If a matrix has 18 elements, what are possible orders it can have? What, if it has 5 elements ?

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3 and 4 .Determinants and Matrices

easy

If a matrix has $18$ elements, what are the possible orders it can have? What, if it has $5$ elements ?

Option A

Option B

Option C

Option D

Solution

We know that if a matrix is of the order $m \times n,$ it has $mn$ elements.

Thus, to find all the possible orders of a matrix having $18$ elements. we have to find all the ordered pairs of natural numbers whose products is $18$ .

The ordered pairs are : $(1,18),\,(18,1),\,(2,9),\,(9,2),\,(3,6),$ and $(6,3)$

Hence, the possible orders of a matrix having $18$ elements are :

$1 \times 18$, $1 8 \times 1$, $2 \times 9$, $9 \times 2$, $3 \times 6$ and $6 \times 3$

$(1,5)$ and $(5,1)$ are the ordered pairs of natural numbers whose product is $5 .$

Hence, the possible orders of a matrix having $5$ elements are $1 \times 5$ and $5 \times 1$.

Standard 12

Mathematics

If $A = \left[ {\begin{array}{*{20}{c}}
3&7\\
1&2
\end{array}} \right]$ then $|A^{2011} -5A^{2010}|$ is equal to

normal

If $R(t) = \left[ {\begin{array}{*{20}{c}}{\cos t}&{\sin t}\\{ – \sin t}&{\cos t}\end{array}} \right],$then $R(s).\,R(t) = $

easy

If $A = \left( {\begin{array}{{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right)$ and $B = \left( {\begin{array}{{20}{c}}{ – 5}&7&1\\1&{ – 5}&7\\7&1&{ – 5}\end{array}} \right)$ then $AB$ is equal to

easy

Let $A$ and $B$ be two symmetric matrices of order $3.$

Statement $-1$: $A(BA)$ and $(AB)A$ are symmetric matrices.

Statement $-2:$ $AB$ is symmetric matrix if matrix multiplication of $A$ with $B$ is commutative.

medium

(AIEEE-2011)

Compute the following : $\left[ {\begin{array}{{20}{l}}
{{a^2} + {b^2}}&{{b^2} + {c^2}} \\
{{a^2} + {c^2}}&{{a^2} + {b^2}}
\end{array}} \right]$ $ + \left[ {\begin{array}{{20}{c}}
{2ab}&{2bc} \\
{ – 2ac}&{ – 2ab}
\end{array}} \right]$

easy

If a matrix has $18$ elements, what are the possible orders it can have? What, if it has $5$ elements ?

Solution

Similar Questions

If $A = \left[ {\begin{array}{*{20}{c}} 3&7\\ 1&2 \end{array}} \right]$ then $|A^{2011} -5A^{2010}|$ is equal to

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If $A = \left[ {\begin{array}{*{20}{c}}
3&7\\
1&2
\end{array}} \right]$ then $|A^{2011} -5A^{2010}|$ is equal to

If $A = \left( {\begin{array}{{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right)$ and $B = \left( {\begin{array}{{20}{c}}{ – 5}&7&1\\1&{ – 5}&7\\7&1&{ – 5}\end{array}} \right)$ then $AB$ is equal to

Let $A$ and $B$ be two symmetric matrices of order $3.$

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Statement $-2:$ $AB$ is symmetric matrix if matrix multiplication of $A$ with $B$ is commutative.

Compute the following : $\left[ {\begin{array}{{20}{l}}
{{a^2} + {b^2}}&{{b^2} + {c^2}} \\
{{a^2} + {c^2}}&{{a^2} + {b^2}}
\end{array}} \right]$ $ + \left[ {\begin{array}{{20}{c}}
{2ab}&{2bc} \\
{ – 2ac}&{ – 2ab}
\end{array}} \right]$