Get Answers to all your Questions

Filter By

Clear Filter

All Questions

The following system of linear equations $7x+6y-2z=0$ $3x+4+2z=0$ $x-2y-6z=0,$ has
Option: 1 infinitely many solutions, $(x,y,z)$ satisfying $y=2z.$
Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

System of Homogeneous linear equations -

$\\\mathrm{Let,} \\\mathrm{a_1x+b_1y +c_1z=0\;\;\; ...(i)} \\\mathrm{a_2x+b_2y +c_2z=0\;\;\; ...(ii)} \\\mathrm{a_3x+b_3y +c_3z=0\;\;\; ...(iii)} \\\mathrm{be \; three\; homogeneous\; equation} \\\mathrm{and \; let\; \Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}}$

If ? ≠ 0, then x= 0, y = 0, z = 0 is the only solution of the above system. This solution is also known as a trivial solution.

If ? = 0, at least one of x, y and z are non-zero. This solution is called a non-trivial solution.

Explanation: using equation (ii) and (iii), we have

$\\\mathrm{\frac{x}{b_2c_3-b_3c_2} = \frac{y}{c_2a_3-c_3a_2}=\frac{z}{a_2b_3-a_3b_2}} \\\\\mathrm{or \;\; \frac{x}{\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}}=\frac{y}{\begin{vmatrix} c_2 &a_2 \\ c_3 & a_3 \end{vmatrix}} = \frac{z}{\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} = k (say \neq 0)} \\\mathrm{\therefore x = k\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}, y = k\begin{vmatrix} c_2& a_2\\ c_3 & a_3 \end{vmatrix} \; and \; z = k\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} \\\mathrm{putting\; these\; value\; in \; equation \; (i), we\; have} \\\mathrm{a_1\left \{ k\begin{vmatrix} b_2 & c_2\\ b_3 & c_3 \end{vmatrix} \right \}+b_1\left \{ k\begin{vmatrix} c_2 & a_2\\ c_3 & a_3 \end{vmatrix} \right \}+c_1\left \{ \begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix} \right \}=0}$

$\\\mathrm{\Rightarrow a_1\begin{vmatrix} b_2 &c_2 \\ b_3 & c_2 \end{vmatrix}-b_1\begin{vmatrix} a_2 & c_2\\ a_3 &c_3 \end{vmatrix}+c_1\begin{vmatrix} a_2 &b_2 \\ a_3 &b_3 \end{vmatrix} = 0 \;\;\;[\because \; k \neq 0]} \\\mathrm{or \;\; \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \; or \; \Delta = 0}$

This is the condition for a system have Non-trivial solution.

$\begin{aligned} &(1)\;\;7 x+6 y-2 z=0\\ &(2)\;\;3 x+4 y+2 z=0\\ &(3)\;\;x-2 y-6 z=0 \end{aligned}$

$\left|\begin{array}{ccc}{7} & {6} & {-2} \\ {3} & {4} & {2} \\ {1} & {-2} & {-6}\end{array}\right| =7(-20)-6(-20)-2(-10)=-140+120+20=0$

so infinite non-trivial solution exist

now equation (1) + 3 equation (3)

10x - 20z = 0

x = 2z

Correct Option 2

View Full Answer(1)

Posted by

avinash.dongre

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$

Elementary row operations -

Elementary row operations

Row transformation: Following three types of operation (Transformation) on the rows of a given matrix are known as elementary row operation (transformation).

i) Interchange of i^th row with j^th row, this operation is denoted by

$\\\mathrm{R_i \leftrightarrow R_j \;\;or\; R_{ij}}$

ii) The multiplication of ith row by a constant k (k≠0) is denoted by

$\\\mathrm{R_i \leftrightarrow kR_i \;\; or \; k\cdot R_i}$

iii) The addition of ith row to the elements of jth row multiplied by constant k (k≠0) is denoted by

$\\\mathrm{R_i \leftrightarrow R_i + kR_j \;\; or \; k\cdot R_{ij}}$

In the same way, three-column operations can also be defined too.

$\\\text { Apply } \mathrm{R}_{1}=\mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\\\Rightarrow f(x)=\left|\begin{array}{ccc}{1} & {0} & {0} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right| \quad \Rightarrow f(x)=1 \\\quad \Rightarrow f(50)=1$

Correct Option (3)

View Full Answer(1)

Posted by

avinash.dongre

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

If for some $\alpha \: and\: \beta$ in R, the intersection of the following three planes $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$ is aline in $R^{3}$ , then $\alpha +\beta$ is equal to :
Option: 1 $0$
Option: 2 $10$
Option: 3 $-10$
Option: 4 $2$

Cramer’s law -

Cramer’s law for the system of equations in two variables :

$\\\mathrm{Let \; a_1x +b_1y = c_1\; and \; a_2x + b_2y = c_2, where} \\\mathrm{\frac{a_1}{a_2}\neq\frac{b_1}{b_2}} \\\mathrm{on \; solving \; this \; equation \; by \; cross \; multiplication, we \; get} \\\mathrm{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}} \\\mathrm{or \; \frac{x}{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}=\frac{y}{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}=\frac{1}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}} \\\mathrm{or \; x=\frac{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}, y=\frac{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}}$

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

We can follow this analogy for the system of equations of 3 variable where third in the numerator of the value of z will be of constant and denominator will be formed by the value of coefficients of the variables.

$\\\mathrm{so\; let \;us\; consider\; the \; system\; of \;equations} \\\mathrm{a_1x+b_1y +c_1z =d_1 \;\;...(i)} \\\mathrm{a_2x+b_2y +c_2z =d_2\;\;\;...(ii)} \\\mathrm{a_3x+b_3y +c_3z =d_3\;\;\;...(iii)} \\\mathrm{then \; \Delta,\; which\; will\; be \;determinant\; of\; coefficient\; of \;variables, will\; be } \\\mathrm{\Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2& b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}, \Delta_1\; numerator\; of\; x \;is:} \\\mathrm{\Delta_1= \begin{vmatrix} d_1 & b_1 &c_1 \\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3 \end{vmatrix}, similarly\; \Delta_2 = \begin{vmatrix} a_1 & d_1 & c_1\\ a_2 & d_2 & c_2\\ a_3 & d_3 & c_3 \end{vmatrix}\; and \;} \\\mathrm{\Delta_3 = \begin{vmatrix} a_1 & b_1 & d_1\\ a_2 & b_2 & d_2\\ a_3 & b_3 & d_3 \end{vmatrix}, if \Delta \neq 0\; then\; \;x=\frac{\Delta_1}{\Delta}} \\\mathrm{y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}$

i) If ? ≠ 0, then the system of equations has a unique finite solution and so equations are consistent, and solutions are $\\\mathrm{x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}$

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

$\begin{array}{l}{\Delta=0 \Rightarrow\left|\begin{array}{ccc}{1} & {4} & {-2} \\ {1} & {7} & {-5} \\ {1} & {5} & {\alpha}\end{array}\right|=0} \\ {(7 \alpha+25)-(4 \alpha+10)+(-20+14)=0} \\ {3 \alpha+9=0 \Rightarrow \alpha=-3}\end{array}$

$\begin{array}{l}{\text { Also }\quad D_{3}=0 \Rightarrow\left|\begin{array}{lll}{1} & {4} & {1} \\ {1} & {7} & {\beta} \\ {1} & {5} & {5}\end{array}\right|=0} \\ {1(35-5 \beta)-(15)+1(4 \beta-7)=0} \\ {\beta=13}\end{array}$

$\alpha+ \beta=10$

View Full Answer(1)

Posted by

avinash.dongre

The number of all $3\times 3$ matrices A, with enteries from the set $\left \{ -1,0,1 \right \}$ such that the sum of the diagonal elements of $AA^{T}$ is 3, is
Option: 1 672
Option: 2 512
Option: 31024
Option: 4 256

Let matrix A be

$\\A=\begin{bmatrix} a &b &c \\ d& e &f \\ g &h & i \end{bmatrix}\\\\\\ A^T=\begin{bmatrix} a &d &g \\ b& e &h \\ c &f & i \end{bmatrix} \\\\\\ \text{trace}(AA^T)=a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=3$

So out of 9 elements, 3 elements must be equal to 1 or −1, and the rest elements must be 0.

Possible cases

$\begin{array}{cc} 0,0,0,0,0,0,1,1,1 & \rightarrow \text{Total possibilities}=^9C_6 \\ 0,0,0,0,0,0,-1,-1,-1 & \rightarrow \text{Total possibilities}=^9C_6 \\ 0,0,0,0,0,0,1,1,-1 & \rightarrow \text{Total possibilities}=^9C_6\times 3 \\ 0,0,0,0,0,0,-1,1,-1 & \rightarrow \text{Total possibilities}=^9C_6\times3 \end{array}$

$\\\\\text{Total number of cases}=^{9}{C}_{6}\times 8=672$

View Full Answer(1)

Posted by

Kuldeep Maurya

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

For which of the following ordered pairs $(\mu ,\delta)$ , the system of linear equations $x+2y+3z=1$ $3x+4y+5z=\mu$ $4x+4y+4z=\varrho$ is inconsistent ?
Option: 1 $(4,6)$
Option: 2 $(3,4)$
Option: 3 $(1,0)$
Option: 4 $(4,3)$

Solution of System of Linear Equations Using Matrix Method -

let us consider n linear equations in n unknowns, given as below

$\\\mathrm{a_{11}x_1+a_{12}x_2+...+a_{1n}x_n = b_1} \\\mathrm{a_{21}x_1+a_{22}x_2+...+a_{2n}x_n = b_2} \\\mathrm{...\;\;\;...\;\;\;...\;\;\;...\;\;\;\;\;\;\;\;...\;\;\;\;\;\;...} \\\mathrm{...\;\;\;...\;\;\;...\;\;\;...\;\;\;\;\;\;\;\;...\;\;\;\;\;\;...} \\\mathrm{a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n = b_n} \\\mathrm{here \; x_{1}, x_2,...x_n \; are \;unknown\; variables} \\\mathrm{if \; b_1=b_2 =...=b_n=0\; then\; the\; system \; of \; equation \; is} \\\mathrm{known\; as \; homogenous\; system \; of \; equation\; and \;if } \\\mathrm{any \; of \;b_1,b_2,...b_n \; is\; not \; known\; then \; it \; is \;called\; } \\\mathrm{non\; homogenous\; system \; of \; equation}$

The above system of equations can be written in matrix form as

$\\\mathrm{\begin{bmatrix} a_{11} & a_{12} & ... & ... & a_{1n}\\ a_{21} & a_{22} & ... & ... & a_{2n}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ a_{n1} & a_{n2} & ... & ... & a_{nn} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ ...\\ ...\\ x_n \end{bmatrix} = \begin{bmatrix} b_1\\ b_2\\ ...\\ ...\\ b_n \end{bmatrix}} \\\\\mathrm{\Rightarrow AX = B, \; where} \\\\\mathrm{A = \begin{bmatrix} a_{11} & a_{12} & ... & ... & a_{1n}\\ a_{21} & a_{22} & ... & ... & a_{2n}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ a_{n1} & a_{n2} & ... & ... & a_{nn} \end{bmatrix} , X=\begin{bmatrix} x_1\\ x_2\\ ...\\ ...\\ x_n \end{bmatrix}, B=\begin{bmatrix} b_1\\ b_2\\ ...\\ ...\\ b_n \end{bmatrix}}$

Premultiplying equation AX=B by A^-1, we get

A^-1(AX) = A^-1B ⇒ (A^-1A)X = A^-1B

⇒ IX = A^-1B

⇒ X = A^-1B

⇒ $\mathrm{X=\frac{adj A}{\left | A \right |}B}$

Types of equation :

System of equations is non-homogenous:
1. If |A| ≠ 0, then the system of equations is consistent and has a unique solution X = A^-1B
2. If |A| = 0 and (adj A)·B ≠ 0, then the system of equations is inconsistent and has no solution.
3. If |A| = 0 and (adj A)·B = 0, then the system of equations is consistent and has infinite number of solutions.
System of equations is homogenous:
1. If |A| ≠ 0, then the system of equations has only trivial solution and it has one solution.
2. If |A| = 0 then the system of equations has non-trivial solution and it has an infinite number of solution.
3. If number of equation < number of unknown then it has non-trivial solution.

$\begin{aligned} D &=\left|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {4} & {4} & {4}\end{array}\right| \\ &=2\left|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {2} & {2} & {2}\end{array}\right| \end{aligned}$

$\\R_1\rightarrow R_1-(R_2+R_3)\\D=\left|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {2} & {2} & {2}\end{array}\right|=0$

Hence, it has infinitely many solutions

$\mathrm{P}_{3} \equiv \alpha \mathrm{P}_{1}+\beta \mathrm{P}_{2}$

$3 \alpha+\beta=4\;\; \&\;\; 4 \alpha+2 \beta=4 \Rightarrow \alpha=2 \;\; \&\;\; \beta=-2$

$2 \mu-2=\delta$

For non infinite solution $2 \mu-2\neq\delta$

Correct Option (4)

View Full Answer(1)

Posted by

Kuldeep Maurya

If the system of linear equations 2x+2ay+az=0 2x+3by+bz=0, 2x+4cy+cz=0, where $a,b,c\epsilon R$ are non-zero and distinct ; has a non-zero solution, then :
Option: 1 $a+b+c=0$
Option: 2 $a,b,c$ are in AP
Option: 3 $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
Option: 4 $a,b,c$ are in G.P.

Cramer’s law -

Cramer’s law for the system of equations in two variables :

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

For non zero solutions D = 0

$\\D=\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c &c \end{vmatrix}=0\\C_2\rightarrow C_2-2C_3\\D=\begin{vmatrix} 1 &0 &a \\ 1& b &b \\ 1 & 2c & c \end{vmatrix}=0\\$

on solving

$\\-bc+2ac-ab=0\\\\\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

Correct Option (3)

View Full Answer(1)

Posted by

Kuldeep Maurya

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

The system of linear equations $\lambda x+2y+2z=5$ $2\lambda x+3y+5z=8$ $4 x+\lambda y+6z=10$ has :
Option: 1 no solution when $\lambda =2$
Option: 2 infinitely many solutions when $\lambda =2$
Option: 3 no solution when $\lambda =8$
Option: 4 a unique solution when $\lambda =-8$

Cramer’s law -

Cramer’s law for the system of equations in two variables :

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

$D=\begin{vmatrix} \lambda &2 &2 \\ 2\lambda &3 & 5\\ 4 &\lambda &6 \end{vmatrix}$

D = ( $\lambda$ + 8) ( 2 – $\lambda$ ) for $\lambda$ = 2

$D_1=\begin{vmatrix} 5 &2 &2 \\ 8 &3 & 5\\ 10 &2 &6 \end{vmatrix}$

= 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30]

= 40 + 4 – 28 $\neq$ 0 No solutions for $\lambda$ = 2

Correct Option (1)

View Full Answer(1)

Posted by

vishal kumar

If $A=\begin{pmatrix} 2 &2 \\ 9 &4 \end{pmatrix}$ and $I=\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$ , then $10A^{-1}$ is equal to :
Option: 1 $6I-A$
Option: 5 $A-6I$
Option: 9 $4I-A$
Option: 13 $A-4I$

Inverse of a Matrix -

A non-singular square matrix “A” is said to be invertible if there exists a non-singular square matrix B such that AB = I = BA, (all matrix are of the same order, they must be for this), then B is called inverse of matrix A.

$\\\mathrm{A^{-1} = B \Leftrightarrow AB = \mathbb{I}_n = BA} \\\mathrm{we \; have, } \\ \\\mathrm{A(adj A)=A\mathbb{I}_n} \\\mathrm{\Rightarrow A^{-1}A (adj A)=A^{-1}\mathbb{I}_n|A|} \\\mathrm{\mathbb{I}_n(adj A)=A^{-1}|A|\mathbb{I}_n} \\\mathrm{A^{-1}=\frac{adj A}{\left | A \right |}}$

Inverse of 2 x 2 matrix

$\\\mathrm{Let\;A\;is\;a\;square\;matrix\;of\;order\;2}\\\mathrm{A=\begin{bmatrix} a &b \\ c & d \end{bmatrix}}\\\mathrm{Then,}\\\mathrm{A^{-1}=\begin{bmatrix} a &b \\ c & d \end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d &-b \\- c & a \end{bmatrix}}$

Multiplication of two matrices -

Matrix multiplication:

Two matrices A and B are conformable for the product AB if the number of columns in A and the number of rows in B is equal. Otherwise, these two matrices will be non-conformable for matrix multiplication. So on that basis,

i) AB is defined only if col(A) = row(B)

ii) BA is defined only if col(B) = row(A)

$\\\mathrm{A = \left [ a_{ij} \right ]_{m\times n}} \\\mathrm{\\B=\left [ b_{ij} \right ]_{n\times p}}$

$\\\mathrm{C = AB = \left [ c_{ij} \right ]_{m\times p}} \\\mathrm{Where\;\; c_{ij} = \sum_{j=1}^{n}a_{ij}b_{jk}, 1\leq i\leq m,1\leq k\leq p} \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}+ ... + a_{in}b_{nk}}$

For examples

$\\\mathrm{Suppose,\;two\;matrices\;are\;given}\\\mathrm{A=\begin{bmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} & a_{33} \end{bmatrix}_{2\times3}\;\;\;and\;\;\;B=\begin{bmatrix} b_{11}& b_{12} &b_{13} \\b_{21} &b_{22} &b_{23} \\b_{31} &b_{32} &b_{33} \end{bmatrix}_{3\times3}}\\\\\mathrm{To\:obtain\:the\:entries\:in\:row\:\mathit{i}\:of\:AB,\:we\:multiply\:the\:entries\:in\:row\:\mathit{i}\:of\:A\:by\:}\\\mathrm{column\:\mathit{j}\:in\:B\:and\:add.}\\\mathrm{given\:matrices\:A\:and\:B,\:where\:the\:order\:of\:A\:are\:2\times3\:and\:the\:order\:of\:B\:are\:3\times3,}\\\mathrm{the\:product\:of\:AB\:will\:be\:a\:2\times3\:matrix.}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:1\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:first\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{11}\\b_{21} \\b_{31} \end{bmatrix}=a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}}$

$\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:2\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:second\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{12}\\b_{22} \\b_{32} \end{bmatrix}=a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:3\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:thired\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{13}\\b_{23} \\b_{33} \end{bmatrix}=a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33}}\\\\\mathrm{We\:proceed\:the\:same\:way\:to\:obtain\:the\:second\:row\:of\:AB.\:In\:other\:words,\:}\\\mathrm{row\:2\:of\:A\:times\:column\:1\:of\:B;}\\\mathrm{row\:2\:of\:A\:times\:column\:2\:of\:B;}\\\mathrm{row\:2\:of\:A\;times\:column\:3\:of\:B.}$

$\\\mathrm{When\:complete,\:the\:product\:matrix\:will\:be}\\\\\mathrm{AB=\begin{bmatrix} a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}\;\;& a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}\;\; &a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33} \\ a_{21}\cdot b_{11}+a_{22}\cdot b_{21}+a_{23}\cdot b_{31} \;\;& a_{21}\cdot b_{12}+a_{22}\cdot b_{22}+a_{23}\cdot b_{32} \;\;& a_{21}\cdot b_{13}+a_{22}\cdot b_{23}+a_{23}\cdot b_{33} \end{bmatrix}}$

$A=\left[\begin{array}{ll}{2} & {2} \\ {9} & {4}\end{array}\right] \text { and } A^{-1}=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]$

$AA^{-1}=\left[\begin{array}{ll}{2a+2c} & {2b+2d} \\ {9a+4c} & {9b+4d}\end{array}\right] =\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]$

On comparing we get

$a=-\frac{2}{5}, \;\; b=\frac{1}{5},\;\;c=\frac{9}{10}\;\&\;d=-\frac{1}{5}$

$\Rightarrow \quad 10 A^{-1}=A-61$

Correct Option (2)

View Full Answer(1)

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

If the system of linear equations, x+y+z=6 x+2y+3z=10 $3x+2y+\lambda z=\mu$ has more than two solutions, then $\mu-\lambda ^{2}$ is equal to _______.
Option: 1 13
Option: 2 17
Option: 3 21
Option: 4 25

Cramer’s law -

Cramer’s law for the system of equations in two variables :

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

x + y + z = 6 …….. (1)

x + 2y + 3z = 10 …….. (2)

3x + 2y + $\lambda$ z = $\mu$ …….. (3)

from (1) and (2)

if z = 0 $\Rightarrow$ x + y = 6 and x + 2y = 10

$\Rightarrow$ y = 4, x = 2

(2, 4, 0)

if y = 0 $\Rightarrow$ x + z = 6 and x + 3z = 10

$\Rightarrow$ z = 2 and x = 4

(4, 0, 2)

so, 3x + 2y + $\lambda$ z = $\mu$ must pass through (2, 4, 0) and (4, 0, 2)

$\mu$ = 14

and 12 + 2 $\lambda$ = $\mu$

$\lambda$ = 1

$(\mu-\lambda^2)$ = 13

View Full Answer(1)

Posted by

Ritika Jonwal

Let $A=\left [ a_{ij} \right ]$ and $B=\left [ b_{ij} \right ]$ be two $3\times 3$ real matrices such that $b_{ij}=(3)^{(i+j-2)}a_{ji},$ where, i, j=1,2,3. if the determinant of B is 81, then the determinant of A is :
Option: 1 $1/9$
Option: 2 $1/81$
Option: 3 $1/3$
Option: 4 $3$

Matrices, order of matrices, row and column matrix -

A set of $\mathrm{m\times n}$ numbers (real or complex) or objects or symbols arranged in form of a rectangular array having m rows and n columns and bounded by brackets [?] is called matrix of order m × n, read as m by n matrix.

E.g for m = 2 and n =3, we have $\begin{bmatrix} 2 & 4 & -3\\ 5 & 4 & 6 \end{bmatrix}$ order of this matrix is 2×3

The m by n matrix is represented as :

$\begin{bmatrix} a_{11} &a_{12} &... & a_{1n}\\ a_{21}&a_{22} &... &a_{2n} \\ ...& ...& ... & ...\\ a_{m1}&a_{m2} &... &a_{mn} \end{bmatrix}$

This representation can be represented in a more compact form as

$\left [ a_{ij} \right ]_{m\times n}$

Where $a_{ij}$ represents element of i^th row and j^th column and i = 1,2,...,m; j = 1,2,...,n.

For example, to locate the entry in matrix A identified as a_ij, we look for the entry in row i, column j. In matrix A, shown below, the entry in row 2, column 3 is a₂₃.

$A=\left[\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right]$

Matrix is only a representation of the symbol, number or object. It does not have any value. Usually, a matrix denoted by capital letters.

Properties of Determinants - Part 2 -

Property 5

If each element of a row (or a column) of a determinant is multiplied by a constant k, then the value of the determinant is multiplied by k.

For example

$\\\mathrm{Let,\;\;\Delta=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\\\mathrm{and\:\Delta'\:be\:the\:determinant\:obtained\:by\:multiplying\:the\:elements\:of\:}\\\mathrm{the\:first\:row\:by\:k.}\\\\\mathrm{\Delta'=\begin{vmatrix} ka_1 &ka_2 &ka_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\mathrm{Expanding\:along\:first\:row,\:we\:get}\\\\\mathrm{\Delta'=ka_1(b_2c_3-b_3c_2)-ka_2(b_1c_3-b_3c_1)+ka_3(b_1c_2-b_2c_1)}\\\mathrm{\;\;\;\;=k[a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)]}\\\mathrm{\Delta'=k\Delta}\\\mathrm{Hence,}\\\mathrm{\begin{vmatrix} ka_1 &ka_2 &ka_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}=k\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}$

Note:

By this property, we can take out any common factor from any one row or any one column of a given determinant.
If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then the determinant value is zero.

$|B|=\left|\begin{array}{lll}{b_{11}} & {b_{12}} & {b_{13}} \\ {b_{21}} & {b_{22}} & {b_{23}} \\ {b_{31}} & {b_{32}} & {b_{33}}\end{array}\right|$

$|B|=\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{1} a_{21}} & {3^{2} a_{22}} & {3^{3} a_{23}} \\ {3^{2} a_{31}} & {3^{3} a_{32}} & {3^{4} a_{33}}\end{array}\right|$

Taking Common $3^2$ from $R_3$ and $3$ from $R_2$

$|B|=3^3\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{0} a_{21}} & {3^{1} a_{22}} & {3^{2} a_{23}} \\ {3^{0} a_{31}} & {3^{1} a_{32}} & {3^{2} a_{33}}\end{array}\right|$

Taking Common $3^2$ from $C_3$ and $3$ from $C_2$

$\Rightarrow 81=3^{3} \cdot 3 \cdot 3^{2}|\mathrm{A}| \Rightarrow 3^{4}=3^{6}|\mathrm{A}| \Rightarrow|\mathrm{A}|=\frac{1}{9}$

Correct Option (A)

View Full Answer(1)

Posted by

Ritika Jonwal

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Animation Courses

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

All Questions

The following system of linear equations has Option: 1 infinitely many solutions, satisfying Option: 2 infinitely many solutions, satisfying Option: 3 no solution Option: 4 only the trivial solution.

Posted by

avinash.dongre

Let If then : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

Crack CUET with india's "Best Teachers"

If for some in R, the intersection of the following three planes is aline in , then is equal to : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

The number of all matrices A, with enteries from the set such that the sum of the diagonal elements of is 3, is Option: 1 672 Option: 2 512 Option: 31024 Option: 4 256

Posted by

Kuldeep Maurya

JEE Main high-scoring chapters and topics

For which of the following ordered pairs , the system of linear equations is inconsistent ? Option: 1 Option: 2 Option: 3 Option: 4

The following system of linear equations $7x+6y-2z=0$ $3x+4+2z=0$ $x-2y-6z=0,$ has
Option: 1 infinitely many solutions, $(x,y,z)$ satisfying $y=2z.$
Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$

If for some $\alpha \: and\: \beta$ in R, the intersection of the following three planes $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$ is aline in $R^{3}$ , then $\alpha +\beta$ is equal to :
Option: 1 $0$
Option: 2 $10$
Option: 3 $-10$
Option: 4 $2$

The number of all $3\times 3$ matrices A, with enteries from the set $\left \{ -1,0,1 \right \}$ such that the sum of the diagonal elements of $AA^{T}$ is 3, is
Option: 1 672
Option: 2 512
Option: 31024
Option: 4 256

For which of the following ordered pairs $(\mu ,\delta)$ , the system of linear equations $x+2y+3z=1$ $3x+4y+5z=\mu$ $4x+4y+4z=\varrho$ is inconsistent ?
Option: 1 $(4,6)$
Option: 2 $(3,4)$
Option: 3 $(1,0)$
Option: 4 $(4,3)$

If the system of linear equations 2x+2ay+az=0 2x+3by+bz=0, 2x+4cy+cz=0, where $a,b,c\epsilon R$ are non-zero and distinct ; has a non-zero solution, then :
Option: 1 $a+b+c=0$
Option: 2 $a,b,c$ are in AP
Option: 3 $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
Option: 4 $a,b,c$ are in G.P.

The system of linear equations $\lambda x+2y+2z=5$ $2\lambda x+3y+5z=8$ $4 x+\lambda y+6z=10$ has :
Option: 1 no solution when $\lambda =2$
Option: 2 infinitely many solutions when $\lambda =2$
Option: 3 no solution when $\lambda =8$
Option: 4 a unique solution when $\lambda =-8$

If $A=\begin{pmatrix} 2 &2 \\ 9 &4 \end{pmatrix}$ and $I=\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$ , then $10A^{-1}$ is equal to :
Option: 1 $6I-A$
Option: 5 $A-6I$
Option: 9 $4I-A$
Option: 13 $A-4I$

If the system of linear equations, x+y+z=6 x+2y+3z=10 $3x+2y+\lambda z=\mu$ has more than two solutions, then $\mu-\lambda ^{2}$ is equal to _______.
Option: 1 13
Option: 2 17
Option: 3 21
Option: 4 25

Let $A=\left [ a_{ij} \right ]$ and $B=\left [ b_{ij} \right ]$ be two $3\times 3$ real matrices such that $b_{ij}=(3)^{(i+j-2)}a_{ji},$ where, i, j=1,2,3. if the determinant of B is 81, then the determinant of A is :
Option: 1 $1/9$
Option: 2 $1/81$
Option: 3 $1/3$
Option: 4 $3$