Matrices and Determinants - SS3 Mathematics Past Questions and Answers - page 1
Let \(A = \begin{bmatrix} 2 & - 4 & 3 \\ 5 & 1 & 0 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 4 & - 2 \\ - 3 & 3 & - 1 \\ \end{bmatrix}\). Find \(B - A\)
\[\begin{bmatrix} - 1 & 8 & - 5 \\ - 8 & 2 & - 1 \\ \end{bmatrix}\]
\[\begin{bmatrix} - 1 & 4 & 5 \\ 8 & 2 & - 1 \\ \end{bmatrix}\]
\[\begin{bmatrix} 1 & 8 & 5 \\ 8 & 2 & 1 \\ \end{bmatrix}\]
\[\begin{bmatrix} 2 & 4 & 1 \\ 0 & 6 & 3 \\ \end{bmatrix}\]
Let \(A = \begin{bmatrix} 2 & - 4 & 3 \\ 5 & 1 & 0 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 4 & - 2 \\ - 3 & 3 & - 1 \\ \end{bmatrix}\)
\[B - A = \begin{bmatrix} 1 - 2 & 4 - ( - 4) & - 2 - 3 \\ - 3 - 5 & 3 - 1 & - 1 - 0 \\ \end{bmatrix}\]
\(= \begin{bmatrix} - 1 & 8 & - 5 \\ - 8 & 2 & - 1 \\ \end{bmatrix}\)
Let \(A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \\ \end{bmatrix}\), \(B = \begin{bmatrix} 0 & - 3 \\ 5 & 1 \\ \end{bmatrix}\) and \(C = \begin{bmatrix} 1 & 2 \\ 6 & - 1 \\ \end{bmatrix}\). Find \(2A - 3B + C\)
\(\begin{bmatrix} 8 & 28 \\ 4 & 0 \\ \end{bmatrix}\)
\(\begin{bmatrix} 3 & 19 \\ - 5 & 2 \\ \end{bmatrix}\)
\(\begin{bmatrix} 2 & - 7 \\ - 7 & 3 \\ \end{bmatrix}\)
\(\begin{bmatrix} 4 & 16 \\ 8 & - 8 \\ \end{bmatrix}\)
Let \(A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \\ \end{bmatrix}\), \(B = \begin{bmatrix} 0 & - 3 \\ 5 & 1 \\ \end{bmatrix}\) and \(C = \begin{bmatrix} 1 & 2 \\ 6 & - 1 \\ \end{bmatrix}\)
\[2A - 3B + C = \ \left\lbrack 2\begin{bmatrix} 1 & 4 \\ 2 & 3 \\ \end{bmatrix} - 3\begin{bmatrix} 0 & - 3 \\ 5 & 1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 6 & - 1 \\ \end{bmatrix} \right\rbrack\]
\(= \ \left\lbrack \begin{bmatrix} 2 & 8 \\ 4 & 6 \\ \end{bmatrix} - \begin{bmatrix} 0 & - 9 \\ 15 & 3 \\ \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 6 & - 1 \\ \end{bmatrix} \right\rbrack = \begin{bmatrix} 2 - 0 + 1 & 8 - ( - 9) + 2 \\ 4 - 15 + 6 & 6 - 3 + ( - 1) \\ \end{bmatrix}\)
\(= \begin{bmatrix} 3 & 19 \\ - 5 & 2 \\ \end{bmatrix}\)
Let \(A = \begin{bmatrix} 2 & 3 \\ 0 & - 2 \\ 4 & 5 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 & 2 \\ - 1 & 0 & 6 \\ \end{bmatrix}\). Find \(BA\)
\(\begin{bmatrix} 10 & 7 \\ 22 & 27 \\ \end{bmatrix}\)
\(\begin{bmatrix} 3 & 6 \\ 3 & 0 \\ \end{bmatrix}\)
\(\begin{bmatrix} 5 & 12 \\ 11 & 5 \\ \end{bmatrix}\)
\(\begin{bmatrix} 9 & 0 \\ - 1 & 5 \\ \end{bmatrix}\)
\(A = \begin{bmatrix} 2 & 3 \\ 0 & - 2 \\ 4 & 5 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 & 2 \\ - 1 & 0 & 6 \\ \end{bmatrix}\)
\(BA = \begin{bmatrix} 1 & 3 & 2 \\ - 1 & 0 & 6 \\ \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 0 & - 2 \\ 4 & 5 \\ \end{bmatrix}\)
\[c_{11} = 1(2) + 3(0) + 2(4) = 2 + 0 + 8 = 10\]
\[c_{12} = 1(3) + 3( - 2) + 2(5) = 3 - 6 + 10 = 7\]
\[c_{21} = - 1(2) + 0(0) + 6(4) = - 2 + 0 + 24 = 22\]
\[c_{22} = - 1(3) + 0( - 2) + 6(5) = - 3 + 0 + 30 = 27\]
\(BA = \begin{bmatrix} 10 & 7 \\ 22 & 27 \\ \end{bmatrix}\)
If \(A = \begin{bmatrix} 1 & 0 \\ 2 & 2 \\ 11 & 5 \\ 4 & - 6 \\ \end{bmatrix}\), find \(A^{T}\)
\(\begin{bmatrix} 1 & 0 \\ 2 & 2 \\ 11 & 5 \\ 4 & - 6 \\ \end{bmatrix}\)
\(\begin{bmatrix} 0 & 2 & 5 & - 6 \\ 1 & 2 & 11 & 4 \\ \end{bmatrix}\)
\(\begin{bmatrix} 1 & 11 \\ 2 & - 6 \\ \end{bmatrix}\)
\(\begin{bmatrix} 2 & 2 \\ 11 & 5 \\ \end{bmatrix}\)
If \(A = \begin{bmatrix} 1 & 0 \\ 2 & 2 \\ 11 & 5 \\ 4 & - 6 \\ \end{bmatrix}\)
\(A^{T} = \begin{bmatrix} 0 & 2 & 5 & - 6 \\ 1 & 2 & 11 & 4 \\ \end{bmatrix}\)
Let \(A = \begin{bmatrix} 7 & 3 \\ 4 & 2 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} - 3 & 1 \\ 2 & 4 \\ \end{bmatrix}\), find \(|A| + |B|\)
\[A = \begin{bmatrix} 7 & 3 \\ 4 & 2 \\ \end{bmatrix}\]
\[|A| = 7(2) - 3(4) = 14 - 12 = 2\]
\[B = \begin{bmatrix} - 3 & 1 \\ 2 & 4 \\ \end{bmatrix}\]
\[|B| = - 3(4) - 1(2) = - 12 - 2 = - 14\]
\[|A| + |B| = 2 + ( - 14) = 2 - 14 = - 12\]
Evaluate \(|B| = \left| \begin{matrix} 2 & 3 & - 4 \\ 1 & 2 & 3 \\ 3 & - 1 & - 1 \\ \end{matrix} \right|\)
Using the third row, \(|A| = a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33}\)
\[C_{31} = {( - 1)}^{3 + 1}\left| \begin{matrix} 3 & - 4 \\ 2 & 3 \\ \end{matrix} \right| = ( - 1)^{4}\left| \begin{matrix} 3 & - 4 \\ 2 & 3 \\ \end{matrix} \right| = \ + \left\lbrack 3(3) - ( - 4)2 \right\rbrack = + \lbrack 9 + 8\rbrack = 17\]
\[C_{12} = {( - 1)}^{3 + 2}\left| \begin{matrix} 2 & - 4 \\ 1 & 3 \\ \end{matrix} \right| = ( - 1)^{5}\left| \begin{matrix} 2 & - 4 \\ 1 & 3 \\ \end{matrix} \right| = \ - \left\lbrack 2(3) - ( - 4)(1) \right\rbrack = - \lbrack 6 + 4\rbrack = - 10\]
\[C_{13} = {( - 1)}^{3 + 3}\left| \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} \right| = ( - 1)^{6}\left| \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} \right| = \ + \left\lbrack 2(2) - 3(1) \right\rbrack = + \lbrack 4 - 3\rbrack = 1\]
\[|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = 3(17) + ( - 1)( - 10) + ( - 1)1 = \ 51 + 10 - 1 = 60\]
Using determinant method to solve the simultaneous equations: \(2x + 3y = - 2\) and \(3x + 4y = - 6\)
\[\mathrm{\Delta} = \left| \begin{matrix} 2 & 3 \\ 3 & 4 \\ \end{matrix} \right| = 2(4) - 3(3) = 8 - 9 = - 1\]
\[\mathrm{\Delta}x = \left| \begin{matrix} 3 & - 2 \\ 4 & - 6 \\ \end{matrix} \right| = (3)( - 6) - ( - 2)4 = - 18 + 8 = - 10\]
\[\mathrm{\Delta}y = \left| \begin{matrix} 2 & - 2 \\ 3 & - 6 \\ \end{matrix} \right| = 2( - 6) - ( - 2)(3) = - 12 + 6 = - 6\]
\[x = \frac{\mathrm{\Delta}x}{\mathrm{\Delta}} = \frac{- 10}{- 1} = 10\]
\[y = \frac{\mathrm{\Delta}y}{\mathrm{\Delta}} = \frac{- 6}{- 1} = 6\]
Thus, \(x = 10\) and \(y = 6\)