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Consider a square matrix A of order 2 which has its elements as 0, 1, 2, and 4. Let N denote the number of such matrices, all elements of which are distinct.
Column AColumn B
A. Possible non-negative value of det (A) is(P) 2
B. Sum of values of determinants corresponding to N matrices is(Q) 4
C. If absolute value of (det (A)) is least, then possible value of $\left| adj\; adj\;\; adj\left( A \right) \right|$(R) -2
(S) 0
D. If det (A) is algebraically least, then possible value of $\det \left( 4{{A}^{-1}} \right)$ is(T) 8


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Last updated date: 13th Jun 2024
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Answer
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Hint: As we can see from the question, we are given a square matrix of order 2 having four distinct elements and we have to evaluate four parts. Since there are 4 elements in each matrix, so we will have 4!=24 such matrices. We will use a formula of determinants to find a solution of given parts, for (A) we will use a combination of any two elements (taking their product) to be the value of determinant because of zero being one of the elements of all matrices. For (B), we will calculate possible values of determinant and hence, add them to find a solution. For (C), we will choose least absolute value from values of determinant calculated earlier and then use the formula $\left| adj\; adj\; adj........adjA\left( r\text{ times} \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{r}}}}$ where n is the order of matrix to find $\left| adj\; adj\; adjA \right|$. For (D), we will choose least value from values of determinant calculated earlier and then use the formula $\det \left( a{{A}^{-1}} \right)={{a}^{n}}\det \left( {{A}^{-1}} \right),\left| {{A}^{-1}} \right|={{\left| A \right|}^{-1}}$ to find $\det \left( 4{{A}^{-1}} \right)$.

Complete step-by-step solution
Since we have 4 distinct elements in each matrix and therefore, there are 4!=24 such matrices. But possible values of determinant can only be six. Therefore, we can get possible values by
\[\begin{align}
  & \left| \begin{matrix}
   0 & 1 \\
   2 & 4 \\
\end{matrix} \right|=\left| \begin{matrix}
   0 & 2 \\
   1 & 4 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & 1 \\
   2 & 0 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & 2 \\
   1 & 0 \\
\end{matrix} \right|=-2 \\
 & \left| \begin{matrix}
   2 & 0 \\
   4 & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   2 & 4 \\
   0 & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & 0 \\
   4 & 2 \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & 4 \\
   0 & 2 \\
\end{matrix} \right|=2 \\
 & \left| \begin{matrix}
   0 & 1 \\
   4 & 2 \\
\end{matrix} \right|=\left| \begin{matrix}
   0 & 4 \\
   1 & 2 \\
\end{matrix} \right|=\left| \begin{matrix}
   2 & 4 \\
   1 & 0 \\
\end{matrix} \right|=\left| \begin{matrix}
   2 & 1 \\
   4 & 0 \\
\end{matrix} \right|=-4 \\
 & \left| \begin{matrix}
   1 & 2 \\
   0 & 4 \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & 0 \\
   2 & 4 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & 0 \\
   2 & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & 2 \\
   0 & 1 \\
\end{matrix} \right|=4 \\
 & \left| \begin{matrix}
   0 & 2 \\
   4 & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   0 & 4 \\
   2 & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & 4 \\
   2 & 0 \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & 2 \\
   4 & 0 \\
\end{matrix} \right|=-8 \\
 & \left| \begin{matrix}
   2 & 0 \\
   1 & 4 \\
\end{matrix} \right|=\left| \begin{matrix}
   2 & 1 \\
   0 & 4 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & 1 \\
   0 & 2 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & 0 \\
   1 & 2 \\
\end{matrix} \right|=8 \\
\end{align}\]
Hence, possible values are $-2, 2, -4, 4, -8, 8.$
A. Non negative values of det(A) are 2, 4, 8. Hence, for $\left( \text{A} \right)\to \left( \text{p} \right),\left( \text{q} \right),\left( \text{r} \right), \left( \text{s} \right)$ are true.
B. Sum of values of determinants corresponding to N matrices is $2+\left( -2 \right)+4+\left( -4 \right)+8+\left( -8 \right)=0$ hence, for $\left( \text{B} \right)\to \text{S}$ is true.
C. Least absolute value determinant of any such matrix as we can see is 2. Hence, $\left| A \right|=2$.
We have to calculate $\left| adj\;\; adj\;\; adj\;\;\left( A \right) \right|$. Since, we know that $\left| adj\;\; adj\;\; adj........adjA\left( r\text{ times} \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{r}}}}$ where n is order of matrix, we can use this to find $\left| adj\; adj\; adj\left( A \right) \right|$
Here, n = 2 because A is a $2\times 2$ matrix.
Also, r = 3 as we can see. Hence, $\left| adj\;\; adj\;\; adj\left( A \right) \right|={{\left| A \right|}^{{{\left( 2-1 \right)}^{3}}}}={{\left| A \right|}^{{{\left( 1 \right)}^{3}}}}=\left| A \right|$
But $\left| A \right|=2$
Hence, $\left| adj\; adj\; adj\left( A \right) \right|=2$
Hence, for $\left( \text{C} \right)\to \left( \text{p} \right)$ is true.
D. As we can see, the least algebraic value of determinant for such a matrix is -8. Therefore, $\left| A \right|=-8$ Possible value of $\left( 4{{A}^{-1}} \right)$ can be calculated as. As we know, $\left| a{{A}^{-1}} \right|={{a}^{n}}\left| {{A}^{-1}} \right|$ here a = 4 and n = 2.
Therefore, $\left| 4{{A}^{-1}} \right|={{4}^{2}}\left| {{A}^{-1}} \right|=16\left| {{A}^{-1}} \right|$
As we know, $\left| {{A}^{-1}} \right|={{\left| A \right|}^{-1}}$ so $\left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}$
But $\left| A \right|=-8$ so $\left| {{A}^{-1}} \right|=\dfrac{1}{-8}$
So $\left| 4{{A}^{-1}} \right|=16\left| {{A}^{-1}} \right|=\dfrac{16}{-8}=-2$
Hence for $\left( \text{D} \right)\to \left( \text{r} \right)$ is true.

Note: Students should carefully determine the values of determinants. They should learn basic formulas for calculating determinants. There is a very high chance of making mistakes in plus-minus signs while calculating determinants. Students should not get confused with the least absolute value; both are completely different. In absolute value, we don't care for signs, hence 2 is the least absolute value. In algebraic value, signs are given importance hence, -8 is the least algebraic value.