Match List - I with List - II.
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List - I |
List - II |
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(A) \begin{bmatrix}0 & -5 & 9\\5 & 0 & -3\\-9 & 3 & 0\end{bmatrix} |
(I) Scalar matrix |
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(B) \begin{bmatrix}5 & 0 & 0\\0 & 5 & 0\\0 & 0 & 5\end{bmatrix} |
(II) Diagonal matrix |
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(C) \begin{bmatrix}5 & 0 & 0\\0 & -5 & 0\\0 & 0 & 7\end{bmatrix} |
(III) Symmetric matrix |
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(D) \begin{bmatrix}3 & -2 & 1\\-2 & -5 & 6\\1 & 6 & 0\end{bmatrix} |
(IV) Skew-symmetric matrix |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
(A) $\begin{bmatrix}0 & -5 & 9\\5 & 0 & -3\\-9 & 3 & 0\end{bmatrix}$
for diagonal = $\begin{bmatrix}a & 0 & 0\\0 & b & 0\\0 & 0 & c\end{bmatrix}$
not a diagonal matrix.
for symmetric = AT ≠ T
$A+\begin{bmatrix}0 & -5 & 9\\5 & 0 & -3\\-9 & 3 & 0\end{bmatrix}⇒A^T=\begin{bmatrix}0 & 5 & -9\\-5 & 0 &3\\9 & -3 & 0\end{bmatrix}$
$A^T=\begin{bmatrix}0 & 5 & -9\\-5 & 0 &3\\9 & -3 & 0\end{bmatrix}⇒-\begin{bmatrix}0 & -5 & 9\\5 & 0 &-3\\-9 & 3 & 0\end{bmatrix}$
$A^T=A⇒\begin{bmatrix}0 & -5 & 9\\5 & 0 & -3\\-9 & 3 & 0\end{bmatrix}$
So its a skew symmetric matrix.
(B) $\begin{bmatrix}5 & 0 & 0\\0 & 5 & 0\\0 & 0 & 5\end{bmatrix}$
its a scalar matrix as its all elements in all principal diagonal are equal to some nor-zero constant.
(C) $\begin{bmatrix}5 & 0 & 0\\0 & -5 & 0\\0 & 0 & 7\end{bmatrix}$
its a diagonal matrix because all its non-zero diagonal elements are zero.
(D) $\begin{bmatrix}3 & -2 & 1\\-2 & -5 & 6\\1 & 6 & 0\end{bmatrix}=A$
$A^T=\begin{bmatrix}3 & -2 & 1\\-2 & -5 & 6\\1 & 6 & 0\end{bmatrix}$
$∴A^T=A$
So it is symmetric matrix.
(A) - (IV), (B) - (I), (C) - (II), (D) - (III)
option 3 is correct.