\text{ and }\text{z}=1$
View full question & answer→Question 622 Marks
Find the adjoint of the following matrices: $\text{C}=\begin{bmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{adjoint C}=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$
$\text{(adjoint C)C}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$|\text{C}|=\cos^2\alpha-\sin^2\alpha$
$|\text{C}|\text{I}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$\text{C(adjoint C)}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$\therefore\ \text{(adjoint C)}=|\text{C}|\text{I}=\text{C(adjoint C)}$
Hence verified.
View full question & answer→Question 632 Marks
If $\begin{vmatrix}3\text{x}&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix},$ find the value of x.
Answer$\begin{vmatrix}3\text{x}&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix}$
⇒ 12x + 14 = 32 - 42
⇒ 12x + 14 = -10
⇒ 12x = -24
⇒ x = -2
$\therefore$ x = -2
View full question & answer→Question 642 Marks
Find the value of x, if:
$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
AnswerGiven,
$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
$\Rightarrow(\text{x}+1)(\text{x}+2)-(\text{x}-3)(\text{x}-1)=12+1$
$\Rightarrow\text{x}^2+3\text{x}+2-\text{x}^2+4\text{x}-3=13$
$\Rightarrow7\text{x}-1=13$
$\Rightarrow7\text{x}=14$
$\Rightarrow\text{x}=2$
View full question & answer→Question 652 Marks
Evaluate the following determinant:
$\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$
AnswerConsider the determinant
$\triangle=\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$
Applying $C_1 → C_1- 4C_3$, We get,
$\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\-3&13&14\\0&11&12\end{vmatrix}$ [Applying $R_3 → R_3 - R_2$ and $R_1 → R_1 + R_2$]
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\0&109&119\\0&11&12\end{vmatrix}$ [Applying $R_2 → 3R_1 + R_2$]
$\Rightarrow\triangle=1(109\times12-119\times11)$
$\Rightarrow\triangle=-1$
View full question & answer→Question 662 Marks
Show that the following systems of linear equations is inconsistent:
$2x - y = 5$,
$4x - 2y = 7$
AnswerConsider,
$2x - y = 5$
$4x - 2y = 7$
$\text{D}=\begin{vmatrix}2&-1\\4&-2\end{vmatrix}=-4+4=0$
$\text{D}_1=\begin{vmatrix}5&-1\\7&-2\end{vmatrix}=-10+7=-3$
$\text{D}_2=\begin{vmatrix}2&5\\4&7\end{vmatrix}=14-20=-6$
Hence, $D_1$ and $D_2$ are non zero. Thus the given system is inconsistent.
View full question & answer→Question 672 Marks
A matrix of order $3 \times 3$ has determinant $2.$ What is the value of $|A(3I)|,$ where I is the identity matrix of order $3 \times 3.$
AnswerLet $A$ be the given matrix. Then,
$|A| = 2 [$Order $= n = 3]$
$|I| = 1 [I$ is an identity matrix$]$
$3(I) = 3$
$|A^3(I)| = |3A| = 3^3|A| [A$ being of order $3]$
$= 27 \times 2 = 54$
$|A^3(I)| = 54$
View full question & answer→Question 682 Marks
Find the value of $\lambda$ so that the points $(1, - 5), (-4, 5)$ and $(\lambda,7)$ are collinear.
AnswerIf the points are collinear, then the area of the triangle must be zero.
Hence,
$\begin{vmatrix}1&-5&1\\-4&5&1\\\lambda&7&1\end{vmatrix}=0$
Expanding along $R_1$
$1(-2)+5(-4-\lambda)+1(-28-5\lambda)=0$
$-2-20-5\lambda-28-5\lambda=0$
$-50-10\lambda=0$
$\lambda=5$
Hence, $\lambda=5$
View full question & answer→Question 692 Marks
If A is a square matrix such that A (adj A) = 5I, where I denotes the identity matrix of the same order. Then, find the value of |A|.
AnswerA (adj A) = 5I (Given)
|A| I = 5I [$\because$ A(adj A) = |A|I]
$\therefore |\text{A}|=5$
View full question & answer→Question 702 Marks
Evaluate the following determinant:
$\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$
Applying $R_3 → 17R_2 - R_3$, we get
$\triangle=\begin{vmatrix}102&18&36\\1&3&4\\0&48&62\end{vmatrix}$
Applying $R_2 → 102R_2 - R_1$, we get
$\triangle=\begin{vmatrix}102&18&36\\0&288&327\\0&48&62\end{vmatrix}$
Thus, $\triangle=102(288\times62-372\times48)$
$\triangle=0$
View full question & answer→Question 712 Marks
Evaluate $\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}.$
Answer$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}$
Expanding along $C_3$, we have:
$\triangle= -\sin\alpha(-\sin\alpha\sin^2\beta-\cos^2\beta\sin\alpha)+\cos\alpha(\cos\alpha\cos^2\beta+\cos\alpha\sin^2\beta)$
$=\sin^2\alpha(\sin^2\beta+\cos^2\beta)+\cos^2\alpha(\cos^2\beta+\sin^2\beta)$
$=\sin^2\alpha(1)+\cos^2\alpha(1)$
$= 1$
View full question & answer→Question 722 Marks
For what value of x, the following matrix is singular?
$\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}$
AnswerIf a matrix A is singular, then |A| = 0
$\therefore\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}=0$
⇒ 4(5 - x) - 2(x + 1) = 0
⇒ 20 - 4x - 2x - 2
⇒ 18 - 6x = 0
⇒ 18 = 6x
⇒ x = 3
View full question & answer→Question 732 Marks
A matrix A of order $3 \times 3$ is such that $|A| = 4$. Find the value of $|2A|$.
Answer$|KA| = k^n |A|$
Here, $n$ is the order of $A$.
Given, $|A| = 4$
$\Rightarrow |2A| = 2^3 \times 4 = 32$
View full question & answer→Question 742 Marks
If $A$ is a square matrix satisfying $A^T A = l$, write the value of $|A|$.
AnswerLet $|\text{A}|=|\text{A}|^{\text{T}} $ [By property of determinants]
Given,
$\text{A}^{\text{T}}\text{A}=\text{I}$
$\Rightarrow|\text{A}^{\text{T}}\text{A}|=1$
Then,
$|\text{A}^{\text{T}}\text{A}|=|\text{A}^{\text{T}}||\text{A}|$ [Since the determinants are of the same order]
$\Rightarrow|\text{A}^{\text{T}}||\text{A}|=1$
$\Rightarrow|\text{A}|=\frac{1}{|\text{A}^{\text{T}}|}$
$\Rightarrow|\text{A}|=\frac{1}{|\text{A}|}$ $\big[\therefore|\text{A}|=|\text{A}^{\text{T}}|\big]$
$\Rightarrow|\text{A}|^2=1$
$\Rightarrow|\text{A}|=\pm1$
View full question & answer→Question 752 Marks
Write the cofactor of $a_{12}$ in the following matrix $\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}$
AnswerGiven,
$\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}$
Here, $\text{a}_{12}=-3$
Cofactor of $\text{a}_{12}=(-1)^{1+2}\begin{vmatrix}6&4\\1&-7\end{vmatrix}$
$\text{a}_{12}=-(-42-4)=46$
View full question & answer→Question 762 Marks
Find equation of line joining (1, 2) and (3, 6) using determinants.
AnswerLet P (x, y) be any points on the line joining the points (1, 2) and (3, 6).
Then, Area of triangle that could be formed by these points is zero.
$\therefore\ \ \text{Area of triangle}=\text{Modulus of}\ \frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=0$
$\Rightarrow\ \text{Modulus of}\ \frac{1}{2}\begin{vmatrix}x&y&1\\1&2&1\\3&6&1\end{vmatrix}=0$
$\Rightarrow\ \frac{1}{2}\left[x(2-6)-y(1-3)+1(6-6)\right]=0$
$\Rightarrow\ \ -4x+2y=0\ \Rightarrow\ \ -2x+y=0$
$\Rightarrow\ \ y=2x$ Which is required line.
View full question & answer→Question 772 Marks
If A is a non-singular square matrix such that $\text{A}^{-1}=\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix},$ then find $(A^T)^{-1}$.
AnswerFor any invertible matrix A.
$(A^T)^{-1} = (A^{-1})^T$
We have $\text{A}^{-1}=\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix}$
$\Rightarrow(\text{A}^\text{T})^{-1}=\begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix}$
View full question & answer→Question 782 Marks
If A is a square matrix such that $\text{A (adj A)}=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix},$ then write the value of $|adj\ A|$.
AnswerGiven,
$\text{A (adj A)}=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$\Rightarrow|\text{A}|\text{I}_\text{n}=5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow|\text{A}|=5$
Now, $|adj A| = |A|^{n-1} = 5^{3-1} = 25$.
View full question & answer→Question 792 Marks
Evaluate the following integrals:
$\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$
AnswerLet $\text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$
Put $\pi+\text{x}=\text{z}$
$\Rightarrow\text{dx}=\text{dz}$
When $\text{x}\rightarrow-\frac{3\pi}{2},\text{ z}\rightarrow-\frac{\pi}{2}$
When $\text{x}\rightarrow-\frac{\pi}{2},\text{ z}\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big[\sin^2(2\pi+\text{z})+\text{z}^3\Big]\text{dx}$
$=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{z}+\text{z}^3\big)\text{dz}$ $\big[\sin(2\pi+\theta)=\sin\theta\big]$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1-\cos2\text{z}}{2}\text{ dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dz}-\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{z dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\Big[\text{z}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{2}\Big[\frac{\sin2\text{z}}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\Big[\frac{\text{z}^4}{4}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\bigg[\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)\bigg]-\frac{1}{4}\big[\sin\pi-\sin(-\pi)\big]+\frac{1}{4}\Big(\frac{\pi^4}{16}-\frac{\pi^4}{16}\big)$
$=\frac{1}{2}\times\pi-\frac{1}{4}(0+0)+\frac{1}{4}\times0$
$=\frac{\pi}{2}$
View full question & answer→Question 802 Marks
Evaluate the following determinant:
$\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$
AnswerLet $\text{A}=\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$
$|\text{A}|=\text{x}(5\text{x}+1)+7\times\text{x}$
$=5\text{x}^2+\text{x}+7\text{x}$
$=5\text{x}^2+8\text{x}$
Hence, $|\text{A}|=5\text{x}^2+8\text{x}$
View full question & answer→Question 812 Marks
Write $A^{-1}$ for $\text{A}=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$
Answer$|\text{A}|=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}=1\neq0$
Let $C_{ij}$ be the cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11} = 3$
$C_{12} = -1$
$C_{21} = -5$
$C_{22} = 2$
$\text{adj A}=\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
$|\text{A}|=6-5=1$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
View full question & answer→Question 822 Marks
If $A = [A_{ij}]$ is a $3 \times 3$ diaginal matrix such that $a_{11} = 1, a_{22} = 2, a_{33} = 3,$ then find $|A|.$
AnswerIf $A = [A_{ij}]$ is a diagonal matrix of order n, then $|A| = a_{11} \times a_{22} \times a_{33} \times ...... \times a_{mn}.$
Given, $a_{11} - 1, a_{22} - 2$ and $a_{33} - 3$
$\Rightarrow |A| = 1 \times 2 \times 3 = 6 [$Applying the above property$]$
View full question & answer→Question 832 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 842 Marks
If $A$ is symmetric matrix, write whether $A^T$ is symmetric or skew-symmetric.
AnswerFor any symmetric matrix, $A^T = A$.
Hence, $A^T$ is also symmetric.
View full question & answer→Question 852 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$
$=1\begin{vmatrix} -1&2\\5&2\end{vmatrix}-(-3)\begin{vmatrix}4&2\\3&2 \end{vmatrix}+2\begin{vmatrix}4&-1\\3&5 \end{vmatrix}$
$=1(-2-10)+3(8-6)+2(20+3)$
$=(-12)+6+46$
$=40$
View full question & answer→Question 862 Marks
Find value of k if area of triangle is 4 sq. units and vertices are:
(k, 0), (4, 0), (0, 2)
AnswerGiven: Area of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=4$
$\Rightarrow\ \text{Modulus of}\ \frac{1}{2}\begin{vmatrix}k&0&1\\4&0&1\\0&2&1\end{vmatrix}=4$
$\Rightarrow\ \bigg|\frac{1}{2}\left[k(0-2)-0+1(8-0)\right]\bigg|=4$
$\Rightarrow\ \bigg|\frac{1}{2}(-2k+8)\bigg|=4$
$\Rightarrow\ \bigg|-k+4\bigg|=4\ \ \Rightarrow\ \ -k+4=\pm4$
Taking positive sign, -k + 4 = 4 $\ \Rightarrow\ \ \ k=0$
Taking negative sign, -k + 4 = -4 $\Rightarrow\ \ \ k=8$
View full question & answer→Question 872 Marks
Evaluate $\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4785&2\\4789&2\end{vmatrix}$ [Applying $C_2 → C_2 - C_1$]
$=2\times\begin{vmatrix}4785&1\\4789&1\end{vmatrix}$
$=2\times(4785-4789)$
$=2\times(-4)=-8$
$\Rightarrow\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}=-8$
View full question & answer→Question 882 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 892 Marks
Write the value of $\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$
$=\sin20^{\circ}\cos70^{\circ}+\cos20^{\circ}\sin70^{\circ}$
$=\sin(20^{\circ}+70^{\circ})$ [trignometric identity]
$=\sin90^{\circ}$
$=1$
View full question & answer→Question 902 Marks
Using determinants prove that the points $(a, b), (a', b)$ and $(a - a', b - b')$ are collinear if $ab' = a'b$.
Answer$\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'&\text{b}'&1\\\text{a}-\text{a}'&\text{b}-\text{b}'&1\end{vmatrix}=\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'-\text{a}&\text{b}'-\text{b}&0\\\text{a}-\text{a}'&\text{b}-\text{b}'&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'-\text{a}&\text{b}'-\text{b}&0\\-\text{a}'&-\text{b}'&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=\begin{vmatrix}\text{a}'-\text{a}&\text{b}'-\text{b}\\-\text{a}'&-\text{b}'\end{vmatrix}$
$=-\text{b}'(\text{a}'-\text{a})+\text{a}'(\text{b}'-\text{b})$
$=-\text{b}'\text{a}'+\text{b}'\text{a}+\text{a}'\text{b}'-\text{a}'\text{b}$
$=\text{b}'\text{a}-\text{a}'\text{b}$
If the points are collinear then $\triangle=0$
$\text{a}\text{b}'-\text{a}'\text{b}=0$
Thus, $\text{a}\text{b}'=\text{a}'\text{b}$
View full question & answer→Question 912 Marks
Use elementary column operation $C_2 → C_2 + 2C_1$ in the following matrix equation:
$\begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$
Answer$\begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$
Applying $C_2 → C_2 + 2C_1$
$\begin{pmatrix} 2 & 5 \\ 2 & 4 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix}$
View full question & answer→Question 922 Marks
Find adjoint of each of the matrix.
$\begin{bmatrix}1&2\\3&4\end{bmatrix}$
AnswerHere $\text{A}=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\ \Rightarrow\ |\text{A}|=\begin{vmatrix}1&2\\3&4\end{vmatrix}$
$\therefore A_{11}$ = Confactor of $a_{11} = (-1)^2 (4) = 4$
$A_{12}$ = Confactor of $a_{12} = (-1)^3 (3) = -3$
$A_{21}$ = Confactor of $a_{21} = (-1)^3 (2) = -2$
$A_{22}$ = Confactor of $a_{22} = (-1)^4 (1) = 1$
$\therefore\ \text{adj.}\ \text{A}=\begin{bmatrix}\text{A}_{11}&\text{A}_{12}\\\text{A}_{21}&\text{A}_{22}\end{bmatrix}=\begin{bmatrix}4&-3\\-2&1\end{bmatrix}=\begin{bmatrix}4&-2\\-3&1\end{bmatrix}$
View full question & answer→Question 932 Marks
In the following matrix equation use elementary operation $R_2 → R_2 + R_1$ and the equation thus obtained:
$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 9 & -4 \end{bmatrix}$
Answer$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 9 & -4 \end{bmatrix}$
By applying elementary operation $R_2 → R_2 + R_1$, we get
$\begin{bmatrix} 2 & 3 \\ 3 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 17 & -7 \end{bmatrix}$
(Every row operation is equlvalent to left-multiplication be an elementary matrix.)
View full question & answer→Question 942 Marks
If $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix}$, find x, y and z.
AnswerHere,
$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix}$
$\therefore\ \text{x}=1,\ \text{y}=-1\text{ and }\text{z}=0$
View full question & answer→Question 952 Marks
If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.
AnswerLet A & B be non-singular matrices of order n.
A ≠ 0 and B ≠ 0 By definition.
Since they are of same order, AB = AB, AB = 0 if either A = 0 or B = 0 But it is not the case here. Thus, AB is non-zero and AB is non-singular matrix.
View full question & answer→Question 962 Marks
Write Minors and Cofactors of the elements of following determinant:
$\begin{vmatrix}a&c\\b&d\end{vmatrix}$
AnswerThe given determinant is $\begin{vmatrix}a&c\\b&d\end{vmatrix}.$
Minor of element $a_{ij}$ is $M_{ij}$.
$\therefore M_{11}$ = minor of element $a_{11} = d$
$M_{12}$ = minor of element $a_{12} = b$
$M_{21}$ = minor of element $a_{21}= c$
$M_{22}$ = minor of element $a_{22} = a$
cofactor of $a_{ij}$ is $A_{ij} = (-1)^{i+j} M_{ij}$.
$\therefore A_{11}= (-1)^{1+1} M_{11} = (-1)^2(d) = d$
$A_{12}= (-1)^{1+2}M_{12} = (-1)^3 (b) = -b$
$A_{21} = (-1)^{2+1} M_{21} = (-1)^3(c) = -c$
$A_{22} = (-1)^{2+2} M_{22} = (-1)^4 (a) = a$
View full question & answer→Question 972 Marks
Let A be a $3 \times 3$ square matrix, such that $A (adj\ A) = 2I,$ where $I$ is the identity matrix. Write the value of $|adj\ A|.$
Answer$\because A(adj\ (A)) = |A|I$
$2I = |A|I ($Given $A(adj\ A) = 2I)$
$|A| = 2$
Also, $|adj\ A| = |A|^{n-1}$
$= (2)^{3-1}$
$= (2)^2$
$= 4$
$|adj\ A| = 4$
View full question & answer→Question 982 Marks
Find the value of the determinant $\begin{vmatrix}243&156&300\\81&52&100\\-3&0&4\end{vmatrix}$
Answer[Applying $R_1 → R_1 - 3R_2$]
$=\begin{vmatrix}0&0&0\\81&52&100\\-3&0&4\end{vmatrix}$
$=0$
View full question & answer→Question 992 Marks
If $I_3$ denotes identity matrix of order $3 \times 3$, write the value of its determinant.
AnswerIn an identity matrix, all the diagonal elements are $1$ and rest of the elements are $0$.
Here,
$\text{I}_3=\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}$
$\text{I}_3=1\times\begin{vmatrix}1&0\\0&1\end{vmatrix}$ [Expanding along $C_1$]
$\text{I}_3=1$
$\text{I}_3=1$
View full question & answer→Question 1002 Marks
If A is a square matrix of order n × n such that |A| = λ, then write the value of |-A|.
Answer$|\text{A}|=\lambda$ [Order of A is n]
$\Rightarrow|-\text{A}|=(-1)^{\text{n}}|\text{A}|=(-1)^{\text{n}}\lambda$
View full question & answer→Question 1012 Marks
Evaluate the following determinant:
$\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta \end{vmatrix}$
Answer$\triangle=\cos^2\theta-(-\sin^2\theta)$
$\triangle=\cos^2\theta+\sin^2\theta=1$
View full question & answer→Question 1022 Marks
If $\text{A}=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix},$ write adj A.
Answer$|\text{A}|=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix}=6\neq0$
A 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
$C_{11} = 0$
$C_{12} = -2$
$C_{21} = 3$
$C_{22} = 1$
$\therefore\ \text{adj A}=\text{A}=\begin{bmatrix} 0 & -2 \\ 3 & 1 \end{bmatrix}^\text{T}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$
View full question & answer→Question 1032 Marks
Find the adjoint of the following matrices: $\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{D}=\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$
$\text{adjoint D}=\begin{bmatrix}1 & -\frac{\tan\alpha}{2} \\ \frac{\tan\alpha}{2} & 1 \end{bmatrix}$
$(\text{adjoint D)D}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$
$|\text{D}|=1+\tan^2\frac{\alpha}{2}$
$|\text{D}|\text{I}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$
View full question & answer→Question 1042 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 1052 Marks
If $C_{ij}$ is the cofactor of the element $a_{ij}$ of the matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix},$ then write the value of $a_{32}C_{32}$.
AnswerIn the given matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix}$
$C_{32} = (-1)^{3+2} (8 - 30) = 22$
Therefore, $a_{32}C_{32} = 5 \times 22 = 110$.
Hence, the value of $a_{32}C_{32}$ is $110$.
View full question & answer→Question 1062 Marks
$A, B, C$ are three non-null square matrices of the same order, write the condition on A such that $AB = AC ⇒ B = C$.
AnswerConsider AB = AC. On multiplying both sides by $A^{-1}$, we get $AA^{-1}B = AA^{-1}$
$\Rightarrow IB = IC$ [Because $AA^{-1} = I$ where I is the identity matrix] ⇒ B = C Therefore, the required condition is A must be invertible or $|\text{A}|\neq0$.
View full question & answer→Question 1072 Marks
Let $A$ be a square matrix such that $A^2 - A + I = 0,$ then write $A^{-1}$ interms of $A.$
Answer$A^2 - A + I = 0$
$Px-$ multiplying with $A^{-1},$
$(A^{-1} A) - (A^{-1} A) + A^{-1}I = 0$
$IA - I + A^{-1} = 0$
$A^{-1} = I - A$
Hence, $A^{-1} = I - A$
View full question & answer→Question 1082 Marks
Show that the following systems of linear equations is inconsistent:
3x + y = 5,
-6x - 2y = 9
Answer$\text{D}=\begin{vmatrix}3&1\\-6&-2\end{vmatrix}=-6+6=0$
$\text{D}_1=\begin{vmatrix}5&1\\9&-2\end{vmatrix}=-10-9=-19\neq0$
Since D = 0 but $\text{D}_1\neq0$
Hence the given system of equations is inconsistent.
View full question & answer→Question 1092 Marks
If $\text{A}=\begin{bmatrix}5&3&8\\2&0&1\\1&2&3\end{bmatrix}.$ Write the cofactor of element $a_{32}$.
AnswerMinor of $\text{a}_{32}=\text{M}_{32}=\begin{vmatrix}5&8\\2&1\end{vmatrix}=5-16=-11$
Cofactor of $\text{a}_{\text{n}}=\text{A}_{32}=(-1)^{3+2}\text{M}_{32}=11$
Hence, the cofactor of the elements $a_n$ is $11$.
View full question & answer→Question 1102 Marks
If the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, prove that $a + b = ab$.
AnswerIf the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, then
$\begin{vmatrix}\text{a}&0&1\\0&\text{b}&1\\1&1&1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1&1&1\end{vmatrix}=0$ [Applying $R_2 → R_2 - R_1$]
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1-\text{a}&1&0\end{vmatrix}=0$ [Applying $R_3 → R_3 - R_1$]
$\Rightarrow\triangle=\begin{vmatrix}-\text{a}&\text{b}\\1-\text{a}&1\end{vmatrix}=0$
$\Rightarrow-\text{a}-\text{b}(1-\text{a})=0$
$\Rightarrow\text{a}+\text{b}=\text{ab}$
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