Matrices Test

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Matrices Test - 26

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Question 1
1 / -0

If $$\mathrm{A}=\left(\begin{array}{lll}
x & 1 & 4\\
-1 & 0 & 7\\
-4 & -7 & 0
\end{array}\right)$$ such that $$\mathrm{A}^{\mathrm{T}}=-\mathrm{A}$$, then $$\mathrm{x}=$$

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$4$$

Solution

Given $$A=\begin{pmatrix}
x & 1 & 4\\
-1 & 0 & 7\\
-4 & -7 & 0
\end{pmatrix}$$
and $$A^{T}=-A$$
So By property of traspose (I), and opertaion of matrixes (4),
$$A^{T}=\begin{pmatrix}
x & -1 & -4\\
1 & 0 & -7\\
4 & 7 & 0
\end{pmatrix}=-A=\begin{pmatrix}
-x & -1 & -4\\
1 & 0 & -7\\
4 & 7 & 0
\end{pmatrix}$$
So, By equality of matrixes,
$$x=-x$$
$$\Rightarrow x=0$$
Question 2
1 / -0

$$\mathrm{If}\mathrm{A}=\left[\begin{array}{lll}
2 & x-3 & x-2\\
3 & -2 & -1\\
4 & -1 & -5
\end{array}\right]$$ is a symmetric matrix then $$\mathrm{x}$$

A
$$0$$

B
$$3$$

C
$$6$$

D
$$8$$

Solution

Given $$A=\begin{bmatrix}
2 & x-3 & x-2\\
3 & -2 & -1\\
4 & -1 & -5
\end{bmatrix}$$ is a symmetric matrix.
So, by property of symmetric matrices
$$A=A^{T}$$
$$\Rightarrow \begin{bmatrix}
2 & x-3 & x-2\\
3 & -2 & -1\\
4 & -1 & -5
\end{bmatrix}=\begin{bmatrix}
2 & 3 & 4\\
x-3 & -2 & -1\\
x-2 & -1 & -5
\end{bmatrix}$$
So, By operation of matrices - equality,
$$\Rightarrow x-3=3$$
$$\Rightarrow x=6$$
Question 3
1 / -0

$$A=\left[\begin{array}{ll}
2 & 1\\
3 & 0
\end{array}\right]$$ then $$\mathrm{A}^{2}+2\mathrm{A}+I=$$

A
$$\left[\begin{array}{ll}

12 & 4\\

12 & 4

\end{array}\right]$$

B
$$\left[\begin{array}{ll}

12 & -4\\

4 & 12

\end{array}\right]$$

C
$$\left[\begin{array}{ll}

4 & 12\\

12 & 4

\end{array}\right]$$

D
$$\left[\begin{array}{ll}

4 & 12\\

-12 & -4

\end{array}\right]$$

Solution

$$A^2+2A+I=(A+I)^2$$

Now, $$A+I=\begin{bmatrix}
3 & 1\\
3 & 1
\end{bmatrix}$$

$$(A+I)^2=A^2+2A+I=\begin{bmatrix}
3 & 1\\
3 & 1
\end{bmatrix}\begin{bmatrix}
3 & 1\\
3 & 1
\end{bmatrix}=\begin{bmatrix}
12 & 4\\
12 & 4
\end{bmatrix}$$
Question 4
1 / -0

$$A=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}$$ then $$A^{2}-A=$$

A
$$\begin{bmatrix} 3 & 0 & 0\\

0 & 1 & -1\\

0 & 5 & 4

\end{bmatrix}$$

B
$$ \begin{bmatrix} 1 & 2 & -1\\

-3 & 1 & 1\\

3 & 5 & -4

\end{bmatrix}$$

C
$$\begin{bmatrix} 3 & 1 & -1\\

-3 & 1 & -1\\

3 & 5 & 4

\end{bmatrix}$$

D
$$\begin{bmatrix} 3 & -1 & 1\\

3 & -1 & 1\\

3 & 5 & 4

\end{bmatrix}$$

Solution

$$A=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}$$

$${ A }^{ 2 }=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}=\begin{bmatrix} 4 & 3 & 0 \\ -3 & 2 & -2 \\ 6 & 4 & 5 \end{bmatrix}$$

Now, $${ A }^{ 2 }-A=\begin{bmatrix} 4 & 3 & 0 \\ -3 & 2 & -2 \\ 6 & 4 & 5 \end{bmatrix}-\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}$$

$$=\begin{bmatrix} 3 & 1 & -1 \\ -3 & 1 & -1 \\ 3 & 5 & 4 \end{bmatrix}$$
Question 5
1 / -0

$$\begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} x & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$$ then $${x}=$$

A
$$10$$

B
$$20$$

C
$$30$$

D
$$40$$

Solution

Given, $$\begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} x & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$$

$$\Rightarrow \begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} x & 0 & 0 \\ 2x & 5 & 0 \\ 3x & 20 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$$

$$\Rightarrow \begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} x & 2x & 3x \\ 2x & 4x+5 & 6x+20 \\ 3x & 6x+20 & 9x+81 \end{bmatrix}$$

On comparing, we get $$x=10$$
Question 6
1 / -0

lf $$ \left[3x^{2}+10xy+5y^{2} \right]=\begin{bmatrix}x & y \end{bmatrix}A\begin{bmatrix} x\\ y\end{bmatrix}$$, and $${A}$$ is a symmetric matrix then $$\mathrm{A}=$$

A
$$\left[\begin{array}{ll}

3 & 10\\

10 & 5

\end{array}\right]$$

B
$$\left[\begin{array}{ll}

10 & 3\\

5 & 10

\end{array}\right]$$

C
$$\left[\begin{array}{ll}

+3 & -5\\

-5 & +5

\end{array}\right]$$

D
$$\left[\begin{array}{ll}

3 & 5\\

5 & 5

\end{array}\right]$$

Solution

Given $$\begin{bmatrix} 3{ x }^{ 2 }+10xy & +5{ y }^{ 2 } \end{bmatrix}=\begin{bmatrix} x & y \end{bmatrix}A\begin{bmatrix} x \\ y \end{bmatrix}$$

We will check using options
If option A is the matrix A, then consider RHS
$$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & 10 \\ 10 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$

$$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x+10y \\ 10x+5y \end{bmatrix}$$

$$=\begin{bmatrix} 3{ x }^{ 2 }+20xy & +5{ y }^{ 2 } \end{bmatrix}$$
which is not equal to given LHS

Now, option B, so let $$A=\begin{bmatrix} 10 & 3 \\ 5 & 10 \end{bmatrix}$$
$$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 10 & 3 \\ 5 & 10 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$

$$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 10x+3y \\ 5x+10y \end{bmatrix}$$

$$=\begin{bmatrix} 3{ x }^{ 2 }+8xy & +5{ y }^{ 2 } \end{bmatrix}$$
which is not equal to given LHS

Now, we will try option C
Let $$A=\begin{bmatrix} 3 & -5 \\ -5 & 5 \end{bmatrix}$$

$$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -5 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$

$$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x-5y \\ -5x+5y \end{bmatrix}$$

$$=\begin{bmatrix} 3{ x }^{ 2 }-10xy & +5{ y }^{ 2 } \end{bmatrix}$$
which is not equal to given LHS

Now, lastly we will try option D.
Let $$A=\begin{bmatrix} 3 & 5 \\ 5 & 5 \end{bmatrix}$$

$$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & 5 \\ 5 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$

$$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x+5y \\ 5x+5y \end{bmatrix}$$

$$=\begin{bmatrix} 3{ x }^{ 2 }+10xy & +5{ y }^{ 2 } \end{bmatrix}$$
which is equal to given LHS
Question 7
1 / -0

$$1\mathrm{f}\mathrm{A}=\left[\begin{array}{lll}
4 & 1 & 0\\
1 & -2 & 2
\end{array}\right],\ \mathrm{B}=\left[\begin{array}{lll}
2 & 0 & -1\\
3 & 1 & 4
\end{array}\right]$$, $$\mathrm{C}=\left[\begin{array}{l}
1\\
2\\
-1
\end{array}\right]$$ and $$(3\mathrm{B}-2\mathrm{A})\mathrm{C}+2\mathrm{X}=0$$ then $$\mathrm{X}$$ is equal to

A
$${\displaystyle \frac{1}{2}} \begin{bmatrix} 3\\ 13 \end{bmatrix}$$

B
$${\displaystyle \frac{1}{2}}\begin{bmatrix} 3\\ -13 \end{bmatrix}$$

C
$${\displaystyle \frac{1}{2}} \begin{bmatrix} -3\\ 13 \end{bmatrix}$$

D
$$\begin{bmatrix} 3\\ -13 \end{bmatrix}$$

Solution

Given, $$ A =\begin{bmatrix} 4 & 1 & 0 \\ 1 & -2 & 2 \end{bmatrix},B=\begin{bmatrix} 2 & 0 & -1 \\ 3 & 1 & 4 \end{bmatrix},{ C }=\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$

Now, $$(3{ B }-2{ A }){ C }+2{ X}=0$$
$$\Rightarrow \left( \begin{bmatrix} 6 & 0 & -3 \\ 9 & 3 & 12 \end{bmatrix}-\begin{bmatrix} 8 & 2 & 0 \\ 2 & -4 & 4 \end{bmatrix} \right) \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}+2X=0$$

$$\begin{bmatrix} -2 & -2 & -3 \\ 7 & 7 & 8 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}+2X=0$$

$$\begin{bmatrix} -3 \\ 13 \end{bmatrix}+2X=0$$

$$\Rightarrow 2X=\begin{bmatrix} 3 \\ -13 \end{bmatrix}$$

$$\Rightarrow X=\displaystyle \frac{1}{2}\begin{bmatrix} 3 \\ -13 \end{bmatrix}$$
Question 8
1 / -0

$$A=\begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}$$ and $$\mathrm{A}^{2}=\lambda I$$ then $$\lambda=$$

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$-2$$

Solution

$$A=\begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}$$

$${ A }^{ 2 }=\begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}\begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}$$

$$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

But given $$ { A }^{ 2 }=\lambda I$$
$$\Rightarrow \lambda=0$$
Question 9
1 / -0

$$A=\left[\begin{array}{lll}
2 & 2 & 1\\
1 & 2 & 1\\
3 & 4 & 2
\end{array}\right]$$ then ($$\mathrm{A}-\mathrm{I}$$) $$(\mathrm{A}-2I)=$$

A
$$\left[\begin{array}{lll}

5 & 6 & -2\\

1 & 0 & 7\\

0 & 1 & 1

\end{array}\right]$$

B
$$\left[\begin{array}{lll}

5 & 6 & 3\\

4 & 6 & 2\\

7 & 10 & 7

\end{array}\right]$$

C
$$\left[\begin{array}{lll}

1 & 0 & 7\\

7 & 10 & 7\\

4 & 6 & 2

\end{array}\right]$$

D
$$\left[\begin{array}{lll}

-1 & 1 & 2\\

3 & 4 & 1\\

1 & 1 & 2

\end{array}\right]$$

Solution

$$A-I=\left[\begin{array}{}
1 & 2 & 1\\
1 & 1 & 1\\
3 & 4 & 1
\end{array}\right]$$
$$A-2I=\left[\begin{array}{}
0 & 2 & 1\\
1 & 0 & 1\\
3 & 4 & 0
\end{array}\right]$$
$$\therefore (A-I)(A-2I)=\left[\begin{array}{}
1 & 2 & 1\\
1 & 1 & 1\\
3 & 4 & 1
\end{array}\right]\left[\begin{array}{}
0 & 2 & 1\\
1 & 0 & 1\\
3 & 4 & 0
\end{array}\right]=\left[\begin{array}{}
5 & 6 & 3\\
4 & 6 & 2\\
7 & 10 & 7
\end{array}\right]$$
Question 10
1 / -0

lf $$A=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}$$ and $$ B=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}$$ then

A
$$\mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}$$

B
$$\mathrm{A}\mathrm{B}=-\mathrm{A}\mathrm{B}$$

C
$$\mathrm{A}\mathrm{B}=2\mathrm{B}\mathrm{A}$$

D
$$\mathrm{A}\mathrm{B}=3\mathrm{B}\mathrm{A}$$

Solution

Given, $$A=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}, B=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}$$

$$AB=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}$$

$$=\begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix}$$

$$BA=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}$$

$$=\begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix}$$

Hence, $$AB=BA$$

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Re-Attempt Weekly Quiz Competition

Matrices Test - 26

Result

Matrices Test - 26

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Question 1

$$-1$$

$$0$$

$$1$$

$$4$$

Solution

Question 2

$$0$$

$$3$$

$$6$$

$$8$$

Solution

Question 3

$$\left[\begin{array}{ll}12 & 4\\12 & 4\end{array}\right]$$

$$\left[\begin{array}{ll}12 & -4\\4 & 12\end{array}\right]$$

$$\left[\begin{array}{ll}4 & 12\\12 & 4\end{array}\right]$$

$$\left[\begin{array}{ll}4 & 12\\-12 & -4\end{array}\right]$$

Solution

Question 4

$$\begin{bmatrix} 3 & 0 & 0\\0 & 1 & -1\\0 & 5 & 4\end{bmatrix}$$

$$ \begin{bmatrix} 1 & 2 & -1\\-3 & 1 & 1\\3 & 5 & -4\end{bmatrix}$$

$$\begin{bmatrix} 3 & 1 & -1\\-3 & 1 & -1\\3 & 5 & 4\end{bmatrix}$$

$$\begin{bmatrix} 3 & -1 & 1\\3 & -1 & 1\\3 & 5 & 4\end{bmatrix}$$

Solution

Question 5

$$10$$

$$20$$

$$30$$

$$40$$

Solution

Question 6

$$\left[\begin{array}{ll}3 & 10\\10 & 5\end{array}\right]$$

$$\left[\begin{array}{ll}10 & 3\\5 & 10\end{array}\right]$$

$$\left[\begin{array}{ll}+3 & -5\\-5 & +5\end{array}\right]$$

$$\left[\begin{array}{ll}3 & 5\\5 & 5\end{array}\right]$$

Solution

Question 7

$${\displaystyle \frac{1}{2}} \begin{bmatrix} 3\\ 13 \end{bmatrix}$$

$${\displaystyle \frac{1}{2}}\begin{bmatrix} 3\\ -13 \end{bmatrix}$$

$${\displaystyle \frac{1}{2}} \begin{bmatrix} -3\\ 13 \end{bmatrix}$$

$$\begin{bmatrix} 3\\ -13 \end{bmatrix}$$

Solution

Question 8

$$0$$

$$1$$

$$\dfrac{1}{2}$$

$$-2$$

Solution

Question 9

$$\left[\begin{array}{lll}5 & 6 & -2\\1 & 0 & 7\\0 & 1 & 1\end{array}\right]$$

$$\left[\begin{array}{lll}5 & 6 & 3\\4 & 6 & 2\\7 & 10 & 7\end{array}\right]$$

$$\left[\begin{array}{lll}1 & 0 & 7\\7 & 10 & 7\\4 & 6 & 2\end{array}\right]$$

$$\left[\begin{array}{lll}-1 & 1 & 2\\3 & 4 & 1\\1 & 1 & 2\end{array}\right]$$

Solution

Question 10

$$\mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}$$

$$\mathrm{A}\mathrm{B}=-\mathrm{A}\mathrm{B}$$

$$\mathrm{A}\mathrm{B}=2\mathrm{B}\mathrm{A}$$

$$\mathrm{A}\mathrm{B}=3\mathrm{B}\mathrm{A}$$

Solution

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$$\left[\begin{array}{ll}

12 & 4\\

12 & 4

\end{array}\right]$$

$$\left[\begin{array}{ll}

12 & -4\\

4 & 12

\end{array}\right]$$

$$\left[\begin{array}{ll}

4 & 12\\

12 & 4

\end{array}\right]$$

$$\left[\begin{array}{ll}

4 & 12\\

-12 & -4

\end{array}\right]$$

$$\begin{bmatrix} 3 & 0 & 0\\

0 & 1 & -1\\

0 & 5 & 4

\end{bmatrix}$$

$$ \begin{bmatrix} 1 & 2 & -1\\

-3 & 1 & 1\\

3 & 5 & -4

\end{bmatrix}$$

$$\begin{bmatrix} 3 & 1 & -1\\

-3 & 1 & -1\\

3 & 5 & 4

\end{bmatrix}$$

$$\begin{bmatrix} 3 & -1 & 1\\

3 & -1 & 1\\

3 & 5 & 4

\end{bmatrix}$$

$$\left[\begin{array}{ll}

3 & 10\\

10 & 5

\end{array}\right]$$

$$\left[\begin{array}{ll}

10 & 3\\

5 & 10

\end{array}\right]$$

$$\left[\begin{array}{ll}

+3 & -5\\

-5 & +5

\end{array}\right]$$

$$\left[\begin{array}{ll}

3 & 5\\

5 & 5

\end{array}\right]$$

$$\left[\begin{array}{lll}

5 & 6 & -2\\

1 & 0 & 7\\

0 & 1 & 1

\end{array}\right]$$

$$\left[\begin{array}{lll}

5 & 6 & 3\\

4 & 6 & 2\\

7 & 10 & 7

\end{array}\right]$$

$$\left[\begin{array}{lll}

1 & 0 & 7\\

7 & 10 & 7\\

4 & 6 & 2

\end{array}\right]$$

$$\left[\begin{array}{lll}

-1 & 1 & 2\\

3 & 4 & 1\\

1 & 1 & 2

\end{array}\right]$$