Matrices Test

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Matrices Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{bmatrix} 2 & -1\\ -7 & 4\end{bmatrix}$$ and $$B=\begin{bmatrix} 4 & 1\\ 7 & 2\end{bmatrix}$$ then $$B^TA^T$$ is equal to?

A
$$\begin{bmatrix} 1 & 0\\ -12 & 1\end{bmatrix}$$

B
$$\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$$

C
$$\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$$

D
$$\begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}$$

Solution
Question 2
1 / -0

If $$A^2-A+1=0$$, then the inverse of A is?

A
A

B
$$A+I$$

C
$$I-A$$

D
$$A-I$$

Solution
Question 3
1 / -0

Let $$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
If $$A=\begin{bmatrix} 1 & n\\ 0 & 1\end{bmatrix}$$ then $$A^{-1}=?$$

A
$$\begin{bmatrix} 1 & 12\\ 0 & 1\end{bmatrix}$$

B
$$\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$

C
$$\begin{bmatrix} 1 & -12\\ 0 & 1\end{bmatrix}$$

D
$$\begin{bmatrix} 1 & 0\\ -13 & 1\end{bmatrix}$$

Solution

$$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}..\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
$$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$$
$$A=\begin{bmatrix} 1 & 13\\ 0 & 1\end{bmatrix}$$
so $$A^{-1}=\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$.
Question 4
1 / -0

If $$\begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}=\begin{bmatrix} 2 \\ 4\end{bmatrix}$$, then the values of $$x $$ and $$y $$ respectively are?

A
$$-3, -1$$

B
$$1, 3$$

C
$$3, 1$$

D
$$-1, 3$$

Solution

On equating given two matrices, we get
$$x+y=2$$ and $$-x+y=4$$

Solving for $$x, y : x=-1, y=3$$.
Question 5
1 / -0

If $$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} ..\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$, then the inverse of $$\begin{bmatrix} 1 & n\\ 0 & 1\end{bmatrix}$$ is?

A
$$\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$

B
$$\begin{bmatrix} 1 & 0\\ 12 & 1\end{bmatrix}$$

C
$$\begin{bmatrix} 1 & -12\\ 0 & 1\end{bmatrix}$$

D
$$\begin{bmatrix} 1 & 0\\ 13 & 1\end{bmatrix}$$

Solution

$$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 1 & 1+2+3+...+n-1\\ 0 & 1\end{bmatrix}=$$ $$\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
$$\Rightarrow \dfrac{n(n-2)}{2}=78\Rightarrow n=13$$, $$-12$$ (reject)
$$\therefore$$ We have to find inverse of $$\begin{bmatrix} 1 & 13\\ 0 & 1\end{bmatrix}$$
$$\therefore \begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$
Question 6
1 / -0

If $$A = \begin{bmatrix} n& 0 & 0\\ 0 & n & 0\\ 0 & 0 & n\end{bmatrix}$$ and $$B = \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$$, then $$AB$$ is equal to

A
$$B$$

B
$$nB$$

C
$$B^{n}$$

D
$$A + B$$

Solution

Let, $$A = \begin{bmatrix} n& 0 & 0\\ 0 & n & 0\\ 0 & 0 & n\end{bmatrix}$$ and

$$B = \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$$.

Now,

$$AB = \begin{bmatrix} n& 0 & 0\\ 0 & n & 0\\ 0 & 0 & n\end{bmatrix}$$ $$ \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$$

or, $$A B= \begin{bmatrix}na_{1} & na_{2} & na_{3}\\ nb_{1} & nb_{2} & nb_{3}\\ nc_{1} & nc_{2} & nc_{3}\end{bmatrix}$$

or, $$AB=nB$$.
Question 7
1 / -0

If $$A = \begin{bmatrix}1 & 2 & x\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$ and $$B = \begin{bmatrix}1 & -2 & y\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$ and $$AB = I_{3}$$, then $$x + y=$$ _____.

A
$$0$$

B
$$-1$$

C
$$2$$

D
None of these

Solution
Question 8
1 / -0

If $$A = \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ a & b & -1\end{bmatrix}$$, then $$A^{2}$$ is equal to

A
null matrix

B
unit matrix

C
$$-A$$

D
$$A$$

Solution
Question 9
1 / -0

If $$A = \begin{bmatrix}3 &-4 \\ -1 & 2\end{bmatrix}$$ and $$B$$ is a square matrix of order $$2$$ such that $$AB = I$$ then $$B = ?$$

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

B
$$\begin{bmatrix}1 &\dfrac {1}{2} \\ 2 & \dfrac {3}{2}\end{bmatrix}$$

C
$$\begin{bmatrix}1 &2 \\ \dfrac {1}{2} & \dfrac {3}{2}\end{bmatrix}$$

D
None of these

Solution
Question 10
1 / -0

If $$ \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}A\begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$ , then A =

A
$$ \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $$

B
$$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $$

C
$$ \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} $$

D
$$ - \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $$

Solution

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

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

$$ \Rightarrow A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}(-1)\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix}$$

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