Matrices & Determinants
Adjoint of a matrix
Grade Class 12
Question:
Matrix A = <table><tr><td>x</td><td>3</td><td>2</td></tr><tr><td>1</td><td>y</td><td>4</td></tr><tr><td>2</td><td>2</td><td>z</td></tr></table>, if xyz = 60 and 8x + 4y + 3z = 20, then A(adj A) is equal to -
<table><tr><td>64</td><td>0</td><td>0</td></tr><tr><td>0</td><td>64</td><td>0</td></tr><tr><td>0</td><td>0</td><td>64</td></tr></table>
<table><tr><td>88</td><td>0</td><td>0</td></tr><tr><td>0</td><td>88</td><td>0</td></tr><tr><td>0</td><td>0</td><td>88</td></tr></table>
<table><tr><td>68</td><td>0</td><td>0</td></tr><tr><td>0</td><td>68</td><td>0</td></tr><tr><td>0</td><td>0</td><td>68</td></tr></table>
<table><tr><td>34</td><td>0</td><td>0</td></tr><tr><td>0</td><td>34</td><td>0</td></tr><tr><td>0</td><td>0</td><td>34</td></tr></table>
Step-by-Step Solution
Key Concept: The property A(adj A) = |A|I, where I is the identity matrix. Calculate the determinant of A using the given values.
We know that A(adj A) = |A|I. The determinant of A is |A| = x(yz - 8) - 3(z - 8) + 2(2 - 2y) = xyz - 8x - 3z + 24 + 4 - 4y = xyz - (8x + 4y + 3z) + 28. Given xyz = 60 and 8x + 4y + 3z = 20, we have |A| = 60 - 20 + 28 = 68. Thus, A(adj A) = 68I = [[68, 0, 0], [0, 68, 0], [0, 0, 68]].
Correct Answer: 3