Question:

Let A=$\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 &t \\ 4&7-t&-6 \\ \end{bmatrix}$, then the values of t for which inverse of A does not exist

Updated On: Apr 15, 2024
  • -2, 1
  • 44622
  • 2, -3
  • 3, -1
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The Correct Option is C

Solution and Explanation

We know that inverse of A does not exist
only when |A| = 0
$\therefore$ $\begin{vmatrix}
1 & 3 & 2 \\
2 & 5 &t \\
4&7-t&-6 \\ \end{vmatrix}$=0
$(-30-7t+t^2)-3(-12-4t)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2(14-2t-20)=0 $
$\Rightarrow \ \ -30- 7t + t^2 +36+12t-12-4t=0$
$\Rightarrow \ \ t^2+t-6=0 \Rightarrow \ \ t^2 +3t \ -2t-6=0$
$\Rightarrow \ \ t(t+3)-2(t+3)=0$
$\Rightarrow \ \ \ \ (t+3)(t-2)=0 \ \Rightarrow \ t=2, -3$
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Concepts Used:

Invertible matrices

A matrix for which matrix inversion operation exists, given that it satisfies the requisite conditions is known as an invertible matrix. Any given square matrix A of order n × n is called invertible if and only if there exists, another n × n square matrix B such that, AB = BA = In, where In  is an identity matrix of order n × n.

For example,

It can be observed that the determinant of the following matrices is non-zero.