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If matrixA=\left[\begin{array}{ccc}{1} & {\;\;2} & {\;3} \\ {3} & {\;\;4} & {\;6} \\ {1} & {-2} & {\;8} \\ {6} & {-3} & {\;9}\end{array}\right] then choose the correct option

Option: 1

A=-6\left[\begin{array}{ccc}{-\frac{1}{6}} & {-\frac{1}{3}} & {-\frac{1}{2}} \\\\ {-\frac{1}{2}} & {-\frac{2}{3}} & {-1} \\\\ {-\frac{1}{6}} & {\frac{1}{3}} & {-\frac{4}{3}} \\\\ {-1} & {\frac{1}{2}} & {\frac{3}{2}}\end{array}\right]


Option: 2

A=-6\left[\begin{array}{ccc}{-\frac{1}{6}} & {-\frac{1}{3}} & {-\frac{1}{2}} \\\\ {-\frac{1}{2}} & {-\frac{2}{3}} & {-1} \\\\ {-\frac{1}{6}} & {\frac{1}{3}} & {-\frac{4}{3}} \\\\ {-1} & {\frac{1}{2}} & {-\frac{3}{2}}\end{array}\right]


Option: 3

A=-6\left[\begin{array}{ccc}{-\frac{1}{6}} & {-\frac{1}{3}} & {-\frac{1}{2}} \\\\ {-\frac{1}{2}} & {-\frac{1}{3}} & {-1} \\\\ {-\frac{1}{6}} & {\frac{1}{3}} & {-\frac{4}{3}} \\\\ {-1} & {\frac{1}{2}} & {-\frac{3}{2}}\end{array}\right]


Option: 4

A=-6\left[\begin{array}{ccc}{-\frac{1}{6}} & {-\frac{1}{3}} & {-\frac{1}{2}} \\\\ {-\frac{1}{2}} & {-\frac{2}{3}} & {-1} \\\\ {-\frac{1}{3}} & {\frac{1}{3}} & {-\frac{4}{3}} \\\\ {-1} & {\frac{1}{2}} & {\frac{3}{2}}\end{array}\right]


Answers (1)

best_answer

 

 

Scalar Multiplication of Matrix -

Scalar multiplication: let k be any scalar number, and A = \left [ a_{ij} \right ]_{m\times n}  be a matrix. Then the matrix obtained by multiplying every element A by a scalar k and denoted as kA.

kA = \left [ ka_{ij} \right ]_{m\times n}

 

-

Now,

A=\left[\begin{array}{ccc}{1} & {\;\;2} & {\;3} \\ {3} & {\;\;4} & {\;6} \\ {1} & {-2} & {\;8} \\ {6} & {-3} & {\;9}\end{array}\right]

Now taking -6 as common,

A=-6\left[\begin{array}{ccc}{1\times-\frac{1}{6}} & {2\times-\frac{1}{6}} & {3\times-\frac{1}{6}} \\\\ {3\times-\frac{1}{6}} & {4\times-\frac{1}{6}} & {6\times-\frac{1}{6}} \\\\ {1\times-\frac{1}{6}} & {-2\times-\frac{1}{6}} & {8\times-\frac{1}{6}} \\\\ {6\times-\frac{1}{6}} & {-3\times-\frac{1}{6}} & {9\times-\frac{1}{6}}\end{array}\right]

A=-6\left[\begin{array}{ccc}{-\frac{1}{6}} & {-\frac{1}{3}} & {-\frac{1}{2}} \\\\ {-\frac{1}{2}} & {-\frac{2}{3}} & {-1} \\\\ {-\frac{1}{6}} & {\frac{1}{3}} & {-\frac{4}{3}} \\\\ {-1} & {\frac{1}{2}} & {-\frac{3}{2}}\end{array}\right]

hence option (b) is correct

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