Question

solve this matrices ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$x=2,y=1,z=-1$

Question :

$Find \left[\begin{array}{ccc}-5&1&3\\7&1&-5\\1&-1&1\end{array}\right] \left[\begin{array}{ccc}1&1&2\\3&2&1\\2&1&3\end{array}\right] and \: hence \: solve\\ x+y+2=1\\3x+2y+z=7\\2x+y+3z=2$

Solution :

$\left[\begin{array}{ccc}-5&1&3\\7&1&-5\\1&-1&1\end{array}\right] \left[\begin{array}{ccc}1&1&2\\3&2&1\\2&1&3\end{array}\right]$

Multiply the matrices,$\left[\begin{array}{ccc}(-5\times 1)+(1\times 3)+(3\times 2) & - (5\times 1) + (1\times2)+(3\times 1)&(-5\times 2)+(1\times 1)+(3\times 3)\\(7\times1)+(1\times3)-(5 \times2)&(7\times1)+(1\times2)-(5\times1)&(7\times2)+(1\times1)-(5\times3)\\(1\times1)-(1\times3)+(1\times2)&(1\times1)-(1\times2)+(1\times1)&(1\times2)-(1\times1)+(1\times3)\end{array}\right]$

$\implies\left[\begin{array}{ccc}(-5+3+6)&(-5+2+3) &(-10+1+9)\\(7+3-10)&(7+2-5)&(14+1-15)\\(1-3+2) &(1-2+1) &(2-1+3)\end{array}\right]$

$\implies\left[\begin{array}{ccc}4&0&0\\0&4&0\\0&0&4\end{array}\right]$$\implies4\times \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] = 4I$

Thus,

$\implies\left[\begin{array}{ccc}-5&1&3\\7&-1 &-5\\1&-1 & 1 \end{array}\right] \left[\begin{array}{ccc}1&1&2\\3&2&1\\2&1&3\end{array}\right] = 4I...…(1)$

$\implies x+y+2z=1\\\implies3x+2y+z=7\\\implies2x+y+3z=2$

$\implies\left[\begin{array}{ccc}1&1&2\\3&2&1\\2&1&3 \end{array}\right] \left[\begin{array}{c}x\\y\\z \end{array}\right] = \left[\begin{array}{c}1\\7\\2\end{array}\right]$

$\implies\left[\begin{array}{ccc}-5&1&3\\7&1&-5\\1&-1&1\end{array}\right] \left[\begin{array}{ccc}1&1&2\\3&2&1\\2&1&3\end{array}\right] \left[\begin{array}{c}x \\y \\z\end{array}\right] = \left[\begin{array}{ccc}-5&1 &3 \\7&1&-5\\1&-1&1\end{array}\right]\left[\begin{array}{c}1\\7\\2\end{array}\right]$

Substitute (1)  

$\implies4I \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}(-5\times1 ) +(1\times7)+ (3\times2)\\(7\times1) +(1\times 7) - (5\times2)\\(1\times1) -(1\times 7)+(1\times 2)\end{array}\right]$

$\implies4I \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}(-5+7+6)\\(7+7-10 ) \\ (1-7+2) \end{array}\right]$

$\implies4 \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}8\\4\\-4 \end{array}\right]$

$\implies\left[\begin{array}{c}x\\y \\z \end{array}\right] =\left[\begin{array}{c}\frac{8}{4}\\ \\ \frac{4}{4}\\ \\\frac{-4}{4} \end{array}\right]$

$\implies\left[\begin{array}{c}x\\y\\z \end{array}\right] =\left[\begin{array}{c}2\\1\\-1\end{array}\right]$

$x=2,y=1,z=-1$

Multiplication Of Matrices :

To multiply two matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix. Consider the multiplications of $3\times2$ and $3\times2$matrices.

$\mathtt{(order\:of\:left\:hand\:matrix\times(order\:of\:right\:hand\:matrix)\implies(order\:of\:product\:matrix).}$

Matrices are multiplies by multiplying the elements in a row of the first matrix by the elements in a column of the second matrix,and adding the results.

The product of AB can be found if the number of columns of matrix A is equal to the number of rows of matrix B. If the order of matrix A is $m\times n$ and B is $n\times p$then the order of AB is $m\times p$