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(C) Statement $-1$: $A$ is a $3 	imes 3$ matrix and $A^{-1}$ exists. $B$ is a $3 	imes 4$ matrix. The product $A^{-1}B$ involves a $(3 	imes 3) 	i

Vedclass · Accepted Answer

$TFFT$

Vedclass · Answer

(C) Statement $-1$: $A$ is a $3 	imes 3$ matrix and $A^{-1}$ exists. $B$ is a $3 	imes 4$ matrix. The product $A^{-1}B$ involves a $(3 	imes 3) 	imes (3 	imes 4)$ matrix,which is defined and results in a $3 	imes 4$ matrix. Thus,Statement $-1$ is $T$. Statement $-2$: If $A$ and $B$ are both $n 	imes n$ matrices,then $A+B, A-B$,and $AB$ are all defined. Thus,it is possible for them to be defined. Statement $-2$ is $F$. Statement $-3$: Consider the matrix $A = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}$. All entries are non-zero,but $\det(A) = 1(1) - 1(1) = 0$. Thus,$A$ is not invertible. Statement $-3$ is $F$. Statement $-4$: An invertible matrix must be square (by definition). If two rows were the same,the determinant would be $0$,making it non-invertible. Thus,an invertible matrix cannot have two identical rows. Statement $-4$ is $T$. The correct order is $TFFT$.

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