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This is the second of our many quizzes on Multiplying matrices. Click on this to access more such competitive quizzes on Algebra.

Multiplying matrices Quiz 2

mathematics

 Your friend challenged you to find the product of the matrices $ \begin{bmatrix}   7 & 10 & 4  \\  5 & 4 & 2  \\\end{bmatrix} $ and $\begin{bmatrix}   1 &   3 \\  5  &   6  \\   8  &  0  \\\end{bmatrix} $. From the matrix calculated all elements are added to find the score. Calculate the final score.

 Check the compatibility between the rows and columns in both matrices use the formula ${{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}}$.

  The product of the two matrices can be found only if the number of columns of the first matrix is equal to the number of rows in the second matrix. Since this condition is satisfied. $ \\ {{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}} \\ $ $4\times 4$ matrix is calculated such that. $ \\ {{C}_{00}}=(1\times 7)+(5\times 10)+(4\times 8)=89 \\  {{C}_{01}}=(3\times 7)+(6\times 10)+(0\times 4)=81 \\  {{C}_{10}}=(1\times 5)+(5\times 4)+(8\times 2)=41 \\  {{C}_{11}}=(3\times 5)+(6\times 4)+(0\times 2)=89 \\ $ So matrix formed is $ \\ \begin{bmatrix}   89 & 81  \\   41 & 39  \\ \end{bmatrix}  \\ $ Adding them up to find the score$=89+81+41+39 =250$ 

 A secret society uses an encryption method to deliver messages in the form of matrices which Sofia broke. As per the code they are multiplied first as AXB and the answer is subtracted from $BXA$. Find the code for matrices, $\begin{bmatrix}   1 & 5  \\   10 & 6  \\\end{bmatrix} $ and $\begin{bmatrix}   5 & 2  \\   8 & 9  \\\end{bmatrix} $ 

 Matrix multiplication is not associative. Multiply both matrices in order and opposite order use the formula ${{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}}$.

  The product of the two matrices can be found only if the number of columns of the first matrix is equal to the number of rows in the second matrix. Since this condition is satisfied $ \\ {{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}} \\ $ The product of the matrices in same order to find $AXB$ is $ \\  {{C}_{00}}=(1\times 5)+(8\times 5)=45 \\  {{C}_{01}}=(1\times 2)+(5\times 9)=47 \\  {{C}_{10}}=(10\times 5)+(8\times 6)=98 \\  {{C}_{11}}=(10\times 6)+(2\times 9)=74 \\ $ And the matrix is  $ \\  A\times B=\begin{bmatrix}   45 & 47  \\   98 & 74  \\ \end{bmatrix}  \\ $ Similarly the product of matrix in opposite order is  $ \\ B\times A=\begin{bmatrix}   25 & 37  \\   98 & 94  \\ \end{bmatrix}  \\ A\times B-B\times A=\begin{bmatrix}   25 & 37  \\   98 & 94  \\\end{bmatrix} -\begin{bmatrix}   45 & 47  \\   98 & 74  \\\end{bmatrix} =\begin{bmatrix}   -20 & 10  \\   0 & 20  \\\end{bmatrix} $ 

 Consider the two figures, where red means the increasing factor of two starting from 2, blue means increasing factor of 3 starting from 3 and green means increasing factor of 5 starting from 5. Find their multiplication, Add all numbers of the matrix to find the answer. $($Increase the factors individually for each matrix $)$  $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d2a9c340-7763-48dc-a541-279e2f80c6f56583291152184132974.png"/>

 Start according to the naming convention order of the matrices such that $<00>$ is before $<01>$. Use the formula ${{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}}$ to multiply.

From the given figure the matrix are devised as  $ \\ \begin{bmatrix}  5 & 3  \\   2 & 4  \\\end{bmatrix} $  $\begin{bmatrix}   2 & 3  \\   6 & 5  \\\end{bmatrix}  \\ $ Let the answer matrix be C. Multiplication of the matrices in the same order as follows, $ \\  {{C}_{00}}=(5\times 2)+(3\times 6)=28 \\  {{C}_{01}}=(5\times 3)+(3\times 5)=30 \\  {{C}_{10}}=(2\times 2)+(4\times 6)=28 \\  {{C}_{11}}=(2\times 3)+(4\times 5)=26 \\ $ $\begin{bmatrix}   28 & 30  \\   28 & 26  \\\end{bmatrix}  \\ $ Hence the addition of all numbers give $112$.

 Sheetal wrote a code to multiply to matrices and add all elements but found the answer to be 1000 in each case. Help her in finding the correct answer by finding the difference between the actual answer and the predicted one for the matrices and $\begin{bmatrix}   1 & 2  \\   4 & 3  \\\end{bmatrix} \text{and}\begin{bmatrix}   5 & 6  \\   8 & 7  \\\end{bmatrix} $  multiply the product to$\begin{bmatrix}   9 & 8  \\   5 & 4  \\\end{bmatrix} $.   

 Calculate the difference in answer by subtracting the actual answer with the predicted one. Use the formula ${{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}}$ to multiply.

 The product of the two matrices can be found only if the number of columns of the first matrix is equal to the number of rows in the second matrix. Since this condition is satisfied. $ \\ {{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}} \\ $ So the product of first two matrices in the same order with $[C]$ and answer matrix is $\\ {{C}_{00}}=(1\times 5)+(2\times 8)=22 \\  {{C}_{01}}=(1\times 6)+(2\times 7)=20 \\  {{C}_{10}}=(4\times 5)+(3\times 8)=44 \\  {{C}_{11}}=(4\times 6)+(3\times 7)=45 \\ $ Multiplying the resultant matrix i.e., $ \\ \begin{bmatrix}   22 & 20  \\   44 & 45  \\\end{bmatrix} $ with the last matrix $ \\ {{C}_{00}}=(22\times 9)+(20\times 5)=298 \\  {{C}_{01}}=(22\times 8)+(20\times 4)=256 \\  {{C}_{10}}=(44\times 9)+(20\times 4)=621 \\  {{C}_{11}}=(44\times 8)+(45\times 4)=532 \\ $ Adding the elements we get $1707$

 Matrices are used in image processing by converting the screen into pixels. The average of the matrix is found and the pixel closest to it is said to be the “prominent pixel”. Geeta wants to make a system for two matrices $\begin{bmatrix}   1 & 8  \\   4 & 10  \\\end{bmatrix} $ and $\begin{bmatrix}   9 & 10  \\   15 & 12  \\\end{bmatrix} $. Find the “Prominent pixel” when these two matrices are multiplied.

 Element closest to the average calculated is the “Prominent pixel”. Use the formula ${{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}}$ to multiply.

 The product of the two matrices can be found only if the number of columns of the first matrix is equal to the number of rows in the second matrix. Since this condition is satisfied $ \\ {{C}_{ik}}=\sum\limits_{j}^{{}}{{{a}_{ij}}{{b}_{ik}}} \\ $ The matrix is calculated such that $ \\ {{C}_{00}}=(1\times 9)+(15\times 8)=129 \\  {{C}_{01}}=(1\times 10)+(12\times 8)=106 \\  {{C}_{10}}=(4\times 9)+(10\times 15)=186 \\  {{C}_{11}}=(4\times 15)+(10\times 12)=160 \\ $ So the matrix formed is  $ \\ \begin{bmatrix}   129 & 106  \\   186 & 160  \\ \end{bmatrix}  \\ $ Hence the average $=\dfrac{(129+106+186+160)}{4}=145.25$ Closest cell is element $1,1$