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Mathematics MCQ

Matrices · ICSE STD 10 (ICSE - English Medium). 14 questions.

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14 questions · timed · auto-graded

MCQ 11 Mark
$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ then $A^2=$
  • A
    $\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$
  • B
    $\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right]$
  • ✓
    $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
  • D
    $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
Answer
Correct option: C.
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Given
$
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& A^2=A \times A=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \times\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1+0 & 0+0 \\
0+0 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] . \\
&
\end{aligned}
$
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MCQ 21 Mark
If $A+B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and $A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$ then $A$ is equal to
  • A
    $\frac{1}{3}\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
  • B
    $\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
  • ✓
    $\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
  • D
    $\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
Answer
Correct option: C.
$\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
$
A+B=\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]
$
and
$
A-2 B=\left[\begin{array}{cc}
-1 & 1 \\
0 & -1
\end{array}\right]
$
$\Rightarrow 2 A+2 B=\left[\begin{array}{ll}2 & 0 \\ 2 & 2\end{array}\right]$ (Multiplying by 2) $\ldots$ (i)
$A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$
Adding (i) and (ii), we get
$
\begin{aligned}
& 3 A=\left[\begin{array}{ll}
2 & 0 \\
2 & 2
\end{array}\right]+\left[\begin{array}{cc}
-1 & 1 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right] \\
& \therefore A=\frac{1}{3}\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right] .
\end{aligned}
$
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MCQ 31 Mark
If $B=\left[\begin{array}{ll}1 & 5 \\ 0 & 3\end{array}\right]$ and $A-2 B=\left[\begin{array}{cc}0 & 4 \\ -7 & 5\end{array}\right]$ then the matrix $A$ is equal to
  • A
    $\left[\begin{array}{cc}2 & 14 \\ -7 & 11\end{array}\right]$
  • B
    $\left[\begin{array}{cc}-2 & 14 \\ 7 & 11\end{array}\right]$
  • C
    $\left[\begin{array}{cc}2 & -14 \\ 7 & 11\end{array}\right]$
  • ✓
    $\left[\begin{array}{ll}-2 & 14 \\ -7 & 11\end{array}\right]$
Answer
Correct option: D.
$\left[\begin{array}{ll}-2 & 14 \\ -7 & 11\end{array}\right]$
$
B=\left[\begin{array}{ll}
1 & 5 \\
0 & 3
\end{array}\right]
$
and
$
\begin{aligned}
& A -2 B =\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right] \\
& 2 B =2 \times\left[\begin{array}{cc}
-1 & 5 \\
0 & 3
\end{array}\right]=\left[\begin{array}{cc}
-2 & 10 \\
0 & 6
\end{array}\right] \\
& A -2 B =\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right] \\
& \Rightarrow A =\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]+2 B \\
& \Rightarrow A =\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]+\left[\begin{array}{cc}
2 & 10 \\
0 & 6
\end{array}\right] \\
& =\left[\begin{array}{cc}
0-2 & 4+10 \\
-7+0 & 5+6
\end{array}\right] \\
& =\left[\begin{array}{cc}
-2 & 14 \\
-7 & 11
\end{array}\right] .
\end{aligned}
$
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MCQ 41 Mark
If $x \begin{bmatrix} 2 \\ 3\end{bmatrix} +y \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 6\end{bmatrix} $ then the values of $x$ and $y$ are
  • A
    $x=2, y=6$
  • ✓
    $x=2, y=-6$
  • C
    $x=3, y=-4$
  • D
    $x=3, y=-6$
Answer
Correct option: B.
$x=2, y=-6$
Given
$x \begin{bmatrix} 2 \\ 3\end{bmatrix} +y \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 6\end{bmatrix} $
$ \begin{bmatrix} 2x \\ 3x\end{bmatrix} +y \begin{bmatrix} -y \\ 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 6\end{bmatrix} $
$ \begin{bmatrix} 2x-y \\ 3x+0\end{bmatrix} = \begin{bmatrix} 10 \\ 6\end{bmatrix} $
Comapring, we get
$ 3 x =6 \Rightarrow x =\frac{6}{3}=2$ and
$ 2 x-y=10 $
$ 2 \times 2-y=10 $
$ \Rightarrow 4-y=10 $
$ \Rightarrow-y=10-4=6 $
$ \Rightarrow y=-6 $
$ \therefore x=2, y=-6 .$
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MCQ 51 Mark
If $\left[\begin{array}{cc}x+2 y & 3 y \\ 4 x & 2\end{array}\right]=\left[\begin{array}{cc}0 & -3 \\ 8 & 2\end{array}\right]$ then the value of $x-y$ is
  • A
    $-3$
  • B
    1
  • ✓
    3
  • D
    5
Answer
Correct option: C.
3
$
\left[\begin{array}{cc}
x+2 y & 3 y \\
4 x & 2
\end{array}\right]=\left[\begin{array}{cc}
0 & -3 \\
8 & 2
\end{array}\right]
$
Comparing, we get
$
\begin{aligned}
& 3 y=-3 \\
& \Rightarrow y=\frac{-3}{3}=-1 \\
& 4 x=8 \\
& \Rightarrow x=\frac{8}{4}=2 \\
& x-y=2-(-1) \\
& =2+1 \\
& =3
\end{aligned}
$
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MCQ 61 Mark
If $\left[\begin{array}{cc}x-2 y & 5 \\ 3 & y\end{array}\right]=\left[\begin{array}{cc}6 & 5 \\ 3 & -2\end{array}\right]$ then the value of $x$ is
  • A
    $-2$
  • B
    $0$
  • C
    1
  • ✓
    2
Answer
Correct option: D.
2
$
\left[\begin{array}{cc}
x-2 y & 5 \\
3 & y
\end{array}\right]=\left[\begin{array}{cc}
6 & 5 \\
3 & -2
\end{array}\right]
$
Comparing, we get
$
y=-2
$
and
$
\begin{aligned}
& x-2 y=6 \\
& \Rightarrow x-2 x(-2)=6 \\
& \Rightarrow x+4=6 \\
& \Rightarrow x=6-4=2
\end{aligned}
$
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MCQ 71 Mark
If $\left[\begin{array}{cc}x+2 y & -y \\ 3 x & 7\end{array}\right]=\left[\begin{array}{cc}-4 & 3 \\ 6 & 4\end{array}\right]$ then the values of $x$ and $y$ are
  • A
    $x=2, y=3$
  • ✓
    $x=2, y=-3$
  • C
    $x=-2, y=3$
  • D
    $x=3, y=2$
Answer
Correct option: B.
$x=2, y=-3$
$
\left[\begin{array}{cc}
x+2 y & -y \\
3 x & 7
\end{array}\right]=\left[\begin{array}{cc}
-4 & 3 \\
6 & 4
\end{array}\right]
$
Comparing, we get
$
\begin{aligned}
& 3 x =6 \\
& \Rightarrow=2 \\
& \Rightarrow- y =3 \\
& \Rightarrow y =-3 \\
& x =2, y =-3
\end{aligned}
$
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MCQ 81 Mark
If $\left[\begin{array}{cc}x+3 & 4 \\ y-4 & x+y\end{array}\right]=\left[\begin{array}{ll}5 & 4 \\ 3 & 9\end{array}\right]$ then the values of $x$ and $y$ are
  • ✓
    $x=2, y=7$
  • B
    $x=7, y=2$
  • C
    $x=3, y=6$
  • D
    $x=-2, y=7$
Answer
Correct option: A.
$x=2, y=7$
$
\left[\begin{array}{cc}
x+3 & 4 \\
y-4 & x+y
\end{array}\right]=\left[\begin{array}{ll}
5 & 4 \\
3 & 9
\end{array}\right]
$
Comparing we get
$
\begin{aligned}
& x+3=5 \\
& \Rightarrow x=5-3=2
\end{aligned}
$
and
$
\begin{aligned}
& y-4=3 \\
& \Rightarrow y=3+4=7 \\
& x=2, y=7
\end{aligned}
$
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MCQ 91 Mark
If $A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]$, then $A^2=p A$, then the value of $p$ is
  • A
    2
  • ✓
    4
  • C
    $-2$
  • D
    $-4$
Answer
Correct option: B.
4
$
A=\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right]
$
and $A^2=p A$
$
\begin{aligned}
& A^2=A \times A=\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \times\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+4 & -4-4 \\
-4-4 & 4+4
\end{array}\right] \\
& =\left[\begin{array}{cc}
8 & -8 \\
-8 & 8
\end{array}\right] \\
& PA = p \left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 p & -2 p \\
-2 p & 2 p
\end{array}\right] \\
& \because A ^2= pA \\
& \therefore\left[\begin{array}{cc}
8 & -8 \\
8 & 8
\end{array}\right]=\left[\begin{array}{cc}
2 p & -2 p \\
-2 p & 2 p
\end{array}\right]
\end{aligned}
$
Comparing, we get
$
\begin{aligned}
& 8=2 p \\
& \Rightarrow p=4
\end{aligned}
$
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MCQ 101 Mark
If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, then $A^2=$
  • ✓
    $\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$
  • B
    $\left[\begin{array}{cc}8 & -5 \\ 5 & 3\end{array}\right]$
  • C
    $\left[\begin{array}{cc}8 & -5 \\ -5 & -3\end{array}\right]$
  • D
    $\left[\begin{array}{cc}8 & -5 \\ -5 & 3\end{array}\right]$
Answer
Correct option: A.
$\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$
$\begin{aligned} & A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \\ & A^2=A \times A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}9+(-1) & 3+2 \\ -3-2 & -1+4\end{array}\right] \\ & =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] .\end{aligned}$
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MCQ 111 Mark
If $A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$, then $A^2=$
  • A
    $\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$
  • B
    $\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]$
  • C
    $\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
  • ✓
    none of these
Answer
Correct option: D.
none of these
Given
$
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& A^2=A \times A=\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \times\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1+0 & 0+0 \\
1+1 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right] . \\
&
\end{aligned}
$
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MCQ 121 Mark
If $A=\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right]$, then $A^2=$
  • A
    A
  • ✓
    $0$
  • C
    I
  • D
    2A
Answer
Correct option: B.
$0$
Given
$
\begin{aligned}
& A=\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right] \\
& A^2=A \times A=\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right] \times\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{ll}
0+0 & 0+0 \\
0+0 & 0+0
\end{array}\right] \\
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& =0 .
\end{aligned}
$
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MCQ 131 Mark
If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$, then $A^2=$
  • A
    $\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$
  • B
    $\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right]$
  • C
    $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
  • ✓
    $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Answer
Correct option: D.
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Given
$
\begin{aligned}
& A=\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right] \\
& A^2=A \times A=\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right] \times\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{ll}
0+1 & 0++0 \\
0+0 & 1+0
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] .
\end{aligned}
$
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MCQ 141 Mark
If $A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j}=i+j$, then $A$ is equal to
  • A
    $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
  • ✓
    $\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$
  • C
    $\left[\begin{array}{ll}1 & 2 \\ 1 & 2\end{array}\right]$
  • D
    $\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]$
Answer
Correct option: B.
$\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$
$A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j}=i+j$, then $A$ is equal to $\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right]$
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