Let A be a 2 times 2 real matrix with entries | vedclass

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Question

(D) Given $A$ is a $2 	imes 2$ matrix with entries from $\{0, 1\}$ and $|A| 
eq 0$.For statement $(P)$: If $A 
eq I_{2}$,then $|A| = -1$.Consider $

Vedclass · Accepted Answer

$(P)$ is false and $(Q)$ is true

Vedclass · Answer

(D) Given $A$ is a $2 	imes 2$ matrix with entries from $\{0, 1\}$ and $|A| 
eq 0$.For statement $(P)$: If $A 
eq I_{2}$,then $|A| = -1$.Consider $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$. Here $A 
eq I_{2}$ and $|A| = (1)(1) - (1)(0) = 1$.Since we found a case where $A 
eq I_{2}$ but $|A| = 1$,statement $(P)$ is false.For statement $(Q)$: If $|A| = 1$,then $\operatorname{tr}(A) = 2$.The possible $2 	imes 2$ matrices with entries from $\{0, 1\}$ such that $|A| = 1$ are:$A_1 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \implies \operatorname{tr}(A_1) = 1+1 = 2$$A_2 = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \implies \operatorname{tr}(A_2) = 1+1 = 2$$A_3 = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} \implies \operatorname{tr}(A_3) = 1+1 = 2$In all cases where $|A| = 1$,the trace is $2$. Thus,statement $(Q)$ is true.Therefore,$(P)$ is false and $(Q)$ is true.

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