# Matrices - Class 12 - Mathematics

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## Extra Questions - Matrices | NCERT | Mathematics | Class 12

If $A$ is a matrix of order 3 and $|A|=8$, then $|Adj|A\rvert=$

A) 1

B) 2

C) $2^{5}$

D) $2^{6}$

The correct answer is **D) (2^{6})**.

To find the determinant of the adjugate matrix ( \text{adj}(A) ) when given ( |A| = 8 ) and knowing that ( A ) is a matrix of order 3, you use the property: $$ A \cdot \text{adj}(A) = \begin{bmatrix} |A| & 0 & 0 \ 0 & |A| & 0 \ 0 & 0 & |A| \end{bmatrix} $$ This implies: $$ |A| \cdot |\text{adj}(A)| = \left|\begin{bmatrix} |A| & 0 & 0 \ 0 & |A| & 0 \ 0 & 0 & |A| \end{bmatrix}\right| $$ The determinant of a diagonal matrix is the product of the diagonal elements. Thus: $$ |A| \cdot |\text{adj}(A)| = |A|^3 $$

Given that ( |A| = 8 ), we then have: $$ 8 \cdot |\text{adj}(A)| = 8^3 $$ or $$ |\text{adj}(A)| = \frac{8^3}{8} = 8^2 = 64 $$ Since ( 8 ) can be written as ( 2^3 ), we rewrite ( 8^2 ) as: $$ |\text{adj}(A)| = (2^3)^2 = 2^6 $$

Thus **( |\text{adj}(A)| = 2^6 )**, which corresponds to answer choice **D) (2^{6})**.

If $A \times A$, i.e., $A^{2} = 1,$ then $A$ is said to be an involutary matrix.

A) True

B) False

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In a factory, three types of toothbrushes are manufactured every day. On a certain day, the production is 570. The production of toothbrushes of the second kind exceeds the production of toothbrushes of the first kind by 100. Also, the total production of toothbrushes of the first and second kind is four times the production of the third kind.

If $x$, $y$, and $z$ denote the production of toothbrushes of three kinds, respectively, then using the matrix method, the algebraic representation of the given condition is:

(a) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ 100 \\ 0 \end{bmatrix} $

(b) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ 100 \\ 0 \end{bmatrix} $

(c) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ 100 \\ 0 \end{bmatrix} $

(d) $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 570 \\ -100 \\ 0 \end{bmatrix} $

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From the matrix equation $AB = AC$, we conclude that $B = C$ provided:

A is a symmetric matrix,

B is a singular matrix,

C is a skew-symmetric matrix,

D is a non-singular matrix.