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Question: From a set of \(2\times 2\) matrices having 0 or 1 in each place, a matrix is chosen. The probabilit...

From a set of 2×22\times 2 matrices having 0 or 1 in each place, a matrix is chosen. The probability that it is a unit matrix is
A. 116\dfrac{1}{16}
B. 216\dfrac{2}{16}
C. 316\dfrac{3}{16}
D. 14\dfrac{1}{4}

Explanation

Solution

To find the probability that the matrix chosen from 2×22\times 2 matrices having 0 or 1 in each place is a unit matrix, we must first find in how many ways the digits can be placed inside a 2×22\times 2 matrix. Since there are 2 possible values, that is, 0 and 1 and there are 4 positions in a 2×22\times 2 matrix, the number of ways 0 and 1 are arranged is given as 2×2×2×2=24=16 ways2\times 2\times 2\times 2={{2}^{4}}=16\text{ ways} . There is only one possibility for a matrix to be a unit matrix. Hence, by substituting the values in the formula P(A)=Number of favorable outcomeTotal number of outcomesP(A)=\dfrac{\text{Number of favorable outcome}}{\text{Total number of outcomes}} the required probability can be obtained.

Complete step-by-step solution
We need to find the probability that the matrix chosen from 2×22\times 2 matrices having 0 or 1 in each place is a unit matrix.
Let us see in how many ways the digits can be placed inside a 2×22\times 2 matrix. Consider a 2×22\times 2 matrix shown below.
${{\left[ \begin{matrix}

  • & * \\
  • & * \\
    \end{matrix} \right]}_{2\times 2}}The∗positionscanbefilledbyeither1or0.Thatiseachpositioncanget2values.Hence,wecanfindthenumberofwaysinwhich1or0canbefilledina The * positions can be filled by either 1 or 0. That is each position can get 2 values. Hence, we can find the number of ways in which 1 or 0 can be filled in a2\times 2matrixbymultiplyingthepossibilities,thatis,2foreachofthe4positions.Thatis,matrix by multiplying the possibilities, that is, 2 for each of the 4 positions. That is, 2\times 2\times 2\times 2={{2}^{4}}=16\text{ ways}Now,wehavetofindthepossibilityofgettingaunitmatrix.Letusseewhataunitmatrixis.Aunitmatrixisalsoknownasanidentitymatrix.Ithasalltheelementsalongthemaindiagonalas1andtheremainingelementsas0.Letusseehowitisdenoted. Now, we have to find the possibility of getting a unit matrix. Let us see what a unit matrix is. A unit matrix is also known as an identity matrix. It has all the elements along the main diagonal as 1 and the remaining elements as 0. Let us see how it is denoted. \left[ \begin{matrix}
    1 & 0 \\
    0 & 1 \\
    \end{matrix} \right]Theabovematrixisa The above matrix is a2\times 2unitmatrix.Wecanseethatthediagonalelementsare1andotherelementsare0.Weknowthataunit matrix. We can see that the diagonal elements are 1 and other elements are 0. We know that a2\times 2unitmatrixcanhaveonly1unitmatrix.Letusnowfindtheprobabilitythatthematrixischosenasaunitmatrix.Wecandenotethisasunit matrix can have only 1 unit matrix. Let us now find the probability that the matrix is chosen as a unit matrix. We can denote this asP(A).Hence,. Hence, P(A)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcomes}}Weknowthatthenumberoffavorableoutcomesis1sincea We know that the number of favorable outcomes is 1 since a2\times 2unitmatrixcanhaveonly1unitmatrix.Wealsofoundthetotalnumberofoutcomestobe16.Now,letussubstitutetheseintheaboveformula.Wewillgetunit matrix can have only 1 unit matrix. We also found the total number of outcomes to be 16. Now, let us substitute these in the above formula. We will get P(A)=\dfrac{\text{1}}{\text{16}}$
    Hence, the correct option is A.

Note: Do not get confused with the term ‘unit’ in unit matrix. You may think that a unit matrix is a matrix in which all elements are 1. This is a matrix of ones, not a unit matrix. You may make an error when writing the formula for probability. Do not write the probability formula as P(A)=Total number of outcomesNumber of favourable outcomeP(A)=\dfrac{\text{Total number of outcomes}}{\text{Number of favourable outcome}}