Using properties of determinants, show that triangle ABC is isosceles if: 
\(\left| \begin{matrix} 1 & 1 & 1 \\ 1+cosA & 1+cosB & 1+cosC \\ \cos ^{ 2 }{ A } +cosA & \cos ^{ 2 }{ B } +cosB & \cos ^{ 2 }{ B } +cosC \end{matrix} \right| =0\)

Using properties of determinants, prove that:
\(\left| \begin{matrix} { (y+z) }^{ 2 } & xy & zx \\ xy & { (x+z) }^{ 2 } & yz \\ xz & yz & { (x+y) }^{ 2 } \end{matrix} \right| =2xyz{ (x+y+z) }^{ 3 }.\)

If \(A=\left[ \begin{matrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{matrix} \right] \), find A^-1 . 
Hence solve the following system of equations:
x+2y-3z=-4,
2x+3y+2z=2,
3x-3y-4z=11.

Determine the product \(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] ,\)  and use it to solve the system of equations: x - y + z = 4, x - 2y - 2z = 9, 2x + y + 3z = 1.

If \(A=\left[ \begin{matrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{matrix} \right] ,\) find A^-1. Hence solve the following system of equations:
x + 2y + 5z = 10, x - y - z = -2, 2x + 3y - z = -11.

Using matrices, solve the following system of linear equations:
3x + 4y + 7z = 4,
2x - y + 3z = -3,
x + 2y - 3z = 8.

If \({ A }^{ -1 }=\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{matrix} \right] and \ B=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right] ,\)find (AB)^-1 .

Two schools A and B want to award their selected teachers on the values of Honesty, Hard work and Regularity. School A wants to award Rs. x each, Rs. y each and Rs. z each for the three respective values to 3, 2 and 1 teachers with a total award money of Rs.1.28 lakh. School B wants to spend Rs. 1·54 lakh to award its 4, 1 and 3 teachers on the respective values (by giving the same award money for the three values as before). If the total amount of award for one prize on each value is Rs. 57,000, using matrices, find the award money for each value.

A school wants to award its students for the value of Honesty, Regularity and Hard work with a total cash award of Rs. 6,000. Three times the award money for hard work added to that given for honesty amounts to Rs. 11,000.The award money given for honesty and hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values namely, honesty, regularity and hard work, suggest one more value which the school must include for awards.

Using matrices, solve the following system of linear equations:
2x + 3y + 3z = 5, x - 2y + z = -4, 3x - y - 2z = 3