If $$A$$ and $$B$$ are two matrices of the same order, then

$$A^2=I\Leftrightarrow (A + I) (A -I) = 0$$

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 6

Given

$A, B, C$ are three matrices such that

$A = \begin{bmatrix}x & y & z\end{bmatrix}$ ,

$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$ ,

$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix}.$ Evaluate

$ABC$ .

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$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$
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$ABC = \begin{bmatrix}ax^2 - by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$
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$ABC = \begin{bmatrix}ax^2 - by^2 + cz^2 - 2hxy + 2gzx + 2fyz\end{bmatrix}$
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$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 - 2hxy + 2gzx - 2fyz\end{bmatrix}$

Explanation

$A = \begin{bmatrix}x & y & z\end{bmatrix}$ ,

$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$ ,

$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix}$

$AB=\begin{bmatrix}ax+hy+gz & hx+by+z & gx+fy+cz\end{bmatrix}$

$ABC=\begin{bmatrix}ax+hy+gz & hx+by+z & gx+fy+cz\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$

$=\begin{bmatrix}ax^2 + by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$
Hence, option A.

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

$B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$ . Find

$AB$ and show that

$AB \ne BA$

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$AB = \begin{bmatrix} 0 & -4 \\ 10 & -3 \end{bmatrix}$
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$AB = \begin{bmatrix} 0 & 4 \\ 10 & 3 \end{bmatrix}$
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$AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$
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None of these.

Explanation

Given,

$A=\begin{bmatrix} 1\quad & -2\quad & 3 \\ -4 & 2 & 5 \end{bmatrix},B=\begin{bmatrix} 2\quad & 3 \\ 4\quad & 5 \\ 2\quad & 1 \end{bmatrix}\\ AB=\begin{bmatrix} 1\quad & -2\quad & 3 \\ -4 & 2 & 5 \end{bmatrix}\begin{bmatrix} 2\quad & 3 \\ 4\quad & 5 \\ 2\quad & 1 \end{bmatrix}=\begin{bmatrix} 2-8+6\quad & 3-10+3 \\ -8+8+10\quad & -12+10+5 \end{bmatrix}=\begin{bmatrix} 0\quad & -4 \\ 10\quad & 3 \end{bmatrix}\\ BA=\begin{bmatrix} 2\quad & 3 \\ 4\quad & 5 \\ 2\quad & 1 \end{bmatrix}\begin{bmatrix} 1\quad & -2\quad & 3 \\ -4 & 2 & 5 \end{bmatrix}=\begin{vmatrix} 2-12\quad & -4+6\quad & 6+15 \\ 4-20 & -8+10\quad & 12+25 \\ 2-4 & -4+2 & -6+5 \end{vmatrix}\begin{bmatrix} -10\quad & 2\quad & 21 \\ -16 & 2 & 37 \\ -2 & -2 & -1 \end{bmatrix}$

Hence

$AB\neq BA$

$\displaystyle \begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix} \: \begin{bmatrix} 2 \\ -2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$ then

$x-y$ is equal to

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$2$
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$-2$
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$4$
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$6$

Explanation

Comparing LHS and RHS gives us

$2x+2y-2y=3$ ...(i)

And

$4x-2x+2y=2$ ...(ii)

from i

$2x=3$ and from ii

$2x+2y=2$

Therefore

$3+2y=3$

$y=\dfrac{-1}{2}$ and

$x=\dfrac{3}{2}$

Hence

$x-y=\dfrac{3}{2}-(\dfrac{-1}{2})$

$=\dfrac{4}{2}$

$=2$ .

$\displaystyle \left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\: \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O$ then

$x$ is

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$2$
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$-2$
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$14$
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none of these

Explanation

$\displaystyle \left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\: \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O$

$\Rightarrow \begin{bmatrix} 16+2x & \quad 6+5x & \quad 4+x \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O\\ \Rightarrow { x }^{ 2 }+16x+28\quad =\quad 0\\ \therefore \quad x=-2\quad or\quad x=-4$

$\displaystyle A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \\ 0 & 6 \end{bmatrix}$ and

$\displaystyle B=\begin{bmatrix} 5 & 4 & 6 \\ 4 & 1 & 2 \\ -5 & -1 & 1 \end{bmatrix}$ , then

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$A + B$ exists
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$AB$ exists
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$BA$ exists
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none of these

Explanation

$\displaystyle A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \\ 0 & 6 \end{bmatrix}$ and

$\displaystyle B=\begin{bmatrix} 5 & 4 & 6 \\ 4 & 1 & 2 \\ -5 & -1 & 1 \end{bmatrix}$

Order of

$A$ is

$3\times2$
Order of

$B$ is

$3\times3$

Order of

$A$ and

$B$ are not same. So ,

$A+B$ does not exists.

Again ,

$AB$ also does not exists as the number of columns of

$A$ are not equal to the number of rows of

$B.$

Here,

$BA$ exists as the number of columns of

$B$ is equal to the number of rows of

$A$ .

$\displaystyle A = \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}$ and

$\displaystyle B = \begin{bmatrix} a^2 & ab & ac \\ ba & b^2 & bc \\ ca & cb & c^2 \end{bmatrix}$ then

$AB$ is equal to

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$[0]$
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$I$
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$2I$
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none of these

Explanation

$\displaystyle A = \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}$ and

$\displaystyle B = \begin{bmatrix} a^2 & ab & ac \\ ba & b^2 & bc \\ ca & cb & c^2 \end{bmatrix}$

Now,

$AB=\begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}\begin{bmatrix} a^{ 2 } & ab & ac \\ ba & b^{ 2 } & bc \\ ca & cb & c^{ 2 } \end{bmatrix}$

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

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

$AB$ =

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$\begin{bmatrix}5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}$
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$\begin{bmatrix}5 & 9 & 13 \\ -1 & -2 & 4 \\ 2 & -2 & -4\end{bmatrix}$
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$\begin{bmatrix}5 & -9 & 13 \\ 1 & -2 & 4 \\ 2 & -2 & 4\end{bmatrix}$
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$\begin{bmatrix}5 & -9 & 13 \\ 1 & -2 & 4 \\ -2 & 2 & 4\end{bmatrix}$

Explanation

Given,

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

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

$AB=\begin{bmatrix}1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix}2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \end{bmatrix}=\begin{bmatrix}5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}$

Hence, option A is correct.

Given

$A = \begin{bmatrix}1 & -1 \\ 2 & -1\end{bmatrix}$ , which of the following result is true?

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$A^2 = I$
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$A^2 = -I$
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$A^2 = 2I$
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None of these

Explanation

Given,

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

Now,

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

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

$\Rightarrow A^{2}=-I$

$A = \begin{bmatrix}4 & x+2 \\ 2x-3 & x+1\end{bmatrix}$ is symmetric, then

$x$ =

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$3$
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$5$
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$2$
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$4$

Explanation

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

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

$A$ is symmetric.

$\therefore\quad A=A^{T}$

$\Rightarrow 2x-3=x+2$

$\therefore x=5$
Hence, option B.

$\displaystyle A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}$ then

$\displaystyle A^2$ is equal to

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$A$
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$I$
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$\displaystyle A^T$
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none of these

Explanation

Given,

$\displaystyle A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}$

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

$\Rightarrow A^{2}=I$

$A$ be a matrix such that

$\displaystyle A \times \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$ then

$A$ is

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$\displaystyle \begin{bmatrix} 2 & 4 \\ 1 & -1 \end{bmatrix}$
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$\displaystyle \begin{bmatrix} -1 & 1 \\ 4 & 2 \end{bmatrix}$
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$\displaystyle \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$
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none of these

Explanation

$\displaystyle A \times \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

We will check using options
Option A
Consider,

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

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

$=\begin{bmatrix} 6 & 12 \\ 0 & -6 \end{bmatrix}$

$\ne RHS$

Hence, option

$A$ cannot be matrix

$A$ .

Option B,
Consider,

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

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

$=\begin{bmatrix} 0 & 6 \\ 6 & 0 \end{bmatrix}$

$\ne RHS$
Hence, option

$B$ cannot be matrix

$A$ .

Option C,
Consider,

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

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

$=\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

$= RHS$
Hence, option C is correct for

$A$ .

$\displaystyle A=\begin{bmatrix}0 &c &-b \\-c &0 &a \\b &-a &0 \end{bmatrix}$ and

$\displaystyle B=\begin{bmatrix}a^{2} &ab &ac \\ab &b^{2} &bc \\ac &bc &c^{2} \end{bmatrix},$ then

$AB=$

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$\displaystyle A^{3}$
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$\displaystyle B^{2}$
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$I$
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$O$

Explanation

Given :

$\displaystyle A=\begin{bmatrix}0 &c &-b \\-c &0 &a \\b &-a &0 \end{bmatrix}$

$\displaystyle B=\begin{bmatrix}a^{2} &ab &ac \\ab &b^{2} &bc \\ac &bc &c^{2} \end{bmatrix}$

Then,

$\displaystyle AB= \begin{bmatrix} 0& c & -b\\ -c & 0 & a\\b &-a &0 \end{bmatrix}\begin{bmatrix} a^{2} & ab & ac\\ ab & b^{2} & bc \\ ac & bc & c^{2}\end{bmatrix}$

$=\begin{bmatrix} abc-abc & b^{2}c-b^{2}c & bc^{2}-bc^{2} \\ -a^{2}c+a^{2}c & -abc+abc & -ac^{2}+ac^{2} \\ a^2b-a^2b & ab^2-ab^2 & abc-abc \end{bmatrix}$

$=O$

$A = \begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$

If

$A^2 - 4A =pI$ where

$I$ and

$O$ are the unit matrix and the null matrix of order

$3$ respectively. Find the value of

$p$

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$p=2$
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$p=3$
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$p=4$
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$p=5$

Explanation

Given

$\quad A = \begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$

$\therefore A^2 = A.A = \begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}\times\begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$

$\quad = \begin{pmatrix}1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{pmatrix}$

$\quad = \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{pmatrix}$

$\therefore \quad A^2 - 4A = \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{pmatrix} - 4\begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$

$\quad = \begin{pmatrix}9&8&8 \\ 8&9&8 \\ 8&8&9\end{pmatrix} - \begin{pmatrix}4&8&8 \\ 8&4&4 \\ 4&4&8\end{pmatrix} = \begin{pmatrix}5&0&0 \\ 0&5&0 \\ 0&0&5\end{pmatrix}=5I$

Let

$\displaystyle A = \begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}$ . The only correct statement about the matrix

$A$ is

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$\displaystyle A$ is a zero matrix
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$\displaystyle A = \left ( -1 \right )I_{3}$
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$A^2=-I$
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$\displaystyle A^{2} = I$

Explanation

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

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

$\therefore A^2=I$
Hence, option D.

$A$ is a square matrix then

$A-{A}'$ is a

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diagonal matrix
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skew symmetric matrix
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symmetric matrix
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None of these

Explanation

Consider,

$(A-A')'=A'-(A')'$

$=A'-A$

$=-(A-A')$

$\Rightarrow (A-A')'=-(A-A')$
Hence,

$A-A'$ is skew-symmetric

Let

$\displaystyle A=\begin{bmatrix}-1\\2\\3 \end{bmatrix}$ and

$\displaystyle B=\begin{bmatrix} -2 & -1 & -4 \end{bmatrix},$ then matrix

$(AB) A$ equals

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$12A$
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$-12A$
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$4A$
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$3A$

Explanation

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

$AB=\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}$

$(AB)A=\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$

$\Rightarrow (AB)A=\begin{bmatrix} 12 \\ -24 \\ -36 \end{bmatrix}=-12\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}=-12A$

Use the method of elementary row transformation to compute the inverse of

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

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$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$
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$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{1}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{11}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{2}{21}\end{bmatrix}$
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$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{16}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{4}{21}\end{bmatrix}$
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$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{2}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{4}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

Explanation

Let

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

$\Rightarrow \quad Write \space A A^{-1}= I$

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

$\quad \begin{matrix}R_{21}(-2)\\ \mbox{~}\\ R_{31}(1)\end{matrix}\begin{bmatrix}1 & 0 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \\ -2 & 1 & 0 \\ 1 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R_2(-1) \\ \mbox{~} \\ R_3(1/3)\end{matrix}\begin{bmatrix}1 & 2 & 5 \\ 0 & 1 & 9 \\ 0 & 1 & 2\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \\ 2 & -1 & 0 \\ \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R_{12}(-2) \\ \mbox{~} \\ R_{32}(-1)\end{matrix}\begin{bmatrix}1 & 0 & -13 \\ 0 & 1 & 9 \\ 0 & 0 & -7\end{bmatrix} A^{-1}= \begin{bmatrix}-3 & 2 & 0 \\ 2 & -1 & 0 \\ -\displaystyle\frac{5}{3} & 1 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R_3(-1/7)\\ \mbox{~}\end{matrix}\begin{bmatrix}1 & 0 & -13 \\ 0 & 1 & 9 \\ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}-3 & 2 & 0 \\ 2 & -1 & 0 \\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

$\quad \begin{matrix}R_{13}(13) \\ \mbox{~} \\ R_{23}(-9)\end{matrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

Hence,

$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

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Both (A) & (R) are individually true & (R) is correct explanation of (A),
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Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
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(A)is true but (R} is false,
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(A)is false but (R} is true.

Explanation

$\displaystyle A=\begin{pmatrix}0 &a &b \\-a &0 &c \\-b &-c &0 \end{pmatrix}$

$\displaystyle \therefore A^{T}=\begin{pmatrix}0 &-a &-b \\a &0 &-c \\b &c &0 \end{pmatrix}$

$\displaystyle A^{T}=-A$

$\therefore$ Assertion (A) & Reason (R) both are true & Reason(R) is correct explanation of Assertion (A).

$\displaystyle A=\begin{bmatrix} 1 & \frac { 1 }{ 2 } \\ 0 & 1 \end{bmatrix}$ then

${ A }^{ 64 }$ is

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$\begin{bmatrix} 1 & 32 \\ 32 & 1 \end{bmatrix}$
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$\begin{bmatrix} 1 & 0 \\ 32 & 1 \end{bmatrix}$
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$\begin{bmatrix} 1 & 32 \\ 0 & 1 \end{bmatrix}$
0%
none of these

Explanation

$\displaystyle { A }^{ 2 }=\begin{bmatrix} 1 & \frac { 1 }{ 2 } \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac { 1 }{ 2 } \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\\ { A }^{ 4 }=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\\ { A }^{ 8 }=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix}$

Similarly,

${ A }^{ 64 }=\begin{bmatrix} 1 & 32 \\ 0 & 1 \end{bmatrix}$

$A =(a_{ij})_{3\times 3}$ is a skew symmetric matrix, then

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$a_{ii}=0\: \forall \: i$
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$A + A^T$ is a null matrix
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$|A|=0$
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$A$ is not invertible.

Explanation

Let

$A=\begin{bmatrix} a & \alpha & \beta \\ -\alpha & b & \gamma \\ -\beta & -\gamma & c \end{bmatrix}\quad \Rightarrow \quad -A=\begin{bmatrix} -a & -\alpha & -\beta \\ \alpha & -b & -\gamma \\ \beta & \gamma & -c \end{bmatrix}$

$A^{ T }=\begin{bmatrix} a & -\alpha & -\beta \\ \alpha & b & -\gamma \\ \beta & \gamma & c \end{bmatrix}$

since A is a skew-symmetric matrix

$A^{T}=-A$

$\Rightarrow a=-a,b=-b,c=-c$

$\Rightarrow a=b=c=0$

$\therefore a_{ii}=0$ for all i

$A^{T}=-A$

$\Rightarrow A^{T}+A=O$

$|A|=0$ (

$\because$ A is a skew-symmetric of odd order)

since

$|A|=0$ A is not invertible. (

$\because$ only non-singular matrices are invertible).

$\displaystyle A=\begin{bmatrix}a &b \\b &a \end{bmatrix}$ and

$\displaystyle A^{2} =\begin{bmatrix}\alpha & \beta \\\beta & \alpha \end{bmatrix}$ then

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$\displaystyle \alpha=a^{2}+b^{2},\beta =2ab$
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$\displaystyle \alpha=a^{2}+b^{2},\beta =a^{2}-b^{2}$
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$\displaystyle \alpha=2ab ,\beta =a^{2}+b^{2}$
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$\displaystyle \alpha=a^{2}+b^{2},\beta =ab$

Explanation

$\displaystyle A^{2}=AA=\begin{pmatrix}a &b \\b &a \end{pmatrix}\begin{pmatrix}a &b \\b &a \end{pmatrix}$

$\displaystyle =\begin{pmatrix}a^{2}+b^{2} &2ab \\2ab &a^{2}+b^{2} \end{pmatrix}$

$\displaystyle = \begin{pmatrix} \alpha &\beta \\\beta &\alpha \end{pmatrix}$

$\omega \neq 1$ is cube root of unity, and

$A=\begin{bmatrix} 1&\omega &\omega ^2 \\\omega & \omega ^2 &1 \\ \omega ^2 &1 &\omega \end{bmatrix}$ is

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symmetric
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skew symmetric
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singular
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orthogonal

Explanation

$A=\begin{bmatrix} 1&\omega &\omega ^2 \\\omega & \omega ^2 &1 \\ \omega ^2 &1 &\omega \end{bmatrix}$

$A^{T}=\begin{bmatrix} 1&\omega &\omega ^2 \\\omega & \omega ^2 &1 \\ \omega ^2 &1 &\omega \end{bmatrix}$

$\Rightarrow A^{T}=A$

i.e

$A$ is a symmetric matrix.

$|A|=\begin{vmatrix} 1&\omega &\omega ^2 \\\omega & \omega ^2 &1 \\ \omega ^2 &1 &\omega \end{vmatrix}=0$

i.e

$A$ is singular.

Hence, options A and C.

Let

$A$ be a symmetric matrix such that

$A^5 =0$ and

$B=I +A + A^2 +A^3 +A^4$ , then

$B$ is

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symmetric
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singular
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non-singular
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skew symmetric

Explanation

Given

$A$ be a symmetric matrix such that

$A^5 =0$ and

$B=I +A + A^2 +A^3 +A^4$

$B^{-1}=(I+A+A^2+A^3+A^4)^{-1}=I'+A^{-1}+(A^{-1})^2+(A^{-1})^3+(A^{-1})^4=I +A + A^2 +A^3 +A^4=B$

$\therefore B$ is symmetric.

$BA=(I +A + A^2 +A^3 +A^4)A=A + A^2 +A^3 +A^4+A^5=B-I$

$\Rightarrow BA=B-I$

$\Rightarrow B(I-A)=I \Rightarrow B^{-1}=I-A$

$B^{-1}$ exists.

$\therefore B$ is non-singualr.
Hence, options A and C.

$A =\begin{bmatrix}\alpha &\beta \\\gamma &-\alpha \end{bmatrix}$ is such that

$A^2 = I$ , then

0%
$1 +\alpha^2+\beta\gamma=0$
0%
$1 -\alpha^2-\beta\gamma=0$
0%
$1 -\alpha^2+\beta\gamma=0$
0%
$1 +\alpha^2-\beta\gamma=0$

Explanation

$A =\begin{bmatrix}\alpha &\beta \\\gamma &-\alpha \end{bmatrix}$

$A^2=\begin{bmatrix}\alpha^2+\beta \gamma &\alpha \beta-\alpha \beta \\\alpha \gamma - \alpha \gamma&\beta \gamma +\alpha^2 \end{bmatrix}$
but,

$A^2=I$

$\Rightarrow \alpha ^2+\beta \gamma = 1$

$\Rightarrow 1 -\alpha ^2-\beta \gamma=0$
Hence, option B is correct.

$A$ and

$B$ are two matrices of the same order, then

0%
$A^2 -B^2 = (A + B) (A -B)$
0%
$A^2=I\Leftrightarrow (A + I) (A -I) = 0$
0%
$(A')' = A$
0%
$A + A'$ is symmetric

Explanation

A and B

$\rightarrow$ Two matrices of same order

Option A:

${ A }^{ 2 }-{ B }^{ 2 }=\left( A+B \right) \left( A-B \right)$

RHS

$={ A }^{ 2 }-AB+BA+{ B }^{ 2 }$

RHS

$=$ LHS if and only if

$AB=BA$

$\rightarrow$ Option A is not true for all cases

Option B:

${ A }^{ 2 }=I\Leftrightarrow \left( A+I \right) \left( A-I \right) =0$

RHS

$={ A }^{ 2 }-A+A-I=0$

$\Rightarrow { A }^{ 2 }=I$

$\Rightarrow$ Option B is correct

Option C:

${ \left( { A }^{ T } \right) }^{ T }=A$

Transpose means we reverse the columns and rows

Let C be columns and R be rows

${ A }^{ T }$ means C becomes rows

${ \left( { A }^{ ' } \right) }^{ ' }$ mean columns and rows are again interchanged

$C\rightarrow$ columns

$R\rightarrow$ rows

$\therefore { \left( { A }^{ ' } \right) }^{ ' }=A$

Option C is correct

Option D Let

$A+{ A }^{ T }=P\quad { P }^{ T }={ \left( A+{ A }^{ ' } \right) }^{ ' }={ A }^{ ' }+A$

$\therefore P={ P }^{ T }\quad \therefore A+{ A }^{ T }={ \left( A+{ A }^{ T } \right) }^{ T }\Rightarrow$ symmetric

Option D is correct

$A =\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$ , then

$A^2$ is equal

0%
$O$
0%
$I$
0%
$-I$
0%
none of these

Explanation

$A =\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$

$A^2=\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$

$=\begin{bmatrix} a^{ 2 }b^{ 2 }-a^{ 2 }b^{ 2 } & ab^{ 3 }-ab^{ 3 } \\ -a^{ 3 }b+a^{ 3 }b & -a^{ 2 }b^{ 2 }+a^{ 2 }b^{ 2 } \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\therefore A^2=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} =O$
Hence, option A.

If A is any square matrix then (1/2)

$\displaystyle \left ( A+A^{T} \right )$ is a _____ matrix

0%
symmetric
0%
skew symmetric
0%
scalar
0%
identity

Explanation

Given

$\cfrac { 1 }{ 2 } \left( A+{ A }^{ T } \right)$

Let

$A+{ A }^{ T }=P$

Then

${ P }^{ T }={ \left( A+{ A }^{ T } \right) }^{ T }=A+{ A }^{ T }=P$

$\because { P }^{ T }=P$

$\Rightarrow { \left( A+{ A }^{ T } \right) }^{ T }=A+{ A }^{ T }$

$\Rightarrow \cfrac { 1 }{ 2 } \left( A+{ A }^{ T } \right)$ is symmetric

OPTION A

Find the value of

$x$ and

$y$ that satisfy the equation:

$\begin{bmatrix} 3&-2 \\3 &0 \\2 &4 \end{bmatrix} \begin{bmatrix} y&y \\x &x \end{bmatrix}=\begin{bmatrix} 3&3 \\3y &3y \\10 &10 \end{bmatrix}$

0%
$x=3/2, \:y = 2$
0%
$x=2, \:y = 3/2$
0%
$x=3, \:y = 2$
0%
$x=2, \:y = 3$

Explanation

$\left[ \begin{matrix} 3\quad & -2 \\ 3 & \quad 0 \\ 2 & \quad 4 \end{matrix} \right] \begin{bmatrix} y\quad & y \\ x\quad & x \end{bmatrix}=\left[ \begin{matrix} 3\quad & 3 \\ 3y\quad & 3y \\ 10\quad & 10 \end{matrix} \right] \\ \Rightarrow \left[ \begin{matrix} 3y-2x\quad & 3y-2x \\ 3y\quad & 3y \\ 2y+4x\quad & 2y+4x \end{matrix} \right] =\left[ \begin{matrix} 3 & \quad 3 \\ 3y\quad & 3y \\ 10\quad & 10 \end{matrix} \right]$

$\therefore 3y-2x=3$ and

$2y+4x=10$

$\displaystyle \Rightarrow x=\frac { 3 }{ 2 } ,y=2$

Let

$A$ be square matrix. Then which of the following is not a symmetric matrix.

0%
$A+A'$
0%
$AA'$
0%
$A'A$
0%
$A-A'$

Explanation

$A$ is a square matrix, then and if

$A'$ represents its transpose, then

$A+A'$ is symmetric and

$A-A'$ is skew symmetric.
Hence matrix A can be written as

$A=(\dfrac{A+A'}{2})+(\dfrac{A-A'}{2})$

Therefore of all the above matrix,

$A-A'$ is not symmetric.

Say true or false:

All positive odd integral powers of skew-symmetric matrix are symmetric.

0%
True
0%
False

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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