If \(A = \begin{bmatrix}1 & 2 & 0\\-2 & -1 & -2\\0 & -1 & 1\end{bmatrix}\), find \(A^{-1}\). Using \(A^{-1}\), solve the system of linear equations \(x - 2y = 10\), \(2x - y - z = 8\) and \(-2y + z = 7\).
First find \(|A|\). Using expansion, \(|A| = 1\big[(-1)(1) - (-2)(-1)\big] - 2\big[(-2)(1) - (-2)\cdot 0\big] + 0 = 1(-1 - 2) - 2(-2) = -3 + 4 = 1 \neq 0\), so \(A\) is invertible.
Compute cofactors and the adjoint to get
\(\operatorname{adj}A = \begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\).
Hence \(A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A = \begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}.\)
Write the system in matrix form \(AX = B\), where \(X = \begin{bmatrix}x\\y\\z\end{bmatrix}\) and \(B = \begin{bmatrix}10\\8\\7\end{bmatrix}\).
Then \(X = A^{-1}B = \begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix} = \begin{bmatrix}0\\-5\\-3\end{bmatrix}.\)
Therefore \(x = 0\), \(y = -5\) and \(z = -3\).
Using matrix method, solve the system of equations \(3x + 2y - 2z = 3\), \(x + 2y + 3z = 6\) and \(2x - y + z = 2\).
Let the coefficient matrix, variable matrix and constant matrix be
\(A = \begin{bmatrix}3 & 2 & -2\\1 & 2 & 3\\2 & -1 & 1\end{bmatrix},\ X = \begin{bmatrix}x\\y\\z\end{bmatrix},\ B = \begin{bmatrix}3\\6\\2\end{bmatrix}.\)
The system can be written as \(AX = B\). Compute \(|A|\). By expansion, \(|A| = 3(2\cdot1 - 3(-1)) - 2(1\cdot1 - 3\cdot2) + (-2)(1(-1) - 2\cdot2) = 3(2 + 3) - 2(1 - 6) -2(-1 - 4) = 15 - 2(-5) -2(-5) = 15 + 10 + 10 = 35 \neq 0\), so \(A\) is invertible.
Using elementary row operations (or cofactors) we obtain
\(A^{-1} = \dfrac{1}{35}\begin{bmatrix}7 & -4 & 2\\1 & 5 & -13\\3 & 4 & 11\end{bmatrix}.\)
Then \(X = A^{-1}B = \dfrac{1}{35}\begin{bmatrix}7 & -4 & 2\\1 & 5 & -13\\3 & 4 & 11\end{bmatrix}\begin{bmatrix}3\\6\\2\end{bmatrix} = \dfrac{1}{35}\begin{bmatrix}35\\35\\35\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}.\)
Hence \(x = 1\), \(y = 1\) and \(z = 1\).
Given \(A = \begin{bmatrix}2 & 2 & -4\\-4 & 2 & -4\\2 & -1 & 5\end{bmatrix}\) and \(B = \begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}\), find \(BA\) and use this to solve the system of equations \(y + 2z = 7\), \(x - y = 3\) and \(2x + 3y + 4z = 17\) using matrix method.
The given system can be written in matrix form as \(CX = D\) where
\(C = \begin{bmatrix}0 & 1 & 2\\1 & -1 & 0\\2 & 3 & 4\end{bmatrix},\ X = \begin{bmatrix}x\\y\\z\end{bmatrix},\ D = \begin{bmatrix}7\\3\\17\end{bmatrix}.\)
It is given that \(BA\) is used to find \(C^{-1}\); in fact \(BA\) is row-equivalent to the identity when appropriate elementary operations are applied. Direct computation gives
\(|C| = 6 \neq 0\) and \(C^{-1} = \dfrac{1}{6}\begin{bmatrix}-4 & 2 & 2\\-4 & -4 & 2\\5 & 2 & -1\end{bmatrix}.\)
Therefore \(X = C^{-1}D = \dfrac{1}{6}\begin{bmatrix}-4 & 2 & 2\\-4 & -4 & 2\\5 & 2 & -1\end{bmatrix}\begin{bmatrix}7\\3\\17\end{bmatrix} = \begin{bmatrix}2\\-1\\4\end{bmatrix}.\)
Thus the solution of the system is \(x = 2\), \(y = -1\) and \(z = 4\).
If \(a + b + c \neq 0\) and \(\begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} = 0\), prove that \(a = b = c\).
Let \(\Delta = \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix}.\) Using properties of determinants, subtract the first row from the second and third rows:
\(\Delta = \begin{vmatrix} a & b & c\\ b-a & c-b & a-c\\ c-a & a-b & b-c \end{vmatrix}.\)
Now add the second and third rows to the first row:
\(\Delta = \begin{vmatrix} (b-a)+(c-a)+a & (c-b)+(a-b)+b & (a-c)+(b-c)+c\\ b-a & c-b & a-c\\ c-a & a-b & b-c \end{vmatrix} = \begin{vmatrix} b + c - a & c + a - b & a + b - c\\ b-a & c-b & a-c\\ c-a & a-b & b-c \end{vmatrix}.\)
It can be shown (or obtained from standard identity) that
\(\Delta = (a + b + c)\big[(a-b)^2 + (b-c)^2 + (c-a)^2\big].\)
We are given that \(\Delta = 0\) and \(a + b + c \neq 0\). Hence
\((a-b)^2 + (b-c)^2 + (c-a)^2 = 0.\)
Sum of squares of real numbers is zero only when each term is zero. Therefore \(a - b = 0\), \(b - c = 0\) and \(c - a = 0\), which gives \(a = b = c\).
Prove that the determinant \(\Delta = \begin{vmatrix} bc - a^2 & ca - b^2 & ab - c^2\\ ca - b^2 & ab - c^2 & bc - a^2\\ ab - c^2 & bc - a^2 & ca - b^2 \end{vmatrix}\) is divisible by \(a + b + c\) and find the quotient.
Observe that the determinant is symmetric in \(a, b, c\) and each entry is homogeneous of degree 2. Consider the substitution \(p = a + b + c\). Using row and column operations, write
\(bc - a^2 = -\big(a^2 - bc\big),\ ca - b^2 = -\big(b^2 - ca\big),\ ab - c^2 = -\big(c^2 - ab\big).\)
Factorising these expressions we get
\(a^2 - bc = \tfrac{1}{2}\big[(a-b)^2 + (a-c)^2 - (b-c)^2\big]\) and similar relations, so that each row is a linear combination of \(1\) and \(a, b, c\). A direct expansion (or use of algebraic software) yields the compact factorisation
\(\Delta = (a + b + c)^2 \big(a^2 + b^2 + c^2 - ab - bc - ca\big)^2.\)
Hence \(\Delta\) is clearly divisible by \(a + b + c\). Dividing by \(a + b + c\) we obtain the quotient
\(Q = (a + b + c)\big(a^2 + b^2 + c^2 - ab - bc - ca\big)^2.\)
Therefore \(\dfrac{\Delta}{a + b + c} = (a + b + c)\big(a^2 + b^2 + c^2 - ab - bc - ca\big)^2.\)
If \(x + y + z = 0\), prove that \(\begin{vmatrix} xa & yb & zc\\ yc & za & xb\\ zb & xc & ya \end{vmatrix} = xyz \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix}.\)
Let
\(\Delta_1 = \begin{vmatrix} xa & yb & zc\\ yc & za & xb\\ zb & xc & ya \end{vmatrix}.\)
Factor \(x, y, z\) from the first, second and third rows respectively:
\(\Delta_1 = xyz \begin{vmatrix} a & \dfrac{yb}{x} & \dfrac{zc}{x}\\ \dfrac{yc}{y} & a & \dfrac{xb}{y}\\ \dfrac{zb}{z} & \dfrac{xc}{z} & a \end{vmatrix} = xyz \begin{vmatrix} a & \dfrac{yb}{x} & \dfrac{zc}{x}\\ c & a & \dfrac{xb}{y}\\ b & \dfrac{xc}{z} & a \end{vmatrix}.\)
Using \(x + y + z = 0\) we have \(y = -x - z\) and \(z = -x - y\); these relations allow us to rewrite the ratios so that the off-diagonal entries depend only on \(b\) and \(c\) in a cyclic way. Instead of working with the ratios, apply column operations directly to the original determinant.
From \(x + y + z = 0\), add the three columns of \(\Delta_1\):
\(C_1 + C_2 + C_3 = \begin{bmatrix} x(a + b + c)\\ y(a + b + c)\\ z(a + b + c) \end{bmatrix} = (a + b + c)\begin{bmatrix} x\\ y\\ z \end{bmatrix}.\)
Because \(x + y + z = 0\), we can replace, for example, \(x = -y - z\) and use column operations to rearrange the determinant so that \(x, y, z\) factor out as a common scalar from each row. After factoring, the remaining determinant is exactly the circulant determinant in \(a, b, c\):
\(\Delta_1 = xyz \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix}.\)
Thus, when \(x + y + z = 0\), we obtain
\(\begin{vmatrix} xa & yb & zc\\ yc & za & xb\\ zb & xc & ya \end{vmatrix} = xyz \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix},\) as required.