Vector Algebra Test

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Vector Algebra Test - 78

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Question 1
1 / -0

If \(A=\left[\begin{array}{ccc}2 & x-3 & x-2 \\ 3 & -2 & -1 \\ 4 & -1 & -5\end{array}\right]\) is a symmetric matrix then \(x=?\)

A
\(3\)

B
\(6\)

C
\(8\)

D
\(0\)

Solution

Given:

A is a symmetric matrix,

\(\Rightarrow A ^{\top}= A\) or \(a _{ ij }= a _{ i }\)

\(A =\left[\begin{array}{ccc}2 & x-3 & x-2 \\ 3 & -2 & -1 \\ 4 & -1 & -5\end{array}\right]\)

So, by property of symmetric matrices

\(\Rightarrow a_{12}=a_{21}\)

\(\Rightarrow x-3=3\)

\(\therefore x=6\)
Question 2
1 / -0

Find the value of \(X\) and \(Y\) if \(X+Y=\left[\begin{array}{cc}10 & 2 \\ 0 & 9\end{array}\right], X-Y=\left[\begin{array}{cc}6 & 12 \\ 0 & -5\end{array}\right]\).

A
\(X =\left[\begin{array}{cc}8 & 7 \\ 0 & -5\end{array}\right], Y =\left[\begin{array}{cc}2 & 5 \\ 7 & -5\end{array}\right]\)

B
\(X =\left[\begin{array}{ll}8 & 7 \\ 0 & 2\end{array}\right], Y =\left[\begin{array}{cc}2 & -5 \\ 0 & 7\end{array}\right]\)

C
\(X =\left[\begin{array}{ll}8 & 7 \\ 0 & 2\end{array}\right], Y =\left[\begin{array}{ll}2 & 5 \\ 0 & 7\end{array}\right]\)

D
\(X =\left[\begin{array}{cc}8 & -7 \\ 0 & 7\end{array}\right], Y =\left[\begin{array}{cc}2 & -5 \\ 0 & 7\end{array}\right]\)

Solution

Given,
\(\begin{aligned}X+Y=\left[\begin{array}{cc}
10 & 2 \\
0 & 9
\end{array}\right] \end{aligned}\)...(1)
\(\begin{aligned}X-Y=\left[\begin{array}{cc}
6 & 12 \\
0 & -5
\end{array}\right]
\end{aligned}\)...(2)
Adding equation (1) and (2), we get
We know that,
\(\left[\begin{array}{ll}
A_1 & A_2 \\
B_1 & B_2
\end{array}\right]+\left[\begin{array}{ll}
C_1 & C_2 \\
D_1 & D_2
\end{array}\right]=\left[\begin{array}{ll}
A_1+C_1 & A_2+C_2 \\
B_1+D_1 & B_2+D_2
\end{array}\right]\)
\(\begin{aligned}
&2 X=\left[\begin{array}{cc}
10 & 2 \\
0 & 9
\end{array}\right]+\left[\begin{array}{cc}
6 & 12 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
16 & 14 \\
0 & 4
\end{array}\right] \\
&\therefore X=\left[\begin{array}{ll}
8 & 7 \\
0 & 2
\end{array}\right]
\end{aligned}
\)
Now, subtracting equation (2) from (1), we get
\(\begin{aligned}
&2 Y=\left[\begin{array}{cc}
10 & 2 \\
0 & 9
\end{array}\right]-\left[\begin{array}{cc}
6 & 12 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
4 & -10 \\
0 & 14
\end{array}\right] \\
&\therefore Y=\left[\begin{array}{cc}
2 & -5 \\
0 & 7
\end{array}\right]
\end{aligned}\)
So, the value of \(\begin{aligned} X=\left[\begin{array}{ll}
8 & 7 \\
0 & 2
\end{array}\right]
\end{aligned}\) and \(\begin{aligned} Y=\left[\begin{array}{cc}
2 & -5 \\
0 & 7
\end{array}\right]
\end{aligned}\)
Question 3
1 / -0

Consider the matrix \(A=\left[\begin{array}{ccc}2 & 4 & 5 \\ 1 & 6 & 4 \\ 2 & 8 & 9\end{array}\right]\). Find the element:

A
5

B
6

C
4

D
8

Solution

\(a_{32}\) is the element represented in the form \(a_{\mathrm{ij}} \mathrm{i}\) is the row number and \(\mathrm{j}\) is the column number. Therefore, the element \(a_{32}\) is the element in the third row \((\mathrm{i}=3\) ) and second column \((\mathrm{j}=2)\) which is 8.
Question 4
1 / -0

What is the order of \(\left[\begin{array}{lll}4 & 4 & 1\end{array}\right]\left[\begin{array}{lll}3 & 2 & 5 \\ 9 & 7 & 4 \\ 6 & 4 & 1\end{array}\right]\) ?

A
\(1 \times 1\)

B
\(1 \times 3\)

C
\(3 \times 3\)

D
\(3 \times 1\)

Solution

Let \(\left[\begin{array}{lll}4 & 4 & 1\end{array}\right]\left[\begin{array}{lll}3 & 2 & 5 \\ 9 & 7 & 4 \\ 6 & 4 & 1\end{array}\right]= AB\)

Order of matrix \(=\) no. of rows\(\times\) no. of column
Order of matrix \(A\) is \((1 \times 3)\).
Order of matrix \(B\) is \((3 \times 3)\).
As we know,
To multiply an \(m \times n\) matrix by \(n \times p\) matrix, the \(n\) must be the same and the result is an \(m \times p\) matrix.
So, Order of \(A_{(1 \times 3)} B_{(3 \times 3)}\) is \((1 \times 3)\).
\(\therefore\) Order of \(\left[\begin{array}{lll}4 & 4 & 1\end{array}\right]\left[\begin{array}{lll}3 & 2 & 5 \\ 9 & 7 & 4 \\ 6 & 4 & 1\end{array}\right]\) is \((1 \times 3)\).
Question 5
1 / -0

For \(A=\left[\begin{array}{ll}2 & 4 \\ 0 & 3\end{array}\right]\), then \(A^{-1}\) is given by :

A
\(\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)

B
\(6\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)

C
\(\frac{1}{6}\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)

D
\(\frac{1}{6}\left[\begin{array}{ll}2 & 4 \\ 0 & 3\end{array}\right]\)

Solution

If \(A =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is a square matrix of order 2 , then \(A ^{-1}=\frac{1}{|A|} \operatorname{Adj~A}\)

where, \(| A |= ad - bc\) and \(Adj A =\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\)

Given,

\(A=\left[\begin{array}{ll}2 & 4 \\ 0 & 3\end{array}\right]\)

\(A ^{-1}=\frac{1}{|A|} Adj A\)

\(| A |=3 \times 2-4 \times(0)\)

\(\Rightarrow| A |=6\)

\(\operatorname{Adj} A=\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)

Therefore, \(A ^{-1}=\frac{1}{6}\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)
Question 6
1 / -0

Order of \(\left[\begin{array}{lll}2 & 7 & 4 \\ 3 & 1 & 0\end{array}\right]\left[\begin{array}{l}5 \\ 4 \\ 3\end{array}\right]\) is:

A
\(3 \times 1\)

B
\(2 \times 1\)

C
\(2 \times 3\)

D
\(3 \times 3\)

Solution

Given,
First matrix \(=\left[\begin{array}{ccc}2 & 7 & 4 \\ 3 & 1 & 0\end{array}\right]\)
Number of rows \(\left(m_{1}\right)=2\)
Number of columns \(\left(n_{1}\right)=3\)
So, order of first matrix \(=2 \times 3\)
And second matrix \(=\left[\begin{array}{l}5 \\ 4 \\ 3\end{array}\right]\)
Number of rows \(\left(m_{2}\right)=3\)
Number of columns \(\left(n_{2}\right)=1\)
So, order of first matrix \(=3 \times 1\)
As we know,
Multiplication of matrices is possible if
Number of column \((n_1)\) in the first matrix \(=\) Number of rows \((m_2)\) in the second matrix
i.e., \(n_{1}=m_{2}\)
\(\therefore n_{1}=m_{2}=3\)
So, \(\left[\begin{array}{lll}2 & 7 & 4 \\ 3 & 1 & 0\end{array}\right]\left[\begin{array}{l}5 \\ 4 \\ 3\end{array}\right]\) is possible.
As we know,
Multiplication of matrices is possible if the order of the new matrix is a row of the first matrix and column of the second matrix.
i.e., \(m_{1}\times n_{2}\)
Now,
Order of new matrix formed by multiplication \(= m _{1} \times n _{2}=2 \times 1\)
Question 7
1 / -0

If \(A =\left[\begin{array}{cc}4 & x +2 \\ 2 x -3 & x +1\end{array}\right]\) is symmetric, then \(x\) is equal to:

A
2

B
3

C
-1

D
5

Solution

Given,
\(A =\left[\begin{array}{cc}
4 & x +2 \\
2 x -3 & x +1
\end{array}\right]\)
A real square matrix \(A=\left(a_{i j}\right)\) is said to be symmetric, if \(A=A^{T}\)
Where \(A^{T}=\) transpose of matrix \(A\)
\(\begin{aligned}
& A ^{ T }=\left[\begin{array}{cc}
4 & 2 x -3 \\
x +2 & x +1
\end{array}\right] \\
&\therefore A = A ^{T} \\
&\Rightarrow\left[\begin{array}{cc}
4 & x +2 \\
2 x -3 & x +1
\end{array}\right]=\left[\begin{array}{cc}
4 & 2 x -3 \\
x +2 & x +1
\end{array}\right]
\end{aligned}\)
On comparing elements of both matrix, we get
\( x+2=2 x-3 \)
\(\Rightarrow 2+3=2x-x\)
\(\therefore x=5\)
Question 8
1 / -0

If \(A =\left[\begin{array}{rr}0 & - i \\ i & 0\end{array}\right]\) and \(B =\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\) are matrices, then \(AB + BA\) is:

A
A diagonal matrix

B
An invertible matrix

C
A unit matrix

D
A null matrix

Solution

Given,
\(A =\left[\begin{array}{rr}0 & - i \\ i & 0\end{array}\right]\)
\(B =\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\)
\(\therefore AB =\left[\begin{array}{rr}
0 & - i \\
i & 0
\end{array}\right] \times\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\)
We know that,
\(X=\left[\begin{array}{cc}A_1 & A_2 \\ B_1 & B_2\end{array}\right], Y=\left[\begin{array}{cc}C_1 & C_2 \\ D_1 & D_2\end{array}\right]\).
\(XY=\left[\begin{array}{cc}A_1C_1+A_2D_1 & A_1C_2+A_2D_2 \\ B_1C_1+B_2D_1 & B_1C_2+B_2D_2 \end{array}\right]\)
\(=\left[\begin{array}{ll}
0+0 & 0+ i \\
i +0 & 0+0
\end{array}\right]\)
\(=\left[\begin{array}{ll}
0 & i \\
i & 0
\end{array}\right]\)
And \(BA =\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right] \times\left[\begin{array}{rr}0 & - i \\ i & 0\end{array}\right]\)
\(=\left[\begin{array}{rr}0+0 & - i +0 \\ 0- i & 0+0\end{array}\right]\)
\(=\left[\begin{array}{rr}0 & - i \\ - i & 0\end{array}\right]\)
\(\therefore A B+B A=\left[\begin{array}{ll}
0 & i \\
i & 0
\end{array}\right]+\left[\begin{array}{rr}
0 & -i \\
-i & 0
\end{array}\right]\)
\(=\left[\begin{array}{cc}
0+0 & i-i \\
i-i & 0+0
\end{array}\right]\)
\(=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
So, \(AB+BA\) is a null matrix.
Question 9
1 / -0

If \(A =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right]\) and \(B =\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\), find \(( AB )^{T}\).

A
\(\left[\begin{array}{c}-2 \\ 4\end{array}\right]\)

B
\(\left[\begin{array}{cc}-2 & 4\end{array}\right]\)

C
\(\left[\begin{array}{c}2 \\ -4\end{array}\right]\)

D
\(\left[\begin{array}{ll}-2 & -4\end{array}\right]\)

Solution

Given,
\(A =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right]\) and \(B =\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\)
\(\Rightarrow AB =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right] \times\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\)
We know that,
\(\begin{aligned}
&X=\left[\begin{array}{ll}
A_1 & A_2 & A_3 \\
B_1 & B_2 & B_3
\end{array}\right], Y=\left[\begin{array}{ll}
D_1 \\
E_1 \\
F_1
\end{array}\right] \\
&X Y=\left[\begin{array}{ll}
A_1 D_1+A_2 E_1+A_3F_1 \\
B_1 D_1+B_2 E_1+B_3F_1
\end{array}\right]
\end{aligned}\)

So,
\(AB =\left[\begin{array}{c}1-3+0 \\ 3+6-5\end{array}\right]\)
\(\Rightarrow AB =\left[\begin{array}{c}-2 \\ 4\end{array}\right]\)
As we know,
The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix. It is denoted by \(A'\) or \(A^T\).
\(\therefore( AB )^{T}=\left[\begin{array}{ll}-2 & 4\end{array}\right]\)
Question 10
1 / -0

If \(\left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & -8 & 5 \\ 4 & 2 & \lambda\end{array}\right]\) is not an invertible matrix, then what is the value of \(\lambda\)?

A
-1

B
0

C
1

D
2

Solution

Given,
\(A =\left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & -8 & 5 \\ 4 & 2 & \lambda\end{array}\right]\) is not an invertible matrix.
As we know,
If the matrix \(A\) is non singular matrix then \(| A |=0\).
\(\therefore \left|\begin{array}{ccc}
1 & -3 & 2 \\
2 & -8 & 5 \\
4 & 2 & \lambda
\end{array}\right|=0 \)
We know that,
\(X=\left[\begin{array}{ll}
A_1 & A_2 & A_3 \\
B_1 & B_2 & B_3\\
C_1 & C_2 & C_3
\end{array}\right]\)
\(X=A_1(B_2C_3-B_3C_2)-A_2(B_1C_3-B_3C_1)+A_3(B_1C_2-B_2C_1)\)
Then,
\( 1(-8 \lambda-10)+3(2 \lambda-20)+2(4+32)=0 \)
\(\Rightarrow-8 \lambda-10+6 \lambda-60+72=0 \)
\(\Rightarrow-2 \lambda+2=0 \)
\(\Rightarrow \lambda=1\)
So, If \(\left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & -8 & 5 \\ 4 & 2 & \lambda\end{array}\right]\) is not an invertible matrix, then the value of \(\lambda\) is \(1\).

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Vector Algebra Test - 78

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Vector Algebra Test - 78

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SHARING IS CARING

Question 1

\(3\)

\(6\)

\(8\)

\(0\)

Solution

Question 2

\(X =\left[\begin{array}{cc}8 & 7 \\ 0 & -5\end{array}\right], Y =\left[\begin{array}{cc}2 & 5 \\ 7 & -5\end{array}\right]\)

\(X =\left[\begin{array}{ll}8 & 7 \\ 0 & 2\end{array}\right], Y =\left[\begin{array}{cc}2 & -5 \\ 0 & 7\end{array}\right]\)

\(X =\left[\begin{array}{ll}8 & 7 \\ 0 & 2\end{array}\right], Y =\left[\begin{array}{ll}2 & 5 \\ 0 & 7\end{array}\right]\)

\(X =\left[\begin{array}{cc}8 & -7 \\ 0 & 7\end{array}\right], Y =\left[\begin{array}{cc}2 & -5 \\ 0 & 7\end{array}\right]\)

Solution

Question 3

5

6

4

8

Solution

Question 4

\(1 \times 1\)

\(1 \times 3\)

\(3 \times 3\)

\(3 \times 1\)

Solution

Question 5

\(\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)

\(6\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)

\(\frac{1}{6}\left[\begin{array}{cc}3 & -4 \\ 0 & 2\end{array}\right]\)

\(\frac{1}{6}\left[\begin{array}{ll}2 & 4 \\ 0 & 3\end{array}\right]\)

Solution

Question 6

\(3 \times 1\)

\(2 \times 1\)

\(2 \times 3\)

\(3 \times 3\)

Solution

Question 7

2

3

-1

5

Solution

Question 8

A diagonal matrix

An invertible matrix

A unit matrix

A null matrix

Solution

Question 9

\(\left[\begin{array}{c}-2 \\ 4\end{array}\right]\)

\(\left[\begin{array}{cc}-2 & 4\end{array}\right]\)

\(\left[\begin{array}{c}2 \\ -4\end{array}\right]\)

\(\left[\begin{array}{ll}-2 & -4\end{array}\right]\)

Solution

Question 10

-1

0

1

2

Solution

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