Matrices and Systems of Equations: math test solutions, Grade 12 – download the PDF

1 Operations / 3 pts

\(2P = \begin{bmatrix}8&-2\\4&0\end{bmatrix}\), so \(2P - Q = \begin{bmatrix}5&-7\\6&-1\end{bmatrix}\) (1 pt).

\(PQ = \begin{bmatrix}4(3)+(-1)(-2)&4(5)+(-1)(1)\\2(3)+0&2(5)+0\end{bmatrix} = \begin{bmatrix}14&19\\6&10\end{bmatrix}\) (2 pts).

2 Determinants / 3 pts

(a) \(7(1) - 3(2) = 1\) (1 pt).

(b) \(1\big(1 \cdot 4 - (-1)(2)\big) - 0 + 2\big(3 \cdot 2 - 1 \cdot 0\big) = 6 + 12 = 18\) (2 pts).

3 Inverse and system / 3 pts

(a) \(\det M = 15 - 14 = 1\), so \(M^{-1} = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}\) (2 pts).

(b) \(\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\9\end{bmatrix} = \begin{bmatrix}12-18\\-28+45\end{bmatrix} = \begin{bmatrix}-6\\17\end{bmatrix}\) (1 pt).

4 Gaussian elimination / 4 pts

Rows: \((1, 2, 1 \mid 4)\), \((2, -1, 3 \mid -3)\), \((3, 1, -1 \mid 6)\).

\(R_2 - 2R_1 = (0, -5, 1 \mid -11)\) and \(R_3 - 3R_1 = (0, -5, -4 \mid -6)\) (2 pts).

\(R_3 - R_2 = (0, 0, -5 \mid 5)\), so \(z = -1\) (1 pt). Then \(-5y - 1 = -11\) gives \(y = 2\), and \(x = 4 - 4 + 1 = 1\). Solution: \((1, 2, -1)\) (1 pt).

5 Cramer’s Rule / 3 pts

\(D = 3(-1) - 2(5) = -13\) (1 pt).

\(D_x = 3(-1) - 2(-8) = 13\), so \(x = -1\) (1 pt). \(D_y = 3(-8) - 3(5) = -39\), so \(y = 3\) (1 pt).

6 A transformation of a square / 4 pts

(a) \((x, y) \to (x + 2y, y)\): \(O' = (0, 0)\), \(A' = (2, 0)\), \(B' = (6, 2)\), \(C' = (4, 2)\) (2 pts).

(b) The area of the square is \(4\) and \(\det S = 1\), so the image area is \(|1| \cdot 4 = 4\) (2 pts).