IB Mathematics Applications & Interpretation Question Type 8: Using the Inverse Matrix, If It Exists, to Solve for a System of Linear Equations Exercises

Solve the system using the inverse matrix method: $$\left(\begin{array}{ccc}4& 1& 1\\ -2& 3& 2\\ 1& -1& 4\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}5\\ 1\\ 0\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ -1& 4& 2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}6\\ -3\\ 5\end{array}\right)$$

Solve the system with fractional coefficients using the inverse matrix method: $$\left(\begin{array}{ccc}\frac{1}{2}& 1& -1\\ 1& -\frac{1}{3}& 2\\ 2& 1& \frac{1}{4}\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}2\\ 1\\ 3\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{cc}3& 2\\ 5& -4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}7\\ -1\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{cc}2& 1\\ 3& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}5\\ 4\end{array}\right)$$

Solve the system of linear equations using the inverse matrix method: $$\left(\begin{array}{ccc}2& 1& 3\\ 1& -1& 1\\ 3& 4& -1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}14\\ 4\\ 7\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{ccc}3& 1& 2\\ 4& 2& 1\\ 2& 1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}5\\ 10\\ 9\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{ccc}1& 3& 2\\ 2& -1& 1\\ 3& 2& -1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}9\\ 1\\ 6\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{ccc}1& 2& 3\\ 2& -1& 1\\ 3& 1& -2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}6\\ 2\\ 2\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{cc}1& 1\\ 2& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 4\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{ccc}1& 2& 1\\ 2& -1& 1\\ 3& 1& -2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}14\\ 3\\ 1\end{array}\right)$$

Solve the system using the inverse matrix method: $$\left(\begin{array}{ccc}3& 0& 1\\ 1& -2& 3\\ 5& 1& 2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}4\\ -1\\ 10\end{array}\right)$$